INTEGRAL CALCULUS

INCLUDING
DIFFERENTIAL EQUATIONS

Recommended by the Universities of Calcutta. Burd wan. Jadavpur, Kalyani,

North Bengal, V idyasagar, Dacca, Patna, Bihar, Bhagalpur, Nalanda,
Sambalpur, Berhampur, Utkal, Cuwahali, Dibrugarh, Tribhu ban, etc.
as a text book for B.A & B.Sc. mathematics.
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LTEGRAL CALCULUS
INCLUDING

DIFFERENTIAL EQUATIONS

BY
B. C. DAS. M. Sc.
'ESSOR OF MATHEMATICS (RETD.)
ESIDENCY COLLEGE CALCUTTA:
TURER IN APPLIED MA THEMA TICS,
CALCUTTA UNIVERSITY
A ND
B. N. MUKHERJEE, M. A.
remchond Roycliand Scholar
ESSOR OF MATHEMATICS (RETD.)
ISH CHURCH COLLEGE, CALCUTTA

FORTY F OURTH EDITION

2O1O

U. N. D1IUR & SONS PRIVATE LIMITED

BOOK-SELLERS & PUBLISHERS
15, BANKIM CHATERJEE STREET, CALCUTrA-700 073
 Published by
RAJENDRANATH DHUR
for U. N. DHUR & SONS PRIVATE LIMITED
15 Banlcim Chatteijee Street, Calcutta-700 073

I St Edition -1938 18th Edition- 1968

19th Edition- 1969 20th Edition- 1971
21st Edition- 1973 22nd Edition 1975
23rd Edition- 1977 24th Edition- 1979
25th Edition- 1981 26th Edition- 1982
27th Edition- 1982 28th Edition- 1983
29th Edition- 1984 30th Edition- 1986
31st Edition- 1987 32nd Edition- 1988
33rd Edition- 1989 34th Edition- 1990
35th Edition- 1991 36th Edition- 1991
37th Edition- 1992 38th Edition- 1992
39th Edition- 1993 40th Edition- 1994
4lth Edition- 1994 42nd Edition- 1995
43rd Edition- 1995 44th Edition- 1936
Revised Edition- 1999
Reprinted, 2005
Reprinted - 2009
I Copjricf lit reserved by the authors I

ISBN-81-85624-91-7

ztdby
Shila Printing Works
6 Rev. Kali [3aneijee Row. Calcutta-7000 06.
 PREFACE TO THE FIRST EDITION

THIS Book is prepared with a view to be used as a text-book for

the B.A. and B.Sc. students of the Indian Universities. We have tried
to make the exposition of the fundamental principles clear as well
as concise without going into unnecessary details and at the same
time an attempt has been made to make the trcatement as much
rigorous and up-to-date as is possible within the scope of this
elementary work.
We have devoted a separate chapter for the discussion of in-
finite (or improper) integrals and the integration of infinite series
in order to emphasise their peculiarity upon the students. Impor-
tant formulae and results of Differential Calculus as also of this
hook are given in the beginning for ready reference. A good nut),-
ber of typical examples have been worked out by way of illustra-
tion.
Examples for exercises have been selected very carefully and
include many which have been set in the Pass and Honours Ex-
aminations of different Universities. University questions of tecent
years have been added at the end to give the students an idea of the
standard of the examination.
Our thanks are due to several friends for their helpful sugges-
tions in the preparation of the work and especiall y t our pupil,
Prof. H. K. Ganguly, M.A. for verifying the answers of all the ex-
amples of the book.
Correction and suggestions will he thankfully received.

CALCUTTA B. C. D.
January, 1938 B. N. M.
 PREFACE TO THE NINETEENTH EDITION
In this edition a few examples have been added here and there and
a set of Miscellaneous Examples has been added at the end. A few
misprirns that had Crept into the previous edition have been corrected.
The generalization of the Rule of Integration by Parts and alternative
proofs evaluating two important integrals
ax COS (Lr
I esinsin S, and have been given in the Section
sin x + cs x
C of the Appendix.
Calcutta B. C. D.
Ju dy, 1969 B. N. M.
PREFACE OF THE TWENTIETH EDITION
THIS edition is practically a reprint of the previous edition : only a
method of finding the C. C. of the Volume and Surface of Revolution
h been given in the section E of the Appendix and some additional
exampleS or various types on C. C. and Moment of Inertia have been
iivcn in the Miscellaneous Examples II of the Appendix.
Calcutta B. C. D.
July, 1971 B. N,
PREFACE TO THE FOPTY-FIRST EDITION
In this edition some rearrangement of the matters have been made
so a. to enable the students to understand the subject more easily,
Mistakes and misprints have been corrected as far as poible.
We thank Sri B. Mahalanabis, M.A. and Sri Malay Chatterjee B.E.
(LI) for their help in identification and rectification of mistakes. Our
thanks are due to the authorities and staff of Messrs U. N. Dhur & Sons
(I'). Ltd. for helping us in bringing out the book within such a short
time. We thank the authorities and staff of Messrs Micromeg (India)
Private Limited for the help extended by them in organising
computerised typesetting of the book. We also thank the authorities
and staff of Messrs Varnakshar for helping in printing the book on
time.
Calcutta
September, 1994 Copyright Holders
 vi INTEGRAL CALCULUS
The University of North Bengal
Indefinite Integral. Definite Integral as the limit of a sum and its
geometric interpretation. Fundamental theorem of Integral Calculus.
Elementary properties of Definite Integrals. Evaluation of Definite
Integrals. Reduction formula for

5 sin h O do, fcOsflOdO and JsiflmOCosflOdo.

RetiIition and Quadrature. Calculation of volume and surface of
solids of Revolution.
D f f rentiul equations Genesis of Differential Equation. Famil y of
curveS repreSented by = J (, y). Solution of first order Difforotitial
Equuovi. Solution of Higher Order Linear Differential Equation with
constant Coefficients. Sirnle applications in Geometry and Mechani
The University of Burdwan
Integral Calculus : Indefinite integral, standard forms. Rules of
Integration. Method of substitution, Integration by parts, partial
fractions. Definite Integral as the limit of a sum, its geometrical
interpretation. Elementary properties of definite integrals. Fundamental
theorem connecting definite and ' indefinite integrals. Certain definite
integrals, viz.
/2/2 /2
Sinn x dx.
sin cos x dx. sinm x cos °x dx.
I0 0 0
(rn, ii positive integers ).
SimpSon s one-third rule for numerical evaluation of definite
integrals. lite of - improper integrals.
Reification of ,i plane curve, Quadrature, Volumi and Surfaces of
solids of revolution. Centreof gravity of simple bodies.
Differential Equation : Cenesi5 of a differential equation. Family of
curves represented by dlv= f , y) - Solution of fi,st order differential

equations. Evaluation of ps'ciaI Solutions passing through a given point.

Linear differential equations with constant cofficients, both
homogeneous and non-homogeneous. Evaluation of special solutions for
given X, y 0 Y'0 of second order linear differential equations with
constant coefficient. Simple applications.
 Syllabus for two-year pass Degree Course
Integ al Calculus
The University of Calcutta
f Integral Calculus (30 Marks) : Integrations of the form
dx I I sin x + iii cos x dx and integration of Rational
-- j n sin x + p cos -
a + p cos x x
functions. Evalua-tion of definite integrals.
Integration as the limit of a sum ( with equally-spaced as well as
unequal intervals).
I I SirlmX dx,
Reduction formulae of J sin"'x cos x dx, J
x
jtan°xdx and associated problems (in and are non-negative integer).
Definition of Improper Integrals : Use or Beta and Gamma functions
(Convergence and important relations being assumed).
Working knowledge of Double Integral.
AppIiations Rectification, Volume and surface areas of solids
tormed by revolution of plane curves and areas ( b y x-axis and y-axis ).
I l'rohlems only I
IJiJjereutial Eq:ia!iou. (20 Marks) Order, degree and Solution uf
an ordinary dirterontial Equation. (ODE) presence of arbitrary
con..tants, formation of ODE.
First order equations (i) variables separable. (it) Homogeneous
equations and equations reducible to Homogeneous forms. (iii) Euler's
and Bernoulli's Equations (I. ii car) (i') E et Equations and thosQ
reducible to such equatiolo.. (v) Clairaut s t'quaion General and
igu Ia r solutions.
Second order linear Fquatins Second order linear fferentiJ I

qMtions with constant co-efficients. Eulers Homogeneous equatns.

Simple applications Orttingontl Trajectories
 CONTENTS
INTEGRAL CALCULUS
PAGE
CHAPTER

I. Definition and Fundamental Properties

Method of Substitution 13
ii.
38
Ill. Integration by Parts
58
IV. Special Trigonometric Functions
V. Rational Fractions
92
Vi. Definite Integrals
VII. Infinite (or Improper) Integrals and
Integration of Infinite Series 150
165
\flfl Irrational Functions
Miscellaneous Examples I 175
IX. Integration by Successive Reduction
and Beta & Gamma Functions 181
Areas of Plane Curves (Quadrature) 225
X
257
xi. Lengths of Plane Curves (Rectification)
xii. Volumes and Surface-Areas of Solids
274
of Revolution
289
xiii. Centrolds and Moments o inertia
310
XIV On some Well known Curves

DIFFERENTIAL EQUATIONS
324
XV. Introduction and Definitions
xvi. Equations of the first order and the
331
first degree
xvii. Equations of the first order but not
 viii INTEGRAL CALCULUS
of the first digree
355
XVIII. Linear Equation with constant coihuients 36
XIX. Applications
393
XX. The Method of Isoclines 40.1
XXL Double and Triple Integrals 41)6
Miscellaneous Examples II 441
Index
cx
448
Past University Papers 4.52

ABBREVIATIONS USED IN THE BOOK

1. I. stands for 'the Integral".
Z C. P. stands for se( in the B. A. and B. Sc. Pass
Examinations of the Calcutta Univrsitv
3. C. H. stands for 'set in the B. A. and ft Sc. Ilonours
Examinations of the Calcutta University,
4. P. P. Examinations of the Patna University.
 IMPORTANT FORMULAE AND RESULTS
of
(A) TRIGONOMETRY
Fundamental relations.
(I) sin 2O + C(,S 20 =1 (iv)sin (-0) = - sinO
(n) sec20 = I + lart20 (v)cos (-0) cosO
(iii) eosec2O = 1 + cot2O (vil tan (-0) = - tat tO
U Multiple an:1es.
-4f1in2A=2 SII1ACQSA.
dos 2A =cos2 A -sin 2fl= I -2sin2 A=2cos2 /l-I.
2 tan A 1- ian2 A
" ii) sin 2A = (iv) cos 2A =I t 2
I + tan A A
-i4 1 -cos 2A 2 sin2A I - cos 2A
'4) 1 + cos 2A = 2c0s2 A (vii) tan2A =
(viii) 1 + sin 2A = (sin A + cos A )2.
(19 1 - sin 2A = (sin A -- cos A )
(x)sin 3A=3 sin A-4 sln3A.
(xi)cos 3A o- 4 COS 3 A - 3 cos A.
3 tanA -- tan 3 A
(xi,) tan 3A =
I - 3 tan A
cot 3A - 3cot A
(xm) cos 3A
=

M. Special angles.
sin 0° = 0 cosec 00
Cos o'=l
tan
I Sec 0° = 1
cot 00 = 00
sin 900 = cosec 90' = I
ens 90° =0 sec. 9O°°° }
tan 90° = 00 col 90' = 0
sin 30° = sin 600
('0€ 30'= Cos 60' )
0
tan 30 = i/JI tan 60°=R
 INTEGRAL CALCULUS
sin 450= ii'I sin 180 0 {)
.os45°= i/'I Cos lSO°=-1
tan 45° = tan 180?=0
sin cos I20°=-.

sin 15°= cos15°='

2'
tan
sin 750 = cos 750
2 2
tan 750 = 2+
sin 18= 1); cos 36°=(4+1).
21=i
sin 24 = 'd
(2-2): cos \(2 + I2).
W . inverse Trigonometric functions.
(i)cosec x= sin cotx=tan71 X = COS71--.

0f)sirr 1 x+cosx= i n. -
(Iii) tan x+ cot 1 x= 1 7L (iv) cosec' x+ sec x= i n.
(v)tan-1 x+ tair1
y = tan-

X y
(vi)tan x- tan y = tan - ' I + xii
(vii)3 sin - ' x= sin - 1 (3x- 4x3).
(xiii) 3 cos - 1 x= cos7 l (4x- 3xj.
3x - x3
(ix)3tan 1 x= tan-
1-3x2
2v1 -x 2x
(92 tan - ' X = Slfl' + 2 = J::-;- = tan1

V. Complex Arguments.
(1) (cos O+i sin 0)" = cos nO + i sin nO.
x24 x2n
(i) cnsx= 1 •-+-—. .+( -1)"-+.. to—
 FORMULAE xi
.iC X'
(iii) sin x = x —+. . .+(- J)'
72u1j!
1 I
(iv)tan' x=x- x)_. . . +(-1)''

(v) c '= cos .v + isin : e= Cos X— isiri x

(vi)cosx= (e+e) :sinx= (e+ e1.

(vii) x" + 2 COS r:x' 1 - - ,j =2i sin n0.

(viii) 2 cos0=cos nO + ncos (n-2) 0

ii(ti — I)
+cos n-4) 0+...
(n being a positive integer)
(ix) (-1) n12 2' ' sin 319
n (n - 1)
co,;= nQ-n cos (ii--2)O+ ------- cos (n -4)9-
(n being an even positive integer).
(x) (I)1h1 1)12 2" sin "
=sin nO-nsin(n-2)9+ 2!"' -4)9--
(n being an odd positive integer).
VI. Hyperbolic Functions.

(I) cosh x= (e+ e'); sinh x=

(ii) eX= cosh x+ slnh x: e= cosh x- sinh
(iii) cosh2 x - sinh2 x 1.
(iv)seeh2 x+ tanh2 x= 1.
(V)COth2 X—COSCCh 2 X 1.
(vi) sinh 2x 2 slnh x cosh x.
(vu) cosh 2x = cosh2 x + slnh2 x
=2cosh2 x- I = I +2sinh2x.
(viii) tanh 2x=----.
2 tanh x
I + tanh-x
Inlegral Calculus (main) -2
xii INTEGRA L CA LCULUS
(Lx) slnh (-x) =- slnhx :cosh (-x) = cosh x.

(x)slnhO=O:cosh I:tanhO=O.
x3
to-

too.

(xiii) sInh x = log jx + ' I( x 2 + I)) for all x.

(xiv) cosh- I X = log {x+ 'I( x 2 - 1)) (x

(xv)tanh1
+'i(1+x2)
NO) cosech1 x=log —
---------(xO).

(xvii) seehx=log

(xv) coth' x = log (x2 > 1 J.

VII. Special series.

I 11 2
(i)+-++...t000=--.
I 1
1 i 1
(Ii)y+-++...t000=-.

(iii)

I I I
(iv) .t000=-.

vm. Logarithm.
W. fli = log, nil logs, a = log,, M log h
 (B) DIFFERENTIAL CALCULUS

1. Fundamental properties.
ci
(I) dx lu ± u ± w ± ..... . ton trns

_
du dv +
d
± dx +to
x - nt-.
ci dv du
(ii) ---f tw)=u--- +v----
dx
du
(iii) (c=c.

du dv
V -U
d (u
(lv) L_J U2

y- dy dz
(v)=
4- --- twherex=f(4 and z= 4) N

H. Standard differential coefficients.

(I)-(c)=O.

(Iii)
(3-) = ( iv) ax)= all ka

(v)')=e'. (vi) fl.x)=

(vii) (logx) = - (viii) (sin x) = cosx

(ix)(cos9=-sthx. (x)(tan x)=sec2x

(xi) (Cot x)=_coec2x

(xI1) (cosec x)=-

cosec x cot

(xiii) dx (see x) = see xtan x

d I
(xlv) j(sIrr14 = (1 • (-1 <x< I)
xlv iNi1X.RAL CALCULUS

(xv) U0 << 1)

(xvi) (tur x) =

ci I
(xvii) -- (COt.! x) - - -
dv

(xvlii) (cose'x)=-
ccv

(xix) --(srv 'x)= x >1)

(xx) (suih x) = cosh x

Lv

(.'oi) - (cash x) = sinh x

d
(xxii) Minh x) = sech2 x.

(x.xili) - (coth x ) = - cosech2 x

d
(xxiv) (scch x) = -- sech x tan

d
(xxv) (cosech x) = - cosec x coth x.

(xxvi) = slnh'x) =

(xxvii) (cosh'x )= --4----- (K> 1)

(xxviii) ftanh'x)= -------. (X< 1)

(xxix) (coth)x

d 1
(x) cosech x)=-

(xxxi) (secW x = -----. (x< 1)

 FORMULAE

III. Important results associated with curves.

(I) Cartesian subtangent =
(ii) Cartesian subnormal = yy1.
(iii) Cartesian normal = y '( I + y

(iv) Cartesian tangent

= dO d:
(v) Polar subtangent = r2=
dr
-du; (u =

(vi) Polar subnormal =

Jr I du (u
=
I).
d d
(vii) tan W = ; COS l = sin 'i' U
ds
dO
LID ; cos =
(viii) tank = r — sin k = r ds
-
dr ds
(jx)ds1=dx+dy'dT'+ r1d01.
.!)2_ + ( 2 idsV
( dx dx,'(j ) =1+I:;).
ds 2'dO (dr
\2 ( AS- 2 =r+
= 1 + r' ' ar/
dr )

I •1 I 'Jr (du '2

(x) p=r sin O;--j -- +
= U2 + )
ds (1+y 1 2 ) 312_ (r 1 + r1 2 ) 3 Jr
(xi) p==
Y2 _r2+2rt2_rT2rdPtdW,
 (C) INTEGRAL CALCULUS
T. Fundamental Properties.
(i) J
(f( x) ±fz (x) ±f) ( x) to n terms } dx

f 1(x)dx±ff2(x)ax±ff3(x)dx+...tonterms

(li) 5 cf(x)dx = cjf(x)dx.

I. Standard Integrals.

x n+1
- (
x"I n *-I
\VIxd

(ii) f dx-- (n -
1

dx = x. (iv)
5
= 2&.

(v)
5 dx
-
x = logt xI (vi)
5
e dx.=e -
e'dx = e l(YHiJ5 adx = a (a > ø.
log, a
cos mx
sin mxdx -
5
(ix)
m
'ç_f sin xdx = - cos x.

(xi)
5 cos mxd x= sin mx
m
\ei1rj cos xdx = sin x.

2 xdx =

cosec 2 xdx = - cot x.

 FORMULAE xvii

sec x tan x dx sec x.

cosec x cot x dx - cosec x.

\li) 5
sinh x dx = cosh x.

cosh xdx = sinhx.

5
tanh x dx = log I (cosh x ) I
(xix)5
cothxdx = log1 (slnhx)I
(xx)5
(xxi) cosech x dx = log I (tanh- x ) I
5
pc5iO
J sechxdx = 2tan (Cx).
sech 2 xdx = tanhx.

jv) cosech 1 x dx — coth x.

5
11. Standard Integrals.

%of f'(x)
f(x)
dx loglf(x)I .

kiiY j tan xdx = log i secxl

\iII1J cot x dx = log I sin x I

dx I x
_ tans - (a * 0).

\J
1 x 2 +a I a a
dx =1 I x - a
.log (I xI > 1 aI ).
x— a2
2
x +a
xviii INTEGRAL CALCULUS

vi) dx I A
f a' - = '- log a -+ x U xl < I al

dx
(vii) f 'I(x'+ a')1091
= x + X2+ 4 2 I = sinh-'--/

dx
(V iii),,J(,_,)logIx +
f 4x ' — a2 l cosh—,!_

idx X
(ix) x') = Sin ' -
a <IaI)

(x)
5 xI(x 2 a 2)
dx
-
a
sec-'
(x
a)
f -.-I) = -
(xi) ,

(xii)
5 (uv)dx = u
5
Integral Of the product of two functions
vdx - t ( du
J .- J vdx) dx.

= 1st function x integral of 2nd

- integral of! Differential coefficient of 1st x integral of 2nd J.

(xiii)
5 e cos bx d C (a cos bx + b
a2 +b
sin bx)

= 4(a 2 + b2 ) COS (bx - tan'

(xiv)
5 e" sin bx dx = C (a sin bx - 1, cos bx)

= fr,) sin
(
bx - tan - iL)
(xv) f 'x iidx = x IT' +a2 +--
a'
log i (x+'Ix' +a' )j
2
' + a2
x -Tx— a' x
1- - sinh-—
= 2 2 a
 FORMULAE xix

= x1x;a
('t) J1x 2 -0 dx log i (x+4x 2 a2 )I

xIx 2_ a 2 a2 x
= --cosh—
2 2 a
Z.1a2_x2 a 2 x
'/a -x 2 dx = in
s
(xvii)J ,2

cosec x dx = log I tan x I

(xviii)5
= log I cosec x - cot xl
Nix) sec xdx = log 1 tan (-n +
5
= log 1 sec x 'i- tan xl.
IV. Definite Integrals.
(A ) Definition.

h f(a + y k) or, Lt h f(a + rh)

h

= L f( x ) dx, where nh = b - a.

(ii) LI + ;: (L_a))
f (
I
/ (-;) = 51 (. ) d

(13) Properties.
b
=r -
f
Ja dx F(x) =
b b
(ii)
fa
f(x)dx
= 5 f(z)dz.
 INTEGRAL CALCULUS

(iii) f(x)dx f(x)dx.

J S - Jb
(iv)Jf(x)dx
f fic
(x)dx
+ J f(x)dx, (a<c<b).
f(x)dx f(a - x)dx.
(v) 50 =5 0

.
f(x)dx = n f(x)dx, if f(a + x) =f(x).
(vi)J0 J0
(vii) f 0 1(x)dx = 2 5f(x)dx. if f(2a - x) = f(x),
= 0,iff(2a - x) = -f(x).

(a) sin x dx = 2 sin x dx.

.1 0 J 0

() cos xdx = 0
.10

f f(x)dx, if f(-x) =f(x),

f S f(x)dx = 2 Jo
(viii) 4_
= 0,if f(- x) = -f(x).
fi sin, x dx
r
cos x dx.
(ix)J 0
=J 0

n - I n-3 n-5
- n
n-I n-3 n-5 42
or .
n n-n-42. 53
according as n is an even or odd integer.
 FORMULAE LU

V. Beta and Gamma Functions: Infinite integrals.

•1 t
(I) B(m,n) x''dx
x"- (I - x)'-' dx -
= J 0 J0 + x)""'
t x' - 'dx
=I (1 + x)""" (m > 0,n > 0).

(ii) 1(n) = 5e'xdx (n >0).

(iii) B(m,n) = B(n,m) = r ( m) r (

r(m+i)
(iv) r ( n + 1) nr ( n ), r ( n + 1) = n !, if n is positive
integer.
(v) r(1) = 1,F(-) = 1n,r(m)r(1 - m) = sin mt
(0< m <1).
sin pOcosq8dO
(vi) J 0.
2___
rp> 1
2 r (P + 2g + 2\1 [q > =1 ]

=! _.±J'
2B (L±J
2 2 1°
R/2 ,
(vii)$ sIn ode
0 f co 'Ode = 2• r (a.2)

(p > - H.
(viii) J e k dx = n) (k < 0>.
a
 ILGRAL CALCULUS

(ix) J e' 2 dx =

(xY $ e ' cosbxdx = a b1 (a > 0).

(x 0ec sin bx dx = (a > 0).

0

r sinbx ac.ording as b > or < 0.

dx = or - . It
(xii)J
r = sin It
(xiii) dx = -
x 2
0

VI. Areas of Plane curves.

1. Cartesian co-ordinates
(a ) If the curve be y = x ), area = Iy dx.
b ) If the curve be x = f ( y ). area = J x dy.
c ) If the curve be x = o(1), y y ( t Para
)( nwiricform ),
the formula for the area is either ( a ) or b.
2. Polar co-ordinates
lf the curve be r = f(e),area = - r2dO.

3. Important areas
(a ) Area of the ellipse x2 la 2 +v 2/b 2 = is mab.

b ) Area of the circle x 2 + y 2 = a 2 is ha 2.

VII. Lengths of Plane curves.

1. Cartesian co-ordinates

dx
Arc length
= 5 ds = 5
 FORMULAE

=5 + dx() dx,

il the equation of the curve be y = f(x);

f (dy dx .

if the equation of the curve be x = f(y).

Arc length
= 5 = 5 ±s-
ds dy.

Arclength ds=54dt
=5
Jj(dx )2
Of )' dt
if the equation of the curve be x = ( I), y = ( t).
2. Polar co-ordinates

(a) Arc length - ds dO

5 5
f)i
5
if the equation of the curve be r
r' + (

f( 0).
do,

=5 =5
(a) Arc length ds dr

=5 4 r +
(J9)2 dr,

dr
if the equation of the curve be r f(0),- , z.e.,1 /
being expressed in terms of r.
3. important lengths:
Perimeter of the circle x i + yi 9 2 is 29a.
 xxiv 1NTERAL CALCULUS

Perimeter of the cycloid, x = a ( 0 + sin 0),

Y = a(l - cosO)is8a.
For the cycloid, x = a (0 + sin 9),
y = a - cos 0),s = 8ay.
Perimeter of the card bide r = a (I - cos 0) is 8a.
VI!!. Volumes of the soiiu 01 re.
1. Cartesian co-ordinatcs
(fl) Volume V = itiy'dx,
if the axis of revolution is the x-axis.
( b ) Volume V = n Jx2dy,
if the axis of revolution is the y-axis.
2. Polar co-ordinates
Put x = r cos 0, y = r sin 0 in (a ) and ( b ).
3. Important volumes
a ) Volume of the solid generated by the revolution of the
circle x1 + y 2 = a 2 ,i.e., the volume of the sphere of radius
a=
(L') Volume of the ellipsoid formed by the revolution of the el-
lipse x2/a2 + y 2 1b 2 = 1,
i) round the major axis = .nab
(ii) round the minor axis = na lb
c ) Volume of the solid generated by the revolution of the
straight line y = x tan (x, about the x-axis,
i.e., volume of a right circular cone of height h, radius a and
semi-'crtical angle a = . rh I tan 2 a
= .lta2h
(d ) Volume of a right circular cylinder of height h, and base
of radius a is =
 FORMULAE XXV

IX. Surface areas of the solids of revolution.

1. Cartesian co-ordinates:

(a) Surface area S = 2n

Jy ds = 2n
5 y Ls dx

= 2n
5 y .\ji+ () dx,.
dx

if the axis of revolution is the x-axis.

(b) Surface area S = 2n

5 = 2,t 5
x ds x Ls

2n 5x ji + (
)2
= Lx dy,
dy

if the axis of revolution is the y-axis.

2. Polar co-ordinates
Put x = r cos 0, y = r sin 0 in (a) and (b).
3. Important surface area
Surface area of a sphere of radius a is 4ita2.

X. Symbolical Operators.
1
(L I (D) e e" , if f (a) * 0.
- f(x)
I
(ii) ) e" V = e- f(D +

where V is any function of X.

1. I sin (ax + b
sin (ax + b ) =
• ( D1 )
if 0(— a 2 0.
xxvi INTEGRAL CALCULUS

(iv )-
I I
cos (ax ± 1')
(— a ) cos (ax +
if (_a2 ) * 0.

(v) (x - f' (D)) -4 v,

where V is any function of x.
 INTEGRAL CALCULUS
CHAPTER I

OFF '! )T1NOAMEN7AL PROPERTIES

1.1. Two view-points of Integration.

There are two distinct view-points from which the process of
integration can be considered. We may consider integration as the
inverse of differentiation and make this as our starting point; or else,
we may start with defining integration as a certain summation and
then proceed to show that the result is identical with the reversal
of a differentiation.
The establishment of an identity of the two view-points is
referred to as the Fundamental Theorem of Integral Calculus. We
shall consider both the points of view, starting first of all with the
former, reserving the consideration of the latter for a subsequent
chapter.

1.2. Integration, the inverse process of differentiation.

If f(x) be a given function of x and if another function F(x) be
obtained such that its differential coefficient with respect to x is
equal to f(x), then F(x) is defined as an integral, or more properly an
indefinite integral of f(x) with respect to x.
The process of finding an integral of a function of x is called In-
tegra1i on
and the operation is indicated by writing the integral sign
.1 before the given function and dx after the given function, the dif-
ferential indicating that x is the variable of integration. The (unc-
dx
tion to be integrated, viz., f(x) is called the lntegrand.

Symbolically, if - F( x) =
then ff(x)dx = F(x),
• 1 listorically this sign is elongated S, the initial letter of the word 'sum',
since Integration was originally studied as a process of summation. (See
Chapter VI.)
integral Calculus (main) -3
2 INTEGRAL CALCULUS

where If ( x ) dx is called an indefinite integral of f ( x ) with

respect to x.
Thus, considered as symbols of operation,

() and () dx are Inverse to each other.

A lternatively, using differentials, the above result may be writ-
on as follows:
Since d(F(x)) f (x ).
f(x)dx F(r).
f
(Thus, since i ( sin x) cos x dx, .. cos x dx = sin r.
5
Again, if f(x ) be a given function, F(x) an integtai off ( x),
and x a and x b be two given values of x, the change in
:he value of the integral function F ( x ) as x chages from a to b,
i.e., the quantity F ( b) - F ( a) is defined as the definite in-
tegral off( x) between the 'limits' a and b, which is denoted•
by the symbol

In other words, if 4-F ( x) =f ( x) ,for all values of x between

X
a and b,

then F(b)_F(a)Jf(X)dX

which is called the definite integral of f ( x ) from a to 1' , and 'b' i

called the upper limit and 'a' the lower limit of the definite integral.

Cor.5f'(x)ax=f(x) and jf'(x)d f(b) - f( a) -

£ _
• provadedf(x) sattshes certain general conditions which will be discussed
later. ( See Chapter VI).
 l UNDAMI NIA L PROP[Jfl1ES

1.3. Constant of integration.

It maybe noted that if F(x) / (x), then we also havi'
- ( F Cx) + C) 1(x), where C is an arbitrary constant. Thus,
if If (x ) dx = F ( x ) , a general value of the indefinite Integral
Jf(x)dx = 1(x) 4 C.
In other words, in finding the indefinite integral of a function
f( a ), an arbitrary constant is to be added to the result to make it
general. This is the reason why the integral is referred to as an in-
definite integral. The arbitrary constant is usually referred to as the
constant of integration.
It is easily seen, however, that in evaluating a definite integral
this constant of integration cancels out and its value is thus definite.
Again, we may see that two functions haoin5 the some derivatirec
dijfer only bya constant. For, it (x) = ' (x), and if
y x V( x),then we got Oalways,and
- thus the rate of
change of y with respect to x is zero everywhere and hence y is
constant . Thus, it is possible to get the indefinite integral of the arne
function in different forms by different processes, but ultimately these
forms can at most differ from each other by constant quantities only.
Hence, an arbitrary constant added to the indefinite integral of a
given function obtained by any process makes the result perfectly
general.
In the following pages, we shall first of all deal with indefinite
integrals. For the sake of convenience the arbitrary constant of in
tegration has generally been omitted but it is always understood to b&'
present In every case, and should be supplied by the students in the
result.
It should also be noted that in case an indefinite intcgr. con-
sists of a sum or difference of two or more integrals, the addition
of one arbitrary constant for each integral is equivalent to the ad-
dition of a single arbitrary constant, denoting their algebraic sum,
in the final result. For illustrations, see Illustrative Examples in
Art. 1.7.
Forarialyial proof see Authors' Diffrrential Calculus, ,4-t. 6.7, Lx 1.
4 INTEGRAL CALCULUS

1.4. General laws satisfied by Integrals

(I) The Integral of the sum or difference of any finite number of func-
tions is equal to the sum or difference of the integrals of the functions
taken separately.
This follows immediately from the known results of the Dif-
ferential Calculus. For we know that
(x) + 12( X ) - f,(r) .....
WX
= f'(x) + 12(1)- f 3 '(X) 4.

(f l, (x) + f,'(x) - f 5'(x) +

J
f(r) + f2(X) — f3 (x) +...
zJf'(x) dx+ Jfa'(x)dx Jf i'(x)dx +

[since .dx— f(x) = f(x), :.f(rr If (x)dx,etc. j.

(ii) The operation of integration is commutative with regard to a con-
g iant, i.e., a factor of the integrand which is constant with regard to
the variable of integration can be taken outside the sign of integra-
tion.
SymbolicalIy. A f ( x) dx = A j ( x ) dx -
This follows immediately from the fact that
[AF(.)) = ia4- ( )} Af ( x), SaY,
TX

so that JAf(x)dX = AF(x) = Aff(r)dx

4 -F( x) f(x).
since, by our supposition, dx
(iii) Combining the above two results, we can write
{Af1 (x)± Bf2 (x)± Cf 3 (x) +
= A Jf(x)dx ± RJfi ( x)dx ± CJf 3 (x)dX +
 FUNDAM EN TAL PROPERTIES S

1.5. Fundamental integrals.

A slight acquaintance with the Differential Calculus will at
once suggest the integrals in many elementary cases. As the first
step towards the facility in integration, the student must be thoro-
ughly familiar with the following fundamental integrals
FUN')AMENTAL INTEGRALS
p
(I) Ix'dx - ---- (n, —1).
J n+1
Cor. J iii I
(n_l)xfl_I^_1>
C dx
= 2'x.
J;

(n) -.-= logi xi.

I
J
etmx dx =
in

Cot. x dx
fe

$ a5dx
=
ax
(a > (, a

(iv)5 sin mx dx
= -
cis mx
m

Cor.
5 sin x dx = - cos 5.

(v)5 cos mx dx
= Sin mx
in

Cor.
$ cos x dx = sin x.

NO
$ seOmx dx
= tan mx
m
 f
1N1EGRAL CALCULUS

Cor. sec 2 xdx = tn x,

Ccot mx
(vu) i coscc 2 mx dx = m
J
Cor. cosec 2 x d = - cot L
5 sec MX
rI sec mx (an mx dx =
(viii) m
J
Cor. sec x tan x dx = sec x
C
cosec mx
(ix) I cosec mx cot nix dx = - m
J
('or. c osecx cot xdx = - cosec x.
5
(x)
f=
C
I sinh mx dx
Cor.
5
co h mx
m
sinh x dx = cash x.

(xi) cosh mx dx = sinhmx

in

Cor. Cosh x dx = sinh x

J
The results can be easily verified by differentiating the right
side in each case.
Other standard results of integration which are also very use-
ful for application will be found in the subsequent chapters.
Note 1. Since the integral (i) is frequently used, for the save of con-
'Incease
vee j,.'oce, we g i ve here a conek.' vrhal statement of the result, viz ,
the index by one and divide it by the increased index".
d
Notc2. Since log xis real when x>O and (OgE) =

So
5 .!
= lug x is defined for x > 0. When x e 0 i.e., - r > 0,
log) .LThCICfOTe when x <oJdx =log(-r)
I hence, both these results will be included if we write L dX = log I x
5
lit
the lorinula and examples where integrals of this type oc urs, i.e., where
 FUNDAMENTAL PROPIR11ES 7

the value of an integral involves the logarithm of a function and the (unc-
lion may become negative for some values of the variable of the function,
the-absolute valuesignI I enclosing the function should be given, but it has
generally been omitted, though it is always understood to be present and it
should be supplied by the Students.
Note 3. Different algebraical symbols a, b, en, n, etc. Occurring in it
tegrands are gneralty supposed to be diffcrQnt unless otherwise stated.
Note 4. In the above integrals (iii), (iv), (v), (x), (xi) it is tacitly 3SSUmec
that on is a non-Zero constant.

1.6. Standard methods of integration.

The different modes of integration all aim at reducing a given
integral to one of the Fundamental or known integrals. As a matter
of fact, there are two principal processes
(i) The method of substitution, i.e., a change of the independent
variable.
(ii) Integration by parts.
In some cases, when the integrand is a rational fraction it may
be broken into partial fractions by the rules of Algebra, and then
each part may be integrated by one of the above methods. (See
Chaptcr V)
In some cases of irrational functions, the method of Integration
by rationalization is adopted, which is a special case of (1) above.
In some Cases, integration by the method of Successive Reduc-
tion is resorted to, which really falls under case (ii).( See Chapter M.
It may be noted that the classes of integrals which are reducible
to one or other of the fundamental forms by the above processes
are very limited, and that the large majority of the expressions,
under proper restrictions, can only be integrated by the aid of in-
finite series, and in some cases When the integrand involves expres-
sions under a radical sign containing powers of x beyond the
second, the investigation of such integrals has necessitated the in-
trod uction of higher classes of transcendental function such as el-
liptic functions, etc.
S INTEGRAL CALCULUS

In fact, integration is, on the whole, a more difficult operation

than differentiation. The Differential Calculus gives general rules
for differentiation, but Integral Calculus gives no such correspond-
ing general rules for performing the inverse operation. Integration
is essentially a tentative process. In fact, so simple an integral in
appearance as
r Jxcosxdx, or rsin — dx
j i
can not be workcdcut; that is, there is no elementary function whose
derivative is 'Ix cos x, or( sin x )/x, though the integrals exist.
There is quite a large number of integrals of these types.
1.7. Illustrative Examples.

Ex. 1. Integrate sin 1xdx.

5
I= cos
- x- J.cos 2x dr
2x)dx= J4a
+C
4J dx -+1 c os 2z x - ••
= sin Zr+C.

E.2. Integrate tan zxdx

J
I =( sec 1 x- I )dx =Jsec1xd(_5dX
tan r - r + C.
5(x - 3)2
Ex. 3. Jnteg rate — dx .
5 x'4x
5x2_30X+45 dx =5 Jx1 dx _305 x2dx
=S _x •Jx
i 2 +C
30.2 'Ix + 45(- 2x)
31
604x - 90x 2 + C.
Ex. 4. Integrate Jsin 3x cos 2xdx.

sin 3x cos 2X =.2 sin 3x cos ZX = ( sin 5x + sin x).

 FUNDAMENTAL PROPERTIES
9
Ex. I

I sin 5xdx + J sin xdx)

= Cos 5x - COS xl + C.
Note. Henceforth the arbitrary constant of integration will be omitted
In the Illustrative examples, as also in the ap swers to the set of examples.

EXAMPLES I

Integrate the following

1. (I)
f( 1+x)3 a (ii) J.ix(x s + .)dx.

2. (I) cos2xdx . ( jj) 5 ! dx.

$
C 1 - tan2X f I + tan 'X dx
- dx. (tv) -
1 + COt2X
1 + tan2x

3. Jsec x( sec X _ tan x)dx.

4. (I) Cos ax dx . ( ii) 5 COt 2 X dx.

5
C 2e ' + 3e 41 + 4 dx. ..() f e 3
+
5. (i)
J e j dx

e sI.. —
(in) J - dx

(e°S + e''°s')dx.
6. 5 dx
x x —-1 'J x + dx (ii) $ -__
7. (I)
5(
8. (i) $ ( a 3 - x3
)3
dx (ii) 5 ( x + 2 )( x + 3 ) d
a + a'+ a" dx. + 4i dx.
9. () j (ii) J 2'
i) C (1 - 2x) dx. .. I C asin'
----x + bcosX dx.
(it) sin 2XCO52X
I 1 --
O.( X'X -)
j
10 INTEGRAL CALCULUS

fcs'x +
dx.
J cosx + sin
sin +tanx
cosec x
1. (t)
f dx. (ii) 5 x ° dx.

12 $cosccx + tan I x+ sin 2x

- sin I —dx.

fx 3 4x?+5x 2
13.(i)
f) x1 -
dx. (ii
2x + 1
x3-6x+9
dx .
r sin + Cos x
14.(i)
J + sin2x) dx. (ii)
5 cosx - Cos 2x
I - cosx dx
(iii) f sec 2x - dx.
J sec 2x + I
(iv)
5 dx
cosh x + sfr}x

15. (I)
5( 1 + sin x ) dx.
LI ± sinx = (sini ± cosfr)21
(ii) 5 'I( I - sin x dx).

16
5 cosx - sinx
cosx + sin (2 + 2sin2x)dx.
17. (i) 5 '1(1 + cosxdx). (ii) 5 Cos
1( I - xdx).

(iii) 5 (3 sin x cos'x - sin 'x) dx.

18 (i)
5 dx
1+sjnx
1
dx
+ cosx 5
I (I) Multiply numerator and denominator by I - sin x

19 (i) -
Cos 2a .. $cos5x + Cos 4x
J cosx - Cosa dx. (ii)
- 2 cos 3x dx.
1(11) Multiply numerator and denominator by sin 3x. I
20 dx
(i)J 2XCOS2X IH.S.'78,'85,J.E '811
 FUN[)AMINTAT. PROPERTIES II
Lx 1

sin'x 4-
dx.
(ii)
S jfl'XCO S
2

cos' x In (he numera t or .1

((i) 7ut sin e x +

(I - COS 4 X -
dx.
2sin2xcOS1x 5
cUSX
dx (iv)
(I)
J -

I. cos x (1 - 3cosX)dX.
21. I
S1fl2X

( ii) sin x dx. 111. S. '85 1

22. (I) cos ' x dr-
5
5
(j) Write co; x = ' ( cos 3r + 3 cos x
[ For another met hod see Art- 4.3.)
23.
(i) 5 sin ,nxsin n.xdx . ( ii)
5
cos 2x cos 3XdX.

(i) 5 sin x cos 2 x

24. dx. (ii) 5 Sin x cos 2x dx.

sin x sin 2x sin 3xdx.

25. 5
ANSWERS
x' +
.X 4- j W) j1j- x + 6x
log r
x.
2. (I) 4x + - stn 2x - (ii) tan x - x - (Ui) .s(n 2x . (Iv) tan x -
sin2ax
3. tan x - x 4 (.)- ,I + -4--- (ii) - cot x - x
(iii) ix'.
5. (i) -2e +3e' - e'. (ii) 14 e
x•1 . (I) !/ - +
a+1 Tog
+ 1. x' + - X + jX.
(ii) x + X + 1 x 3 + x' +
. (I) a1x -
2,4/3 x + a x -
(ii) -z(3x 2 i. 32x' + 126x + 216).
12 INTEGRAL CALCULUS

9. (I) -
i r a
- I + + -
iogaL 3 2
GOTo4 2 [22_

10. (1) - 3x - If) - x 513 + x it

(ii) a Sec r - It cosec x.
(iii) x + Cos 2x . 11. (i) sin - cosec x. !. sin x
12. - cot x + Sec r - cos r . 13. (1)x 2 - 2x. (ii) fx 3 - +3
14. (i) x . (ii) x + 2 sin x. (iii) tan r - r. (iv) sinh x - cosh r.
15. (i) 2 (sin f x - cos -l x) or ( 2 11- sin x)

(ii) 2 'Il + sin x.16. Sin 2x. 17. (i) 242 sin fr.
(ii) -2 42 Cos +r. (iii) - Cos 3x.
18. (I) tan x - sec r. (ii) - cot x + cosec x
19. (i) 2(sjn x + x cos a).
(ii) - (sin x +2x). 20. (I) tan x - cot x.
(ii) tan x - cot x - 3r. (iii) - ! sin 2x (iv) i-.. x

21. - cosecx + 3 cot x + + sin 2x.

22. ( j) sin 3x + ! sin x (ii) 32 sin 4x - sin 2x +

23. (i) sin (m - Ii ) r sin (m + n) x
2(m-n) 2(m + n)
(ii) sin 5x + sin x. 24. (I) x - sin 4r.
(it) (sin 2x - x - ' sin 4x ).
25. - cos 2x - Cos 4x + - Cos 6x.
 CHAPTER II
METHOD OF SUBSTITUTION

21. Change of variable.

Let I f (x) dx, and let x = 0 (z),.
=
J
Then, by definition, f( anda)

by definition, I
= 5 f( (z)) '(z) dz.
we put x = (z) we are to
NOte 1. Thus, if in the integral If( x ) dx
replace x by 0 ( z) in the expression f ( x) and also we ar.-
, to replace dx
by ( z ) dx and then we have to proceed with the integration with z as
the new variable. After evaluating the integral we are to replace z by the
equiv!ent express ion in x
' ( z) lit making
Note that though from x = 0(z) we can write =
our substitution in the given integral, we generally use it in thedifferential
It really means that when x and z are connected by
form dx = 0z ) dz.
the relation x = 0 ( z), I being the funetion of x whose differential coeffi-
it is, when expressed in terms of z, iden-
dent with respect to x is f (x),
tical with the function whose differential coefficient with respect to z
is f((z)) 0, z) which later, by a proper choice of 0(z), may possibly
be of a standard form, and therefore easy to find out.
Note 2. Sometimes It is found convenient to male the substitu-
= z where corresponding differential form will
tion in the form '41 ( x )
be w'( x ) a di; by means of these two relations, f( x) dx Is transformed
Into the for F ( z) dz.
2.2. IllustratIve Examples.
Ex. 1. Integrate J(a + bx)dx.
.. = dz ....dx = (1/)dZ.
Put a + bx = z.
•1 I I '' I
1 =1 z&dZ b ZdZ
It n T1 i7iib .(i +
 14 J'flEGRA L C'ALCL)LLS

Ex 2 Integrate

Put T
J dx
x4(x2..a1)
a sc 0. .. dx = a sec 0 taxi 0 dO.

= •fasecetanodo laO =!
.- JasccO.atano aJ a a a

Cor.
f 'l(x
dx •ec - X.

s_ I
Ex. 3. Integrate '1(1in
-

Put sin x z. Ox )dx dz.

I = I z dz = + z = f ( sin - x ) 2

if
Lx. 4. Show that
tan x dx log I sec x I

(ii)
J cot xdx - log I sin 2d

(i) Put cos x = z then - sin x dx = dz.

t dx
1 = l-dx = - I - =log
- z
JCos x Jz
- log cos x = log ens x = log I sec x I

(ii) Similarly, by substituting sin x = z this result follows

Oth e rwi s e
(i) 5 tan x dx =
x
see x tan - dc = log s'c
sec r

GO fcot xdx
=J _E!!
sin r dx= log[ sinxl ISee Ex. 5 below j
 Miflh101) OF SUI3SITIUTION IS

F.!. Show that

t'
P(X)
—de - logs fix)I
f(s)
J

p,, f ('-) =

Hence,
I
=5 -!- log lf(x)I
log z t'.

,
If the Integrand be i fraction such that Its numerator is the differenial
coefficient of the denor;Inator, then the Integral is equal to log I ( deno-
minator)!
cosx - ,inx
--dx = log! (stnx + cos x)I
Thus
J sins + r osx

t 2ax+t
= log! (.4x bx r)I.
Jax 2 bx+C

The principle is also illustrated in Ex 4 above.

i
sn
Es. 6. Integrate X - dx
f 3 + 3cosx

21 -3 sin x
l can be written as
_J 3-dx.
cosr

Now since the numerator of the integrand is the differenhal coeffi-

cient of the denominator,
I = -logI (5 + 3 cos x)I

dx
Ex. 7. Integrate
'(x + a)+ 'J(x + b)
Multiplying the numrator and denominator by - - we
have

i == 4__[J' 4x-
. a dx- Jr b dr}

Putting x + a = z, so that dx z.
16 INTEGRAL CALCULUS

J 'Idx = 5 Jzdz = 1 z" = 1 (x + a)".

Similarly, 5 + b dx = 1( z + b) 112.

I = - 2 1 x + a)" - (x + b)3fl1
-s [ (
(a + br)'
(a
Ex 8 Integrate dx
(a' + I" )'
z - a'
Put a' + b'x = z, or, x = - .'. dx = , dz.

Now the given integral becomes

b
a + ,(
- z - a'
I dz
-=vJ
(bz + ab' - a'b )'
dz
J{ 1'

—_J_ [b 2 f+2b(ab'_a'b) 5-4+ (,,b'—a'b)2 f dz]

b2 2b (ab' - a'b) I (ab' - a'b ) 2
= log z - - 2b''
b''
152 ab' - a'b ) (ab' - a'b ) 2
= log (a' + b'x) --lb-(
b'' (a' ' b'x) - 21.,' (a' + b'x)'

+ b)"
Note. By the same process we can integrate
J(+
where m is a positive integer, n being a rational number. (Cf. § 9.13
dx
Ex 9 x'(a+bx)''
Integrate

Put a+bx=zx, u, -.+ bz. Then --dX=dZ.

12
T1* given integral thcj x

- i( dz — 1 dzfz-b\'
aj x.z 2x 1a J a /
 METHOD OF SUBSITI1J7ION Il
Ex. 110)

I b']
-
._3bz+3blo
z
1 (a+bx ) 2 ( elbx)
[ 1 _
x
3b I log x a+bx/j
dx
Note. By the same substitution the integral bx) can be
xm J
obtained where m and n are positive Integers, or even when they are Inc.
tions such that m + n Is a positive integer greater than unity. For another
method see § 9.13.
EXAMPLES 11(A)
Integrate the following

i (I) 'z dx (ii)

1 +.x2 Ji - x2)
J
cos (log) dx.
_,4ij) dx. (Iv)
-J sin4x
J
cos 4x - cos 2x dx
(v) .2dx
sin 4x - (vJ cosec 2x —cot 2x

2. (i) JxTi+ 1 dx. (10 1 x 3 dx.

J x''t: — +

3. cos4xdx.. (u) .1 1 + cosx

1 (tan x) SIflX
4 sinx Cos x
dx. cos s x
J
1 +cosx ( 1_+ cosx
. dx (11)1 dx.
5. (')J ( + sinx) J x + Sin x
tan (log x) f dx
6. dx.
(ii)J x log x
cosxdx cosxdx
7
J + SjflX) (a + bsinX)2

Inlegral Calculus (main) -4

18 IN11CKA I. CM.Ctfl.US Lx 11(A )

S. 0) fT . X)
(fl)f dx

I Pit x = sin 0.
dx
-J (1- x )/(1-
ax x1 2
+ x )I(1 +x2)

- I
(.)Jee + I '
---dx. e?+ i 111. S. '85

I Multiply the numerator and denominator of (I) by e , and that of

(ii) by e 'j
tanx
10 --- dx. (n) --- dx.
(i)I log cos x
J Sec x cosec x r
lf
og s.nx
sccxdx
log tan x
dx (ivi I --______________
j )og(sc'cx 1 tan x)

sin 2x dx
JI
fasin1x + bcoc .Ai1ifa tan x dx
+ btan2x
sin 2x dx
('11J(a2cos2x+b1sin2x)7
tan x scc2x
(iv)J a ( 2 s. b2tan2x)Z dx .
fl. A'6'Jx sin x2dx. (11)5 sin xdxcos x
3x— I x dx
13
(i)S J(3x 1 -2x + 7) dx. f(x'-a)

14. 0)5 (1 +x 2 )(dxtan 1 x dx

(ii)5W((nx )

15. .if: b+1 dx. dx

fe - 1)2
dx
I H. S. '87 1
j (e - 1)
 MimloD OF suI3sTrrUTlON 19
Ex 11(A)

Jer(l +x) dx. 1 Put xe =

16 cos' (Xe'
dx
17.)J7TY(xT)
dx

5T
(ii)

dx
-3) + (x - 4))(x - 3)(x - 4))
J -
dx J xdx
('i)
)8
I Put 'Jx=z.) -
dx dx.
19. 0)jf

O (i)J(3X + 2)2X+ ldx.

dx. dx.
on 5---.
f2x + 3 dx dx .
22. (I) \.1bx

f 2x -t
(j) J.)dx.
I
23. S , ( a + L'x) dx
\p1-:' 5 4 a_+_X
a-i
dx . I Put x = a cos 28.
-
J2x3+3x2+4X+5.
dx
25 -t 1

'Ix x2
_______
26. (I) f ,)dx. x') dx
f
I Pu! x' =a sin 2 0 in (I) and a'sin e in (ii).]

x'dx
(iii)
dx1
. (iv)J -: -i .-;:-- j
20 INTEGRAL CALCULUS Lx . 11(A )

,2(
J\/ x dx. (Put = a six 2O.I

e dx six
28. (S)(al xl)4 (I - x)'I(l - x2)
J
r+Zd X2 + I
4Uf( i - .) e - (ii)
( (x' - I)' dx.

30.a Cos x - ksinx dx.

f
a sin x + cos x + c
+dx ((log sec x)2
4jyI • 1a blogx) J cot x
dx.

x2+1
32
• S 3d(x'+3x1-6)
dx.

' jwfsxcos(sinx)dx. (ii)f sinxcotxdx.

(iii) Jtan x tan 2x tan 3x dx.
-1r
5 1 + xi
dx. t Put x = Cos O.)

cot a cot x dx.

COS X -
dx.
35 i)
(f cos(x + a) 5 cot cot
dx f x'dx
x 2(a bx)2
- (ii)J (I - r') 2
x
37. ()
5_ dx. 1 + x/4
fPutx z'. j
\
J(1 x4
dx.

____ dx 2x ax
_____
38•
(i)S x(r - I) (iiJ(1 - x1)(x 4 —I)
Putx secO.J

,( f{f(x)'(x)+(r)f'(x))dx.
 METHOD OF SURS1TI1JTION 21

40. Integrate . f'( x) with respect to x 4 where

f(x) =tan - 'x + Iogh—
+ x- log4i.

ANSWERS
tan 'x asin -Ix. ( Iii)sin( log x ) .
1. (1) e . e
(ii) a -,
(iv) - cot' x. (v) o4 l g cos 3x. (vi) log s1

2. (j) 4 (x + 1) 3 / 2 . (ii) 3 tx 3 )312

3. (I) 2 sin 'Ix. (Ii) tan P.
4. (1) 2'1i. (11)4 tan "2x.

S. (I) 4 (x s sin x ) 213 (ii) log (x + sin x)

6. (1) log sec (log x). (ii) log (log x) . 7. Ci) 2)f' 1 + sin x.
_'Il_- -
+ X2
(ii) b(a +- b sin x) 8.(i) x . (F) 'Ii
x
(ill) x / I - x'. (iv) x / +I
9. (l) 2 log (e' 2 + e 12 ). (ifl- log (1 + efl.
10. (I) - log( log cos x). (ii) log( log sin jr )
(iii) log( log tan x). (iv) logi log ( secx + tan x )I

11. Ci) — log (a sin 2 x + b cos2x)

(II) ) log(acosx + bsin2x).

2(b
I I I
(a s - b') 1. g2 cos 2 x s b2sin2x
11 1
(IV)_1ta2+b2tan:

12. (i) _4cosxl. (ii) log tanx. 13. ( i )I32 - 2x + 7.

(ii) 'Ii -
- a2 .14. (I) 2;3 + tan -x. (ii) 2 tan"2 x + !tan 2x
22 INTEGRAl. CALCULUS

15. (i) e - log (e' + 1). (ii) tan (e

(iii) x - log ( e' - 1) - Ce' - 1) -' . (iv) 2 tan -1 (4(e 1>)

16. tan (xe).

17. 4 ((x + 1) 3/ 2 + Cr-i P12)

(ii)((2x+5)"2-(2x-3)312).

(iii) se 1 (2x - 7).

18. (i) 2 log (1 + 4r)

(ii) - (4x + 1)1(8( 2x + 1) 2). 19. (i) 2 4x + 2 log( 4x - 1).

(ii) 4 x 3 / 2 — x + 2 4x — 2 log ( 4x + I).
20. (I) (2x + 1) 5 12 + 4 (2x +

(Ii) (x + a)"3 - q(x + a )4I3

21. (I) — x -2log(1 - x).

(ii) +xl+
1 x' + 4x 3X + + ri log(x — I).

22. (i) 4 x + ! log (3r + 4). (ii) b -2 1 (a + bx) -a log (a + bx)l

23. (1) (3x + 2 ) 3/2 - (3x + 2)h/2

(ii) p bx) 5 '3 4ab 1 (a + br)2'3

24. - a cos l ( rim) —

25. r3
+4 xI + x + log(2x + I).

26. (i) 4sin - '( ri g ) 1 (ii)4sin I (rig)3

Ui') 'x + JIX1 _1).

(iv) -1 - + I- x2 )3/2

27. ain -1 (—
'I /) - rr(a- x).
 METHOD OF SUIIS11711T1ON 23

28. (j) al(a 2

— x2)

29. (I) e r* (II) -r/(x2

30. log (a sin x + b cos x + c).

31. (i) b- log (a + b log x). (ii) .(logsecr)3.

32. .. (x 3 + 3x + 6)1/3
33. (i) SIfl ( sin x (Ii) - (sin r + cosccx)
(iii) . log sec 3x - . log Sec 2x — log sec x . 34. -

35.. (i) x cos a - sin a log cos (r +a).

(ii) r cos 2x - sin 2a log sin (x + a
r(a — 2bx)
36. (i) lba 2alog- br -a 2x(a - br)
.! (log (1 - x
--- }
37. (i) ( r 311 .- log (1 + x " )) (ii) - J( 1 + r 2)3 / 3x

38. (I) .scc ..2 (ji) , (x2 f1)1I(x — 1). 39. f(xQ(x).
40. - log( I - xe).
2.3. Standard Integrals.
I dt I X
(A) =-tan- — .(a*O)
I t +a 1a
jx a

Proof. l'ut x =a tan 0; then dx = a sec 0 dO

1=1
P a
Sec 0 dO
2 sccO
I I
=-ldG
aj
I
=-O=-tan-'-
a
I
a
x
a
• d
(or.
1 + x2 tan x.
f
Note. Putting x = a cot 0 the above integral takes up the form
- ( I / a) cot (x / a ) which evidently differs from the previous form
by a constant. Usually
I dx 1 Cot
i s written in the form — —x
I -J a'+x 2a a
24 INTEGRAL CALCULUS
' dx I logs
IXx_a (lxi >Iai ).
(B) J x 1 - at +a
dx 11(1
ProoI.Jx2a2Ji; i - a x+aJ1dx
- dx fdxl
x+.aJ
{Jx_a
= {iogi (x -a)! -log! (x +a)! )
since the numerator is the differential coefficient of the denomi-
nator in each case, (See Ex. 5, Art. 2.2. 1,

= Ilx
log —
1 aI
x+ai
Note. The above Is an example of integration by breaking up the In.
tegrand into fractions. (See Chapter V.)
t dx = — I log a+x (ixi<iai).
(C) IJ a 2 —x 2 2a a-x
The proof is as before, noticing that
I I1.1 1
a 2 -x 2 2a'.a+x a — x
= log I (x + Ix' t al)I
J2 a1)
Proof. PutI(x 2 ±a 2 ) z - x, or,z = x +4(x2±a2).
2x
dz =(i+ 2,4( X l±a2)) x = z
f dx (dz = logz
J(x±a) Jz
= log! (x +.I(x'±a)!
Note. Students acquainted with hyperbolic functions may work out the
integrals (D) by substitution x = a sinh z, or x = a cosh z according
as the denominator Is 4( x 2 + i a ), or 4(x2 - a2).
 METhOD OF SUBSTITUTION 25

Thus, putting x = a sinh z, we have dx a cosh z dz.

dx I acosha
(1) Hence, Jr
C
_) = J a 4(1 + sinh 2 z) dz j dz = z = sInh ' -
.I(x l +0
2 +42 )
and this is shown in Trigonometry = log x + ,/(X g

=log (x +4(x2 + a2) —loge.

This form differs From the result given above by a constant (Cf Remarks
in Art. 1.3 1 . In the result, - log a being a constant may be dropped, since
it may be supposed to be included in the constant of integration.
In that case the forms given In (D) and here are the sonic. Similar
remark applies to the results of (2) and (3) below.
Similarly, by putting x a cosh z, we have
I dt x Ix+.I(x1_a7
2 J1(x 1 - a2) cosh'— =log , a
In many text-books for the sake of brevity, inverse hyperbolic forms
are used in preference to the logarithmic forms.
The first of the integrals (1)) can also be evaluated by putting x = a tone,
so that 4x = a sec 2 edo. Thus,
= a cc29dO sec 0 (cc 0 + tan 0 )
d'
jscc OdO=
J (x2+a2) J a sec O = 5 secO + tail O
- .g ( sec 0 tan by F, 5, Art. 2.2 I
= log(4(1 +1on 2 0) + tanO)
II x2 x\ log x+'I(x'+a2)
= loyt4l+—j . ;)= a
Similarly, put x = a sec 9, in the other integral.
Similarly the integrals Of ( B) and ( C ) can be obtained by hyperbolic
substitution.
(13) Thus, putting x a coth 49, dx = - a cosceb 2 0 JO and
2
(x 2 - a 2) = a I (coth 2 o - 1 ) a cosech
2

See Des & Mukherjees Higher Trigonomnetry, Art. 12.9.

 26 INTEGRAL CALCULUS

= —a cosech2O
dO Ifdo 1
a coth-'!a
= - I
5 a cosech 2 - ; = - a
(C) Similarly, putting x = a tanh 0,
fth20tb0 = de
-!f = 0 = _!. tanh1 ..!
a2sech2e a a a a

(E) q( dx
— x2) =
x
sin-' —
a (I x I Ia

Put x = a sin 0; then dx = a cos Ode.

facos8de C x
• acos0 =jdO =0 = S1fl'

Cor. dx
SIfl' x
I

Note I. Pulling r = a cos 0, the integral dx

be put in
f.J(2 x')
I
the form cos - — instead of - sin - i —
I
a a

Note 2. In (A) to (E), it is tacitly assumed that a 0.

2.4. C1 dx
ax' + bx + c .(a^O)
The above integral can be written as
If dx If dx
aJ x2 + - x •i--•
c aj I/x +— j
= b \
+
4ac - b'
a a \ 2ai 4a'

and putting x + -2a

k- z, this reduces to the forn I I' dz
[di Z. ,..

according as 4ac - b 1 is positive or negative, and this is of the form

(A) or (13) of Art. 2.3
If a be negative, we may take - (1/n), which is positive, out-
side the integral sign and it then reduces to the form (C).
g
2.5.I ax' — + bx +c
dx. (a* 0, p * 0)
 METHOD OF SUBSTITUTION 27

= (2ax + b )dx,
Here, noting that the differential of ax' + bx + c
the given integral can be written as

2ax +
jA
a
(2ax +b) +
' dx
dx= hI
2aJax 2 +bX+C 2aJ ax'+bx+c

2a -
.2- (1 12ax + b) 1_dx
Jaxt +b x+c
2a IJaxi+bX+C p

since the numerator

The first integral = log (ax 1 + x + c ) ,
of the integral is equal to the differential coefficient of the deno-
minator. The second integral is evaluated as in the previous article.

Note 1. To express px + q as the sum of two terms we might also

proceed thus
+ m ,where
Let px + q = I (differential coefficient of the denominator)
the constants I , m are to be determined by cor.iparirg the coefficients.

Note 2. In Ails. 2.4 and2.S, if an I + bx + c breaks up into two real

linear factors, then instead of proceeding as above, we may breakup the in-
tegrand into the sum of partial fractions, and then integrate each separate-
ly. (See Chapter V.1
dx
26 (a ^ 0) .
J Tax + bx +c

If abe positive, we can write the integral as

1C dx
b ' 4ac - b2 1
x +
m-J.j f( ) + 4a2 J

and substituting z for x + this reduces to the form

I( dz
I(zi k1)'
j
according as 4ac - b2 is positive or negative, and this is of the form
(D) of Art. 2.3.

(fa be negative, say, =- a' , we can write the integral as

 28 INTEGRAL CALCULUS

'I
dx
- f- (4a'c + b' ' b '
I 4a' J

and putting x - b , = z, this reduces to the form

I
J'J
f dz
I(k ' - z')
which is of the form (E) of Art. 2.3.
Note. If the quadratic under the radical In theabove case breaks up into
two real linear factors, we may, instead of proceeding as above, substitute z2
for one of the factors and then proceed. The method is Illustrated in
Exs. 2 and

2.7.J
3 of Art. 2.9.
p x+g
ax' bx
dx. (a*O,p^O)
c)

As in Art. 2.5 the above integral can be written as

(2ax + b) + - b
U dx
2a Rax' +bx + c)

...r( 2axi-b 2ag-pbr dx

dx +
2a tJI(ax 2 +bx -
+ c) p J4(ax2 + bx +c)
The first integral, on putting ax' + bx + c = z, reduces to
f= 2Iz 2'5x' + bx + C.

The second integral is obtained as in Art. 2.6.

2.8. (A)
5 (ax
ux
+ b) 'I(cx + * 0, c * 0)
Put cx+d=z 2.. cdx=2zdz.
Wc may also put ax' + bx + C =
 MIiTHOD OF SUBS17M11ON 29

The integral then reduces to

2t zdz dz
7J (a z2_d+ b).z --25 azt+(bc_ad)
which reduces to the form (A), (B) or (C) of Art. 2.3.

(B)
5 dx
(px+q)I(ax'+bx+c) .(a^O,p*O)

iere, put p: + q -,so that

pdx = --i- and
( 1z - q).
The given integral then reduces to

if I I(a ( +—( -q
\
-- q)2 b/Iz )

IC dz
z) 2 + —Pbz (1 - qz)
which when simplified takes up the form
f dz
4(Az 2 + Br + C)
when we proceed as in Art. 2.6.
2.9. Illustrative Examples.
E.1. Integrate
7r-9 dr.
- +
f
= f (2x-2)-2
Jz - 2x + 35

7 1• 2x-2
1Jx2_2x+35 dx 25
-
dx
x2-2rf35
30 INTEGRAL CALCULUS

alit
= log(x2- 2x + 35)- 25 (x_ 1) 2
t 34

= - Iog(x' - 2x + 35)- tan' (!! )

Es 2
' J u
dx
J2 +3x - 2x2)
dx
Here, I -
- 'I((l +2x)(2 - x)J
Put 2-x=z'. . -dx2:dz

and 1 + 2r = I + 2(2 -z) = 5 - 2z2.

The integral reduces to
r
- J 2zdz
4((5- 2z2)z') =
dz
'12 J ,T2)

(See (E) Not, Art. 23.1

= 42 cos ' ( 4 . z)

- 42cosJ 2(2 - x)
- 5
dx
F. 3 > a).
Integrate
- a)(13 - x))

Put x - a = 2 2 . .. dx = 2zdz; and tl - x =13 - a- z2.

• S =
2zdz
(j1(13 - a- z)) 2
dz
f T(k2- whore k2

= 2 sin = 2 sin '

Note. This integral can also be cv3luated by putting

X = u cos 8 + 0 sin2e.

dx
F' 4 (2x + 3)4(x 2 + 3x + 2)
Integrate
The integral is of the form (B) ,f Art. 2.8.
I dx
Put 2x+3=-. :.2d2=--1-dz.:.d
z2
 METHOD OF SUBSTITUTION 31
And x(i_3); z
2x 3
dz
I
5Z11T3)TflIT3):21
dz
J
Aliernalively
-z) = - -si n
= i (-23)

_______ dx dx
= Jx^s -(4x2+12x+8)) f(2x+ 3)((
Put 2x + 3 = z. •. 24x = dz.dx =dz.
dz
I) =secz = sec-' (2x - 3),
Although apparently the forms of the two results are different, it can
be easily shown (by using the properties of inverse circular functions) that
one differs from the other by a constant.
Note. It can be easily seen that the linear expression is the sum of the
two linear factors of the quadratic expression under the radical ; also the
linear expression is the derivative of the quadratic expression.

E. S. 1nterasef_II dir. (H. S. '81 1

Rationalizing the numerator, we have
' t
dx r dx I xdx
=J X) _ j (1 -x2) J
Pi I -x z' in the 2nd integral, so that - 2x dx 2zdz
2nd integral then = - J dz =-Z=_ 'I( I - x').
I = sin 'r -'(l -x').
Note. Integrals of the type f.\J +
dx ( 'a 0, c x 0) of which
the 3bove is a particular case can be evaluated exactly in the same way.
Ex. 6. Integrate x2 s 4
+
^ 3 dx .
32 INTEGRAL CALCULUS Ex. JI(B)
2x -1
Here, 1(1 - _+_2_x + 3 ) dx

=
J
+ 2-
2nd integral = C Q" +
3
- dx
dx
_f 2x-I

+3
J
2r+2 dx-3 dx
- + 2x + 3 J(x ++(
I ,i
2)2
5
= log (x 2 + 2x + 3)tan - (!_ jy!)
3
=x_ log (x2+2x+3)+tafl(Ix+1
I

Note. In order to evaluate integrals of the type + 5 as

(of which the above Is a particular case), where 1(x) is a rational function
of x of second or of higher degree, divide the numerator by the deno-
minator till the numerator is of the first degree; then the Integral reduces
to the sum of integrals of the type JAm xm dx Cm > or = 0) and of the type
px+g dx.
fax 2 + bx + c

EXAMPLES 11(B)
Integrate
3x2
1. dx.

xdx xdx
2• (1) (ii)5x4-1
S x4 -s- I
dx x3 dx
______
I H. S. '78,86
3 (i) J 5 j( a' - x')
(Put e' z.I (Put z.j
+ sin SIItXdX
4•
x'
se Oxdx . lo f ________
3 4-sintx
J •
 E ir 11(13) METHOD OF SUBSTI71JflON 33

XI - I
' ka +x) f (ii)5 ri...7)dr.
I (ii) Put x + x I = Z. I
6
f
1Pux'
xdx
a1)(h2
=
-))
(b2 > a)

1. (h f I 'll 111S'SOj (i)fj_ dx

4y

dx dx
(ii'If
S 1 + .1 -x1 6x2 + 7r+2

xdx
j
f r 1 ^2 +2

10.
coxdx
f sin2x + 4 sir, x+ 3
edx
11.
f 2e + 5
12 dx
fV(T - 1 2 )11 + (sinj)TJ

f r' -
x1dx
6x + 5
13. J

14.
S _-.-__
xii
dx
+ 7 tgr+
dx
-

I5.(i) ii)J 3x+1

+ x 2. + 1
- 2dx.
2x -
16, to  I X+xTi
x+1
dx. 2x 3
j f 4x2 + 1 dx.

. f17 (4x + 3)dx

3x2 3iTi (Wf
xdx
2- 6x - xt
X2
18. ( —dx.
-4

Inlegral Calculus (main) -5

34 NTEGRAL CALCULUS Ex. II(B)

r (1)1X2_
19. (I) I
j X + 2x + 2
dx. Z_4i
.i x+ x + I
dx.

f 3 .i.
20. XI + 2x —1dx.
xx+1
j
dx
21
J .I(xt+x_2)
dx 01)5
dx
22 • (i
.I I_x_x2r

dx
2L I
j
dx
24 •
.I I(x2 -7x + 12 )
IPWx-4z2.I

25. J
dx
Tix1Ox2)

cos x dx
(5 sin 2x - 2s1nx + 4)
26. 5
dx
21-
• 1 Jt(x - a)(x -
dx (ii)J dx
2a
• (i)J I(2ax -x2) 'I(2ax + x2)

lq 0)-x
+b
I
dx . 11. E. '83 1 (ii) f
2x+3
1T 2 . + 1.
dx
1. Li
J

30 dx.
jx2_8x+5)
J
(2x-1 )dx
31
() J
(x + fl
+ 8x -5x2)
dx. (ii) J X2 +4x + 2
dx.

dx dx
32.
() J-T(T77) (jI)J (2x +1)I(4x l-3)

dx
-x 2 +-,)
x. 11(8) METIIOI) 01 SUBSTITUTION 35

2x +1) dx.
34..dx. Cii)f JI.3x +
dx
(..)J 'Ixdx
35 (.)I
I Put r z'.J
dx dx
J XJ(X1 ±a 2 ) 5 (1 + x ) 4(1 -x2)

dx dx
(iv'J
(iii)fx'J(9x + 4x + 1)' ( +x)4(l +2x - x2)
[C. P. '86 1
(v)j
dx
xJ(x 2 + 2x - 1) ()f
dx _____
(1 +x)'I(T +x - x)
dx
(vii)J (x - 3)I(x 2 -6x + 8)
J (Q2 - X2)

x dx. (ii)
fx+x-1r dx

dx
J x '1(1 . I Put I + x' = ., C. P. '81 1

.. -
x_ a dx.
M 'lx
+ dx. (ii) x
40. If a < x'z b, show that

dx 2 fT
5 (x — a)((x - a)(b - x)J b x- a

ANSWERS

1. tan (x ) . 2. (i) tan ' (x 2) () . log x 2+

I
3. (I) tan (e'). (ii) - sin

x
4. (i) tan x - tan x. (it) - log 2 - cos
2 + csx
 36 INTEGRA:. CAI.CiJ US

S. (I) f log (x' + [+a'). (ii) kg ((1 ,. + '11 + x )/x)

6.s + I \
211 .7.(i)tn' (2x-).iii) tan'(+!)

(I) VSI log ,j45+2x-1

- 2r (ii)lcg2r+l
3T"• 9 tan - I).
+I

IO. I-log 1+ sin x

inr ii. Pan'((e' .- I)}.12. tan - '(sin 'x)

13. Iog X3_5

3 log 5+kg
1 1. (i) log(x + I) +
12 1 x+1
(il) '- log( x - 3) . 16. (i) log jr 2 + 4r + 5 ) tan -, ( x e 2
(ii). log (..x2 + 1) stan '(2x).
17. (i) f log (3x 3r I) ^ 2 tan 1 (43( 2x + I ))

(00log ---
____
t log ( 2 - 6x - x1),
- -

18. x log' _-2 1. (i) i - 2 tan - l

+
x+2 + 1)
(ii) x log ( 2 I'
x f 1) + - 2tan •' (, 2xfl
-- )

2x I
20.fxs-log(-z I) tan'' ( '-
r +2 +
)
21. 2 log ( '/x+2 + ITT T) . ,f 2x +
22. (I) sin ---p--- )
(ii) log 2x + 3 + 2 9-;-+,)
23.log (x +.+ x1 +.4x+2).
24. 2log('Ix _3 +'/x -4), 25..\J -+4
19
26. - '- log ( 'T2
_ -5 sin r + ' -
5( 2 - sin x1
27. 2 tog I + . 28. (I) s i ft -' (!f)
(Ii) log (x + -'/ $ 2nr).
 MPThOI) OF SUBSTITU11ON 37

29. (1) a2 + b log ( r . 12)

,—2 +x + I - 2 log (x ++4-;F-+—

(ii) 2 1x x+ 1).

30. 4T 1 8x+5 .31. () sin' ___ ._)_(4 +8x-5x2

/Sx-4

(ii) —+ 2. - log ( 2x + I
-'Tx 4- 71—
+4x +2). +4

1 /'14x +— 3 - I
32. (1) 2 jj— \/
tan ( 2Icgt(43))
\

33. log
2 + 1 )+
34.
(i) 4-(7-
- 3>(x-4) + log ( TTi + r; - 4).
J

(ii) 112 )(3x 2 --41 log ( 3 '12x+1 + 2 4Tx+2

_

35. 0) log !
I +x
(. i log
____
x+1

I x
36 (l) log - sec - - . (ii) - - -
2a og 4 a a a
(iii) log -- log (1 * x+I +—
4x+ 1).

(iv) T
2 sinn (v) sin tXl
+x -vr )

(v1) sin
.1 3x* I \
- (vii) Sec (x-3

-
3'. (I) 47 - + a log x2
—
x
(ii) log x + x -
- 2
- -j- tun
/2'Jx-1+1
)
38. log ('/1—
+x - 1) - log r -

39. (i) a log ( 'x + 'x +a) + 'Ix( — Y+ a)

(u) 2'I--a - 2'a tan1

(J' 1 )
 CHAPTER III
INTEGRATION BY PARTS

3.1. Integration of a product 'by parts'.

We know from Differential Calculus that, if u and v, are two
differentiable functions of x,
d dudv ,
- v1 + u -
dx dx dx
integrating both sides with respect to x, we have

= —duv,)
t( dvl)d
u-
UVI
I( OX
dx+ J dx
J( dv,,' dx = uv,, - f(du v, ) dx.
or, u dxi
- ) x ,
dv,
Suppose v, = fvdx.
dx- v, then
Hence, the above result can be written as
t I du f
5(uv)dx = ufvdx vdx)dx.
-J tiJ
The above formula for the integration of a product of two func-
tions is referred to as integration by parts.
It states that
..e
he inhgra
..j product of two funriion.c
1st function (unchanged)x integral of 2nd
- integral of [differential coefficient of 1st x integral of 2nd J.
3.2. Illustrative Examples.

Ex. 1. Integrate Jxe I dx.

I = xfedx -f f- Jeadz} dx,

=xe'_fi.e'ax. =xe'-c'.
 INTEGLtTIO14 BY PAR1 19

Note. In the above integral, instead of taking x as the first function artl
e 9 as the second, if we take C, the first function and x as the second, then
applying the rule for integration by parts we get

The integral 11 e wx 2 dx on the right side Is more complicated than the

ohe we startad with, for it Involve, x 2 instead of x.
Thus, while applying the rule for integration by parts to the product of two
functions, care should be taken to choose properly the first functicn, i.e., the func-
tion not to be integrated.
A little practice and experience will enable the student to make the right
choice.
E.2. Integrate Jios xdx.

I =J log ;. I dx,
= logxjdx (log r).JdX} dx,
_5(
= togx.x — dx,

= x log x - I dx,
= xlogx-- X.

E. 3. 1ntegrae Jtan x dx.

I = tan'x.ldx,

= tan tx.Jax (tanx) fax) dx,

_J(
= tan 'X.X -J1 +1xt .xdx,
r 2:
xtan x -
= xtan lr_tog(I + : 2). I By Ex. 5, Art. 2.2 1
 - INFEGRAL_CALCULUS

Note. Very often an integral

involving a single logarithmic function or a
single Inverse circular funci ion can be evaluated by the application of the rule for
integration by parts, by considering the Integral as the product of the given
function and unity, and taking the given function as the first function and
unity as the second.

This princp1e Is illustrated In Exs. 2 and 3 above and E. 4 below.

Ex. 4. intesratefiog (x + ' X 2 4 a 2 ) dx.

I =flog(x +Jx2+a2).jdx,

= log (x+ Ix 2 +a 2 )fdx_f [ {ig(x+ rX T7)) .fdx]d.

= log (X +4 T al).x -f--.----.T).x dx,

= (x ../X2 a )
xlogf -t
itz
'(r
-+ a 2

Put x 1 +9 2 ,so that xdx =zdz.

C xdx fzdz C
'I
J e(x'+a2) j11d2z=1X2+a2
z j

I = x log (x +Ix 2 t a1) - x ta2.

E.5. Integrate fx3edx

I = x' e' - 3 Ix 2 e 'dx, Integrating by parts
=r'e_3(xle_
- 31x 1 c' -2(xe'-Je'dr)J
- 3[x 2 e' -2 (xc'- el)),
- 3x 2e' + 6xe' -6.e',
(r 3 - 3x 16x - 6)e'.
 INTEGRATION B( PARTS 41

3.3 Standard Integrals.

e co g bx dx =
e "(a cos bx + b sin bx
(A)5 a1 + bi

e
(bx - tan-1^)
= J(a2 + b1)5

= e" (a sin bx - b co g bx
(B) Je sin hx a 2 + b2

sin ( bx - tan-'
Here a * 0)
Proof. (A) Integratir.g by parts,
sinbx t sin bx \
e"cosbxdx = e'-'.----- (ae.__g—_)dx.
$ -J
e" sin bxa
- - Je o sin bxdx.
- b
Now, integrating by parts, the right side of this integral
e 'si rtb.t
cosbx\ cesbx f 1
- b - a
{e b ae ( - b -j )'J
e' sin bx a a1 f
* -- e" cos bx .- - e cos bx dx
- ----i-- J
• transposing,
02 \ t e(acosbx + b s i n bx)
I Cos bx dx
( j )
2 a2+b2
Now, dividing both sides by I + i.e.,

we get fi e os LX 1x
e"(a cos bx + b sin- bx)
J
Again,puttinga =rcosa,b =rsina,sothatr=J(a2+b1)
and a = tan ( b / a ) on the right side of this integral, we have
the right side
 .2 INTEGRAL CALCULUS

e''rcos(bx —a) c" b\

Co.
= a2+12 4(a2 +b') ( bx— tan-'
Integral (B) can be evaluated exactly In the same way.
3.4. Alternative method

Let P =fe u cos bxdx

and Q= sin bxdx

... P + iQ=fe ( cos bx + i sin bx)dx _feu e O, ix

e15'' a - ib
dx =
=Je',' a+ib a'+b2
- C''
- ib)(
- a2 + bl (a cos bx + isinbx).
Equating real and imaginary pita we got the value, of F
and Q.
Note 1. The above integrals can also be obtained thus:
Denoting the integrals (A) and (B) by I, and I and integrating each
by parts, we shall get
bJ1 +a12 = e'st hbx
and all - b12 = a'' cos bx,
from which I, and f, can easily be deterniln.d.
Note 2. Exactly in thesame way the integrals f, coo (bz +cUx and
Je"sin (br + c) dx can be evaluated.
3.5. Standard Intergals.

x7a 7 T at
(C)fr + d
2 +.j. l.gI lx +xt +

X,(Z*+lt II x
0I'
- 2
 INTEGRATION BY PARTS 43

dx - x X2 a2
- j-log I (x+x2_I2)I
2
(Th J x - 

or - x 4a'a cosh
2 • 2 a

xIx a1 -
- x 1 dx a
2
(E)Ja2

Proof. (C) Integrating by parts,

2x
x dx,
+ is1dx =-Tx 2 + a2.x J1? X2
5
X2
XT2 
= dx. ...(1)
-
S
x 2 + a2
Also, —
+ a 2 d
j(x t +

x1
=
f f dx
T)dJ(x2+a2  . ( 2)

Adding (1) and (2) and dividing by 2,

xTx 2 + a2 a' dx
+ . I d x
Tx 2 —
J = 2 (x1+a2)

x1x 2 4 a 2
=
a2
2 - +-- IogI (x +J x2 2)I +a
By A rt. 2 3 (D) I

Proof. (D) 5[xz - a2dx

2x
IX 1 -
= 42 x dx,
.
S
2,J(x2_a1 )

'
X IX 1 -
- a2
- 4(x2 -
$
 INTEGRAL CALCULUS
(x 2 -0) +a'
= xix a -
t( x 22 - a) dx,

jr XG dx a l $
=xx2_a2_(22) dx
-

= xJx a -a 1 _ jix t - a 1 dx _a25 I( x dx

2 - a')

Now transposing J '( x 2 - a') dx to the left side and divid-

ingby 2,

5 slx l - a'dx = XX22_a2 - flog! (x +'Ix' - - a2

I By Art, 2.3 (V)
Note, The integral (C) can be evaluated by the method of evaluating
the integral (D), and The integral (D) can also be evaluated by the method of
evaluating the integral (C).
(E) Although this integral can be easily evaluated by either of
the methods employed in evaluating the integrals (C) and (D)
above, yet another method, the method of substitution, may be
adopted in evaluating this integral.
Putting x = a sin 0.sothat dx =a cos (IdO, we get

'ia' - x2dx =a2$c0S20d0

5
= a'..J(1 + cos2O)dE3,
- .a 1 [fcos 20 dO +JdQ],
= 'a l l sin 20 +0],
=' a' . sin 0 cos 9 + . a'O,
I x2 . X
=-a' +-a2sin
a— ', j - -
1
2
a2 2 a
x[a 2 x 2a2 . x
= 2 ^—
2 -in -I a
 INTEGRATION BY PARTS 45

Note. The integrals ( C ) and ( 0) can also be evaluated by putting

x = a sinh z and x = a cosh z respectively.

3.6. J.rxi + bx + c dx. (a * 0)

' 10 integ aLe this, e.prcss a L'a + c as the sum or difference
of two squares, as the case may he ; that is, express ax 2 + hx + c
in either of the forms a(( x + 1)2 ± m 2 1 or a'( m 2 - ( x + 1)1
nd then substitute z for x + 1. Now tho integral reduces to one
of the forms (C), (0) or (E) discussed above. This is illustrated in
Ex. 3 of Art 3.10.
3.7.5(px+q)Iax2+6x+cdx.(a;e0)

To integrate this, put px + q = -( 2ax + b) + ( q b

- -. )
then the integral reduces to the sum of two integrals, the first of
which can be immediately integrated by putting z = ax 2 + bx + c,
and the second is of the form of the previous article. This is il-
lustrated in Lx. 4 of Art. 3.10.
3.8.$e(f(x) + f'(x)] dx.
integrating by parts e If ( x ) dx, we have
Je'f(x)dx=Jf(x)e'dx=f(x)e_(f'(x)e.dx,
transposing, Je'(f(x) + f'(x)}dx = ef(x).
Alternatively, we may integrate by parts I e 'f'( x ) dx, and derive
the same result
Note. f e$(r)dr, when 0(x) can be broken'up as the sum of twofunc-
tions of x such that one is the differential coefficient of the other, can be easily in-
tegrated as above.
3.9. Generalized rule for Integration by parts.
Let u and v be two differentiable functions of x (differenti
abie n times), and let us denote

(u) byu' andf vdx by t

 46 INTEGRAL CALCULUS

(u') by u" and Jo1 dx by v

Now,
f
uv dx = uv1 -
f u'v dx (by integrating by parts) (1)
;ai,
f u'v1 dx = u'v -
J u11v111 dx, ... (2)

5 u"v 2 dx = u"v, -
f u' v3 dx, ... (3)

where u I denotes u".

5 u"o dx = u"v4 dx -
f u 4 v4 dx, ...

Combining (1), (2), (3) and (4) we get

5 uv dx = u p 1 - u'v 1

And generally
+ Li "v - u"v 4 + (- 1) 4
5 u4v4dx
(5)

5 uv d x u p 1 -

+ (- 1) Juv.
u'v 2 + u " 03 -u" V4 +

dx
(- u

. .. ( 6)
where u denotes u with n dashes.
qn.
5x
II luetra Li

bite8 rate 4 cos x dx.

I =x' sinx-4x 3 (-COS x)+ 12x 1 (—sinx)-24x(c05x)42451nx
xlsinx+4xi COS x_12x 2 sin x_24xsx+24 sin r
3.10. Illustrative Examples.

Ex. 1. JnIesra:efe sin 3xcosxdx.

I = . Je'.2 sin ,3x COS xdx,

=jIc (sin 4x + sin 2x)dx,
..IIe'sIn4xdx +Je'sin2xdxJ,
e2l Sin (2r
= I
[ ON
. ) sIn(4x_ tan-',)
 INTEGRATION BY PARTS 47

ri - tan 2) + sini*
t2x-
-r L 7) ( 2

COSIX
Ex. 2.
Integrate
I - It kcosazdx .4Je-'(coa3x+3 cos x)dx
4(Jeco83xix + 3Ieco.xdxL

. [! j _coex + 3sin3x)+3 ! (-3co,x + .4nx)j.

1

!.{1(ain3x - co( ) + (smx— 3cosx))

Ex. 3. Integrate 514 + 8x-dx.

I =fT(T+!x-x2)dx,
= 1514 " 3!-!r + x')dx,
S

J5f )2..(x_±) 2 dx,

ITT
15f1' -z'_ dz. (putting x -4 and a =4)

z4az g z
+ IBAr.34(E)I
2 -},
r(Sx -4)14 +8x - 51 218
+ in -1 15x- 4
on restoring the values of a and z and simplifying
118 /Sx-4
= 4)'4 + 8x- 5x2 +— 5- sin -'

Ex. 4. 1nteraIe '3r- 2)4x 1 -x + I dx.

J
Since 3x -2(2x -1) f,
I =41(2x -1) ,[ -x +1 dx _ -if Ix2 -x + ldx.
48 INTEGRAL CALCULUS
To evaluate the 1st integral,
Put z=x2-x+i. dz.(2x_1)dx
7 3
Ist integral =f'Izdz=4z 1- x+ 1)2;

2nd integral = I 'I - - f) 1 +I dx

= JI(z 2 +al) dz, putting z - X -j-and a)
+a 2 ) a2
= 2 +.- log (z+z-• +a 2),

- 1) T TTl + . log(x -_+ T71)

3
I = (r 2 -r + 1) 2 _!I 12x
'
-

- log I (x-+ •i'; H

E.5. x2 + x +. I
, +2r +dx.
Integrate
f 4(

( x 2 + 2x + 3) -(x ) +
-
-f- 1(x _'+ 2x +2)3 dx,

Jx 2 i-2r+3
-- 'J(r+2x+3) dr-
r+2
JI(x2+2x+3)dx

=j 2 +2x + 3 d
-f 112 (2x + 2) +
( x1 +2x+3)

(2x+2)dx dx
S

2x+3) I ((x+1 ) +2)

Denoting the right -Side integrals by 11 , 12, 13,
f 1 — I3 = f'z2 +a 2 dx -
-

= .j. z 'Iz+ j
f
+}al!og(z
dz
+a 2) (where z=x+1, a 2 = 2)
+ '/' +a 2 )-log (z + T2)
Ex. 111 INTEGRATION BY PARTS 49

( z + I) ,fx 2 + 2x +3, on restoring the value or z and a1.

Putting x 2 +2x + 3 = z, so that (2x + 2)six dz,

12 2 'Iz = 2 T. +2x + 3.

I Cr + 1'Jx +2x + 3 -'/x + 2x + 3,

.(x -1) 'Ix +2x + 3.
Integrate (
Ex. 6. dx I1.E.'80
( (x + 1 )88 -e' X—)__.j
1 J (x+I)1 _J(x+l_P dX
Integrating by parts the first integral
f +l x+ I j (x+I)2

•
x+F
Ex 7. Prove thai (4 a b),AS
(I) 5e' sinh bx dx a2 -b2 (a sink bx b coal, bx).

(ii) 5'" cosh bx dx 2 - b2 cosh bx - b sink bx).

(I) Integrating by parts.
1= t" cosh bXJ cash bx

e ll co,h bx a
fe cosh bxdr. ... (I)

Again integrating by parts,

fecoshbxfr S"1

- b

nIegraI Calculus (main) -6

30 INTEGRAL CALCULUS Ex. III

From (1) and (2),

.Jzcoah br a
a,tnhbx +jI.
b
Tranap*ing.
(i i -1(b cosh bx -' ':Th tx).
I. (a .lnh bx - b cost, bx)

(ti) This integral can be evaluated In the same way.

A lternatively, we can use the exponential values of ainh x and cosh x
to evaluate these integrals.
Thus, esInh bxdx
1 It ' + - - Xj dx,
+' - -

I
2 4+L7 a
- J• _!.. -
2 La+b a — bJ'

I (a— b0l' (a
- -
S ta(e'— e s') - b(eb +e)J,

a sinh bx - b comb bx J.
EXAMPLES III
1. Integrate the following with respect to x :-
(1) x sin x (ii) x 8 cos X. ( iii) xc"

(iv) x' log x. 10 xe'.tH.S,'831 (vi) x sec 2x.

(tfl) sin - ' X. (viii) cos- 'x .1 H. S. '80 1 'ix) cosec- I x.

(x) (xi) cot - ' X.(xii) Cos- , ( I /x

sec 'x
Will x s1n'x. (xlv) x tan-'x. (xv) x cos nx
Lx 11) INTEGRATION BY PARTS 51

(xvi) (log x ) . xvi) x log x .1 H. S. '79, '86 1 (xviii) sin-' 'ix.

(xix) log (l + x)'".

(xx) log ( x + 1)
(x +1)2
(xxi) log (1 + 2x' + x) . (xxii) log( x2 + 5x + 6)

(xxiii) x' cos 2x. (xxiv) x 3 (log X )2.

Integrate :-
2. (1)Jx sin 2 xdx. (ii)fxsin x cos xdx.

$. () log( x - 'iX 2.) dx (ii) log :2 - x + 1) dx.

5 5
4, (i) I 1--
log (logx) l
dx. (ii) I I + cosx
dx.
j i .j
r sin x log ( sec x ^ tan dx (H.S.'86 1
(, x)
J
(ii)5 cos x log (cosec x + cotx)dx.

6. (i) cos 2x log (1 + tan x ) dx.

5
(ii)J cosec zx log sec xdx.

7. (1)5 sin - (3x - 4x 3 ) dx. (11)5 (sin - 'x)ldx.

S. (1)j' cos -1 11 dx. (ii) Jtan-ij_J.L_2 dx.

+ X2

9. (1)5 sin-' 2x .
x (ii)5tan-13dx.

.(i)j dx. (11)5 tan- 1LL..!

xsin-'x .. I. sin-ix
11.(i) _x2)X (ii)
f j - x2 )311 dx.

12.(1) 5e sin x dx. (ii) je' cos x dx.

(iii) f 2' sin x dx. (iv) f3 cos 3. dx

 52 INTEGRAL CALCULUS

(v)fesinhxdx. (vi)fexcosh xdx.

13. (I) fe sin 2 x dx. (ii) 5e sin x sin 2x dx

f (1
.fx+sinx
+ x')2
d..
log 1)
t'ut tan x = z. I
15. (i)
1 + cosxdx . (ii)
fI(x+1)
(1 + cos x ) dx.
16.
5 tan i
' sin [C. P. '88

5 sin;' \I
V.
x+a
X
dx. [Put x = a tan 2 0 1 1!. E. E '79, '86

18. fe r (cosx + sinx)dx. C. P. '811

9. (i) - (1 + xlog x) dx.

I C. P. '82, H. S.87 I
(ii)
f log xdx
(1 ^ log x)'
20. (i) fe (tan x - log cos x )dx

(ii)
5 e' sec x. (1 + tan x ) dx.

(HO
5 e (log ( Sec x +. an x) + sec x I dx

21
(i) J
(iii) 5e (x
r
( x + 1)2 dx.
+ 1
(ii)
5 (1 +
x)2 dx..
x2)2

i. 1) dx.

I 22 (fr
+cosx
s x
1sin
-
5
dx. ii) e'
$ 1 - sin
I — cosx dx.
C'
I - cos2x dx.
J 2 - sin 2x
4'v)
J
' 2 +
e' sin 2x
I + cos 2x dx.
23. f'125 _9xldx.
24. (1)
$ '15 - 2x + x' dx. 00
5 q 10 - 4x +4x' dx.
 Fr . III INTEGRATION BY PARTS 53

25. (i)5418x
- 65 - _
xldx. (ii)J4T 3x - 2?dx.

f .isx l
:
+ 4dx.

x +'I(x -1)

28. J'I2ax -xdx.

29.fI(x -cx)(13- x)dx. (Put x = acos9+sin2O.I

30.(i)j' (x - 1) 'T-i dx. (ii) f(x + b) 4x 2 + a2dx.

.31. (i)$ (x - 1)Jx 2 - x + Jdx.

(11)5 (x + 2)2xT1dx.
.f x 2 +x+1 •.. x2+2x+3
32.(0 j_ dx .
( I _x 2 )
tu) I dx
J j f '.'( x 1 + x + 1)
x 3 + 2x' + x - 7
33.
f '(, X +.2x + 3) dx.

34.u)
SV
+X
dx. I C. P. '85 ) (ii)j'x \f a dx.

JI '(x-2)
_1) ' dx

36. If u = 5 e ll cos bx dx, o = 5 e- sin bx dx,

prove that
(I) tan' + tan-i = bx.
(ii) (a 2 + b2 )( u l + v ? ) =
54 INTEGRAL CALCULUS

ANSWERS
1. (i) - i cos x + sin x. (ii) ( x ' - 2 ) sin x + Zr cos x.
(iii) --(ax - I). (iv) ,.--j [log x -
J
(v) e' (x Z -2x + 2). (vi) x tan r + log cos r.
(vii) x sin - l r + '11
-x -2 (viii) r cos -' x
(ix) xcosec-'x + log(x+ 4X2 — I).

(x)xscc r- log (x +,1r1

(xi) rcot'x +log(I + x2).
(xii) xsec 1 x — log (x + ,1r 2 -1).
(xiii)+x 2 sin x — !-sin 'x + + x 'I l -xe. -
(xiv) x 3 tanlx _ r2 log(I +r').

(xv) x sin ,tx + —j—

cosnx
. (xvi) x (log x)' — 2x log x + 2x.
(xvii) j.r 1 (2logr -1). (xviii) Cx -!.)sin-'4x + 'Ix(1 -x)
(xix)
f(l +x) 2 109(I+ x) -+x(x + 2).
(xx) -(I + x )'[ Iog(i +1) + II.
(xxi) 2(xlog(l +x 2 ) -2x + 21an-'x).
(xxii) (r + 2)log(x + 2) + (x + 3)log(r + 3) -2x.
(xxiii) x ( 2x 2 - 3) sin 2r +4(2x' - ) cos 2x.
(xxiv) r4l(logx)1 - . logx +.}.

2. Ci)( 2x 2 - 2x sin 2x — co g 2x) . (ii) — - x cos 2x + sin 2x.

3. (2) xlog(x- 'x 2 -1) + 4r 2 -1
(ii) (r _4)log(x2_ z + 1) -2x + 43tan-' ( -2x
-1
—T3 ) -
4. (I) x (log x ) -' (ii) x tan I x + 2 log cos x..
S. (I) x - coax log (sec x + tan x ) .
(ii) sinxlog (ccsec r + cot r) +x.
 INTRCRATION BY PARTS 55
6. (I) sin x cos x log (1 + tan x) - fx + ofl g ( sin r + coax).
(ii) - cot x log (Sec x) + x
7. (I) 3(x sin 'x + 41i1).
(Ii) x(sin'x)3 +3I 1_x 2 (s1n l x) 2_ 6(x5ifl_ I x +'Il—x').
8. (I) 2x tan - l x - log (1 + x 2 ). (Ii) Same as (I).
9. (1) Same as 8 (I). (ii) 3x tan -' x - log (1 + X2).

10. (I) r41 —r - COSX .(ii)

2x z f fx Cos 'x —4(1 —x2)1,

11. Ci) r —41 - x 2 sin -Ix. (ii)4l + flog (I - r2).

12. (1) . e (sin r - cos x ) (U) + cx ( sin x + cos x)
2'sin{r—cot-'(log2))
4(1 + (iog2)2)
3xj + (1013)cos3rl
(iv)
9 + (log3)
(v) f(cosh2x + sinh2x)— f x.(vl)f(cosh2x + sinh2x)+ f.
13. (1) 1 e (1 - f (cos 2x + 2 sin 2x)}
(ii) . e'((cosx + sinx)—f(coi3x + 3 sin 3x)).
e' tan r r I I I I —x' + 4.
I — +-------
m 2 +4Lm I+x 2I+
14.
2 Lm x2

15. Ci) x tan fx. (U) 2 'TTfl log( x + 1) - 4 '1 I.

16. —.x2. 17. (x +
18.. e sin x 19. (1) c I log x. (ii) x/( I + log x).
'. e log S" . e l Sec X (iii) P. ' log C sec x .....
21. (1) '----?• ',,L..,..
22. (i) ei .. -' (Ii) _CXC .y

(iv) e tan x 23.25-

4- .. ,.iT. -I
6 INTEGRAL CALCULUS

24. (1) f(x-1)45-2x+ x 2 + 21c'g(x-.1 +45_2x+xz).

(ii) (2r - 1)410 —4x T 4x

+ ' log (2x - 1+410
- -4x + 4xZ).

25. (I) f(x -. 9)418x -65 - x'+ 8sip -'(x - 9)

41'12
(ii) }(4x + 3)44 - 3x —2x' + -- sin-' 4x
441+3

26. (5x+4)45x2+8r+4

+ log ((5x + 4) + 45(5x 2 +8x + 4)).

27. .(x(x —4x —1) + log (x +4x 2. 1)).

28.

29. -1 1

30. (0 +(x' 311_+xlr-

2- 1 +flog(x + ,&T_
1).

(ii) :" + + 4 bxTx -

2 + aI

+ 4 2 b log (x + i72)
31. 00 4(x2 -x + I )T_4(2x - i )x 2 - x
- log(x -4+ .1x2 -X + 1).
(ii) 4(2x2 +2x + 1)+4(2x + I)42x2 +2x +

+ log ((2r + 1) + 42( 2x2 + 2x + I )

32. (0 sin-'x —4(x +

0i) 4(2r + 5)Ix 2 -

+ x+ I +- log ((r +4) +'
+x -
+
 INTEGRATION BY PARTS 5

33. +(x2 +2x + 3) r ..! ( X + 5) , x 2 + 2% + 3

- 6 log (x + 1 + ,1x 2 + Zr + 3).

34. () asin l._ Ia 2_ x 2 . ( ii) (I x _a)Ia2_x2_f2sjnI.!

+ 6)1r2 - 4+ 4Iog(x +4r2 4).

 CHAPTER IV
SPECIAL TRIGONOMETRIC FUNCr!QNS

4.1. Standard Integrals.

(A) cosec x dx = log! tan I

f
dx
0
C C C
Proof. t cosec x dx = I dx = I
j sin j 2 sin x cos -I x
= f2 sec2xd
J tan —2
on multiplying the numerator and denominator by sec 1 +x)
= log tanxI
since the numerator is the differential coefficient of the denomi-
nator.

(B)fsedx = log! tan( . + -

= log! ( sec x + tan x

dx = f dx
Proof. Sec X dx
f =f- J sin(-!+ x)
dx
)2sin(+-x)cos(.a+x),

sec1(3it+x)dx
=- f
= logi tan(-In +-x)I asin(A).
Note. A lternative Methods:
osec x ( cosec x - cot x)
J
cosec x di -
5 cosec x - cot x = log I ( CO5€C X - cot x)

r dx slnx
j )sec x dx
= J ;:i;— = J dx
 SPECIAL TRIGONOMETRIC FUNCTIONS 59

t d(cosx) dx
, where z = cosx
= -J - =- 5I z'
I I-z =-- COS x
=-log----- IIog I-
2 I+z 2 I+ Cos x

(sec x (sec x + tan x)

dx -. logk see x+ tan x)l
jsec. -,• tan x
;ince the numerator is the derivative of the denominator.
cosx d(sinx)
j-- dx =
5 sec x dx
=— S - sinIx
dx ogI+z
= - log! -z
+z where z = sin x
f-
= j
I II - -sinxl
= .log i
-
dx dx
Jsecxdx
=5____ = - fc0s2

- I-tan
-
sec 4 xdx
2x
=2
dx
5I -z2 here z tan
j
I+z i+ tan 4x
= tog 1— = log
IT - tan
It should be noted that the different forms in which !he integrals
cosec x and of sec x are obtained by different methods can be easily showr
to be identical by elementar y trigonometry.
Thus,
log
i log 2 sin 24x Hi log I tan 2 4 x I
2 Ti--x -
= log I tan 4 x ;etc.
dx
4.2. 1
J a + b Cos x

The given integral

I dx
- Ja(cos-x + sin'-ix) + b(cos 2 x - SIfl2.x)
 INTEGRAL CALCULUS

( sec2 -xdx
= J (a + b) +(a - tan' -1 x
on multiplying the numerator and denominator by sec 2 .- x).
Case 1. a > b

j'ut 'va-b tanx = z. :. .'[a-b sec 2 ,xdx =dz.

The given integral now becomes

2 f dz
'(a- b)J (a +b) + z'

(See (A),ArL 2.3)

(a2_b2) tan 'I(a+b)
2 x
,a2_b)taJ1\Ja+b2
I /b+a cos x'
i(a2 _b2)coS (sa+ b cos x)
Case II, a < b.
Put tan - x = z ; .. - sec - x dx dz.

As before, the required integral becomes

2 f dz
-a ) (a + b)
'(b- - z2
2 .loI 1
ib + a -+ z

F Sec (C), Art. 2.3. I

- I f Jb+a+1b-a tan x
-
Tt T7 7- ) l ° 11b +a) - -a)tanx
Note 1. Here it is assumed that a > 0 b > 0 if a < 0, h > C
or a > () h < 0 or a < 0, b < 0, then the integral can be evaluatce
exactly in the same way.
 SPECIAL. TRIGONOMETRIC FUNCflui'43 01

Note 2. (I) If b = a, the integrand reduces to( 112a )sec fx, the in-
tegral of which is(1/a ) tan x
(ii) If b = -a, the integrand reduces to( I /2a ) coscc 1 i-. the
i . ;ra!'f which is (l/)cut

Note 3. By an exactly similar process, the integral f dx

+ b sin x
dx
or more generally fa + b cos x -+can c sinbex eviji,atd by breaking
sin rand cos r in terms of x and then multiplying the numerator and the
dcnominator of the integrand by sec' and substituting z for tan -- This
is illustrated in Examples 3 and 4o1 Art. 4.8 below.
In fact, any rational function of sin x, cos x can be easily integrated by
expressing sin x and coax in terms of tan 1. x, i.e., by writing
sin = I t2 tan -}x and cosx = 1 _ tan l+r
lan'x T
T + tan 2Tx
and then putting tan = z.
Similar integrals involving hyperbolic functions can be evaluated
by an exactly similar process.
4.3. Positive integral powers of sine and cosine.
(A) Odd positive index.
Any odd positive power of sines and cosines can be integrated
immediately by substituting cos x = z and sin x = z respectively
as shown below.
Ex. (i). fsr. 3 xdx =fsinhxsinrdx =_5(1 - cos2r)d(cos)
)dz t putting z forosr)
= - 1(1 - z 2
= '- (z -=- (cosx
EX, (ii).fcos s xdx =Jcos4xcosxdx
=5 (1 - sin 2 x) 2 d(sin x)
111. 5-. '81 I
= 1(1 - z) 2 tputtingzforsinxl
dx
= 5(1 - 2z 1 + z 4 )dz = z -z +
= slnx -slnx + six.
 62 INTEGRAL CALM US

(B) Even positive index.

In order to integrate any even positive power of sine and

cosine, we should first express it in terms of multipc angles by
means of trigonometry and then integrate it.

iii. Integrate
Ex.
f cos xdx

cos 4 x = ( I (] + cos 2x )} f] + 2 cos 2x cos 2 2x 3

i ll + 2cos2r + 1 (1 + cos4x)]
=- + cos 2x 1- 1 cos 4x
Jcosxdx =f( + jcos2x + 'cos4x)dx
= -x + 1 sin 2x + "sin 4x.

Note 1. It should be noted that when the index is large it would be

more convenient to express the powers of sines or cosines of angles in terms
of multiple angles by the use of Dc Moivrc's theorem, as shown below.

Ex. (iv). Integrate

f sin 5 x dx
Let cosx + isirix = y .. cosnx + isinnx = y"

then cos x - sin x = 1 cos nx - I sin nx

Y + 1 = 2 cos x y I +2cosnx

Y -- 2i sin x y" - = 2i sin ,x.

Y
2'isinx
1\'
= ( Y- )

= +_i) +28 (y4_)_56 (y2_2)+70

+)_
= 2 cos Sx - 8.2 cos 6x + 28.2 os4x - 56.2 cos2x +70.
sin a x = 2 ( cos 8x - S cos 6i + 28 cos 4x - 56 cos 2x +35)+
I sinxdx = 21(cos8x-8c(6x+28co54x-56cos2x+15)d
 SPECIAL TRIGONOMETRiC FUNCTIONS 63

8 sIn 6x sin 4x
I ["It
=V 8 6 +28—i-- _56 !35J

= sin 8x - sin 6x + 7 sin 4x -28 sin 2x + 35x

Ate 2. When the Index Is a large odd positive integer, then also we
can (ir3t express the function in terms of multiple angles as above and then
intgate but in this case, it is better to adopt the method shown above
In (A), when the index is small.
Tnus I sin 3 x dx I .( 3 sir. x - sin 3x ) dx = - 4 cos + 12 cos 3x.
4.4. Products of positive integral powers of sine and cosine.
Any product of the form sin x c os q x admits of immediate in-
tegration as in Sec. A, Art. 4.3, whenever either p or q is a positive
odd integer, whatever the other may be. But when both p and q are
positive even indices, we may first express the function as the sum of
a series of sines or cosines of multiples of x as in Sec. B, Art. 4.3,
and then Integrate it.
EL M. Integrate
f sin 2 x I
COS X dx

I = Isin2xcos4xcosxdx
=J sin x(I_ sin 2 x) 2 d (sin x)
= 12 2 (1 -z 2 ) 1 dz, [putting z sinxl
= 1(z2 -20 +z')dz
= -4x + 4rZ'
OK
+ sinx.

Lx. (Ii). Integrate sin 4 x co s I x dx.

5
Let cos x + i sin x y.. cos nx + i sin nx = y

then cos r - i sin x = cos nx - i sin nx

:.y+ 1 =2 cos x y"+=2 cos nx.

y- 1 =2i sin x y!=2isinnx.

Y.
f4 INTEGRAL CALCULUS

2i4sjnxcos2x
I 1\' I 1\2 IV
= Y — ;) iY^ =/;) Y !)'(Y'- 2)
/ 1\I I
= 2 — 2+
+ .-) ky'

(y'+ ._!) _2(y 4 + _!4 ) _(y2 +_i)+4

= 2 cos 6x- 2.2cos4x •- 2 cos 2x + 4.

sin 4 x cos 2 x = 2 [eQs 6x - 2 cos 4x - cos Zr + 2 1
I sin4 x cos x4x = 2J( cos 6x- 2 cos 4x - cos 2x +
I [ .!in 6x 2 sin 4x sin Lx

Note. The expression sin P x cos x also admits of immediate integration

terms of tan x or cot x if p + q be a negative even integer, whatever p
and q may be. In this case, the best substitution Is tan x or cot x = z
For other case of sin 'x cos I r, a reduction formula is generally required.
(See § 8.14— 8.17.)

Ex.(iij). integrate
Here. p + q =
5
2- 6
Cos Ix

= - 4
dx:

;. put tan x = z, then see I x dx = dz.

Now, I = J tan2x.secxdx
= Jz 2 (1 + z 2 )dz =z +z
= + tan3x +Itansx.
Ex. 60. Integrate ax
jnii2 xco 1x

Here, p + ' q = — -L — Z = 4. .. put tan x = z, then see 2 x dx = dz.

Now, t sec 4 xdx ( I +z2
dx
= J tan" x

=I(z_h/5+z3/2)dz=2zl2+1z5,m
= 2 tan "x +tan" x.
 SPECIAL IRIGONOMF1R1CFUND1OS

4.5. Integral powers of tangent and cotangent.

Any integral power of tangent and cotangent can be readily in-
tegrated. Thus,
(I) I tan I x dx = I tan x. tan 2 z dx = I tan x (sec 2 x - 1 ) dx
=Jtanxd(tanx)-Itanxdx =tan 2 x -logsecx.
(ii)Jcot4xdx=Icotzx(coseclx_1)dx
=Jct 2 xc0sec 2 zdx -jcot'xdx
= I cot xd (cot x)- J(cosec 2 _ I )dx
= --cotx +cotx + x.
4.6. Positive integral powers of secant and cosecant.
(A ) Even positive index.
Even positive powers of secant or cosecant admit o' immediate
integration in terms of tan x or cot x Thus,
sec xdx =J ( +tan2x),ec2rdx
= fsecaxdx +Jtanlxd(tanx)
= tan +tanx.
(ii)J cosec 'x dx =f cosec x. cosec 2 x dx
= 1(1 +cot2x)2cosec2xdx
= -i (I + 2cot 2 x i- cot 4x)d(cotx)
- cot r-cotx-cotx.
(B) Odd positlue index.
Odd positive powers of secant and cosecant are to be integrated
by the applicati—i of the rule of integration by parts.
(iii) I sec xdx -4sec x. see xdx = sec x tan x .. Jsec x tan lxJx
= sec r tan x _Jsec x( sec lx -'I)dx
= secxtanx + Isecxdr -Jsec'dx.
transposing I sec x dx to the .elt side, writing the value off sec x dx,
and dividing by 2, we get

Inlegral Calculus (main)

66 INTEGRAL CALCULUS

1 /A
Jsec 3 xdx =sec x tan x + 1 log tan s X-

(iv)JsecSrdx =Jsecxsexdx

= secxtanx--• j3sextarsxdx

= sec 3 x tan x -3 . .x( scc 2 x -1) dx

= SCC 3 x ton r 3 eclx dx -3j cc x dx

Now, transposing 3 5 sec x dx and writing the value of I sec x di we

et ultimately

Lan x se c 3 r t 3 La n xsex31
*- x\
log tan
4 2 +

(v) Jcoscdx =Jcowcrcosec2xdx

= - csecxcotx -Jcosecxcotxdx
= - cuse,c x cot x 5 cosec x ( cosec x -- 1 ) dx

= cosec x cot x + 5 cosec x dx - 5 cosec I x dx

transposing cosec 3 x dx and writing the value of! coscc x dx,

4.7. Hyperbolic Functions.

(I) Jlinhxdx =J(e - e -)dx =-(e' +e)=

cosh x .

(ii) cosh xdx =J(9 +e)dx =-•(e - e)= sinhx.

j

i tnh x dx = C sinhx dx
I
(iii) fj log I ( cosh x iI
cosh
t coshx
coth x dx dx = log I ( sinh x)
(iv)5 = J sinh
I
dx = dx
(v) fcoseciixdx = I
sinhx 25
e -e
 S1L( 1; kL ( I RIC I UNCI1ONS 67

• 2 I

.1. ( • ) 0

i •- 1
• I c
- Icg tih
o- d l\'iiing t I., nuner,itor and d'norn,iiato by C ,2 )
&'

r -,
(vi) ech x d' -- ---- - 2 ---------- dx
) coshx j1 4

I. ' ( ,' )
2I
j 1 + -- 2 tan ' (e' )
-- 2 tan C cosh x + sinh x ).
Olhe'rwe
Itdx
sechxdx - ----- r dx
J j cosh j cosh . x + sinh 2

1' -scch2-x
= 2 i----------------
j I + anh-.1x dx
F dz
j-1 on putting z = tanh - 1 x I

= 2tan'z = 2tan-(fjnh1x)
$ sech 2 x d tanh x
(vii)

J cosech I x clx = - (0th x

(viii)
(ix) fsech xtanhxdx sech x

(x)J cosech x coth x dx = — cosech x

 L 68 INTEGRAL CALCULUS

4.8. Illustrative Examples.

C dx
Er. 1. Integrate I
JSInx + COsX

dx
I
f c +
2sinijxc's4x ox - s.n2

--Jf_ sec4rdx
2tan4r + 1 - (an 2 '-x
on multiplying the numerator and denominator by sec x)

-. 2 putting tan r = z
=
2dz dz
.:J2 —(z 2 - 2z + IT J(2)2 - - 1)2

=2 J.-r L.1 where a '12. =z

f_tI
= 1¼., '12+( tan r1)
2a 8a—y
dx -
f a sin x + b co x
Er. 2. Integrate

Put a =rcosO.b = rsinthcnaslnr + bosx =rsln(x +0).

Here. , = 4 T Tr2 and 9 = tan

i =f__-
rsin(x -rj1cosc (
-1 9)
x+ 0 ) dx

= - 5 cosec z dz, where z = x + 0

zI
log1 tan -i- I

= J(a1
1
IOj1 tan (r s. I- tan a
Note. Since, as above, sin x + lol x =' r2 sin (- +
 SPECIAL 1RIGONOME1C.RN (1' ! 69,.

.J +
dx
= - 5 cosec (1x x+ dx

1
=-log I tan I-+1) I.
'\ I

f dx
Ex. 3. Integrate a2 2x x + b cos
_
Multiply the numerator and denominator by sec" x and put tan x = z
dz i az
j z 1 + k2 where a
5 a'z'+ b2 = a2
1! = tan - z tan (a
- tan x )
k & a?, b
C
EintegraIe
J
I 5( sin jx + cos 2 ..7.. sinjXCOSX
Multiplying the numerator ario denominator by sec 2 x, this
see 2xdx
- 5 5( tan 2 .x + 1) - 26 tan -x
-
2dz
= 52 a _2& + Lputting tan x =zI
J
2 f dz
=
(z_.)'-()

du
2r
= •gJ ; - . where u =z -v-and a

= 21
-f- = I log z-5
-logu-a

_I 1 5 tan x-25
- og 5 tan lx - I
on restoring the value of z.
dx
7nsexratej
13 + 3 cos x + 4 sin x
 10 INTEGRAL CALCULUS

C ______ dx ___________________________
j
ji
iT7+ -cos x )+ x)+ 4.2 sin x COS I

Multiplying the numerator and denominator by see I - r, this

sec 2xdx
-- f
J1O tan' xtan T +

C 2dz
Lputhngz = tan rl

I f dz 1 do
5J (r +1)2 + sJ u
5 5

wlwr' u = z + 2, 91=

I I u I 3z 2 3 5 tan x 2
---tan - -tan1--- =— tIn
a 6 6 6
2 sin x 3 cos I
Ex 6 tntegra'e
5 T sin x + 4 cos x

Let 2sinx 3cosx

= I ( denominator ) n differential coefficient ot J,noniinat,c

r H 3srx • 4 cos r 4m(3cos 4,jr)

= (31-4m)izir (41 + 3rn) cos .

Now comparing the coefficients of sin x and cos r of bothsides, we
;et 3i - 4m = 2 and 41 .,n = 3, whence I --'n =-.
2sinx + 3 cos x is 3 sin x + 4cosx)+-(3 cos x - 4 sin x).
1l (
I -- I -- ( 3 cos •- 1 sin x
I
2, 2 3 sin x + 4 cos x

=x • i. - log 3 sin x 4 ens x ) Iis

Note. Genera lly

f sin x+ bci:x
---v------ d - - ------ dx can be r.ited in the same way.
t pn g+ - co.x

Ex. 7. IeaeI - ---- dx.

Sill S x - a ) $ifl ( h
 SPECIAL TRIGONOMETRIC FIJNCIIONS 71

1 1 int(x_b)_(X1))
,i n - ),in (z - b) sin(a - b) ,ln(r - ,)81n(x - b)
- I[coe(x-a) co(x-b)]
sin(a1 - b) LSifl(X - - sjn(x - b)

cO(r- a) dx b)
F f cos (z -
J
- •1 t log sin (X -a)log
- sin (x --
n(a -b)
Stfl(x- a) I
= log I sin(x - b)
sin(a -b)

tan drjb >a.

Ex. 8. Integrate J 4-bla,12x)
dx sinxdx
' =s2x+bgjfllX
Sill S (b.(b_a)COS1X
sin xdx I 5dz
I
= _ _____ - J('
k -

2
putting z = co x and k =
[

I z I cos - [[T coax].

cos k V]-a)
b-a)
I See Art. 23 () Note. I
dx
Ex. 9. Integrate j(I 3 + 4 cosh x
dx
+ 4(cosh _x+sinh21r)
J 3( COSh 2 X_1I1h l X)

- C sechl+x
= J 7cosh !xdx+ sinh 2 lx = 7 + tanh2 dx
2 J 1

on multiplying the numerator and denominator by sech x

Put tanh x = z; then - sech 2 x dz = dz
72 INTEGRAL CALCULUS Ex. I

d-,
1 2 f =
2z
tan 2
= W tan 1(tanh!x)
EXAMPLES IV
Integrate with respect to x the following functions:
1. (j) cosec 2x . (ii) cos 3 x (iii) sin 4x
(iv) sin 5 x . (v) sin 2 x cos 2 x
(vi) sin'x cos 3 . ( vii) sin 4 x cos 4x
(viii) sin z x cos 3 x (ix) cos 2 x sin 3x
(x) sin 2x cosx. (xi) sin 3x cos 3x.
(xii) sec 2 x cosec 2 x . (xiii) sin SX sec'x
2. (i) cot 3 x . (ii) tan I . ( iii) sec'x

(iv) cosec 4 x . (v) cosec 5 x . NO tan x sec4x

Evaluate the following integrals:

J
cos Zr
sin dx.
sinx
. t cos 2x
(ii) I cosx dx.
J
. l'cosx
(ili)S
.2dx. (iv) - dx.
, cos 2x
(V)J(tan.r. f xcosx
cosx ) dx. (vi) .
sln2x dx.

4. (i)JIslnxcos 3 xdx. (ii) J dx

,J( is?)
dx
( sin x + cos x )
I + sin 5 dx

6
• J dx
3sinx-4cosx
dx
• I (3sinx + 4cosx)2
f sin
snx i• sinx
8. dx. (ii) •,I sin 3x dx.
 Ex. IV SPLCRLTRIGONOMIITRIC FUNCTIONS 73

dx dx
9 60
(I)J s i x -flX S I — sin4X
cot 2 x
o + dx .
S C0t 2 x -
dx

5 Tosx— Scosx
s1n3x SIEL5X
dx.
r
12. >5 di
(ii)J COSIX

dx dx
13
i)J Sin XCO52X (ii)J sin x C0S3x
sinin2x dx
(Iii) (iv)
SII1
SXSIfl3X 5 cos
CO53X COSX

Put sin 2 x + cos 2 x in the numerator of U) and (ii)

dx • d
••
S
SjflXCOS2; J sinxcos'x
Put tan x = z in (i) and VO. I

.r-::;
J. E. E. '8 )
15. (1)
J ix cos x di .
(ii)5_,( sin- 2) dx.

•: r dx
16. (j) ) (I1)J
— 5sin2x +

17.(i)f- sin 2x
+ cos4x
dx. H. S. '86 J

dx
(ii) J s1n 4 x + coc4x
J(H) Write sin 4 x + COS4 X = cos' 2x + sin 2x

I- +
dx. (ii) 5 sin 2x dx
(sinx + cosx)2
 74 INTEGRAL CALCULUS Ex. IV

sin x
19 Cos x)2dX
• f (1 +

dx dx
20 (I)
J I s- tan x 5 1+ cos a cos x
cos x
21. (1) dx.
5 + Cos x

cos x dx
(ii) C. P. 87
J 2sinr + 3 Cos x
1
Numerator = ^ ((sin x + cos x) + (cos x - x

5 / ( cosec x - cot x ) sec x

dx.
\ \ cosec x + cot x 'J( I + 2 sec x)
sec
22 Ja + btan x
dx.

dx
23 • dx.
S a + b tan
dx
24. dx.
S a +bsinx
dx dx
• S 5^ 4sinx 5 ^ 5 sin x
dx dx
(iv)
S 4 + 3sinhx 5 4 + 3 cosh x
dx dx
[C.P.86,'88) (ti)
• (i)S 5+4cosx 5 3+5cosx
dx cos x dx
________
27• .rJ.E.E89 (ii)J
WS Cosa + cos x 5 . 3cosx

dx
28
J a 2 — b2cos2x
sinxdx
29 • b2sin2x)
J 2COS2X +
 SECIA!. TRIGONOMETRIC FUNCTIONS 75
Lx IV

sin 2x dx
30.
--;--- bcosx)2
cos x - 16 sin X
31 • dx.
$ + 5sinx
dx dx
32
•0 'S cosx
I— + sin (ii)S 3 + 2 sin r + cos x

r + 3sinx + l4 cos x
33. I6------ -----dx.
3 + 4 sin x + 5 cos x

34. JIi + secxdx. [ Put 42 s i n j x = z.1

35. (ill ------- dx. I C. P. 'S5 I

) sec .- cosec

r dx
0') I C. P. '89 I
sin x 4- tan x

36. tan x _

Il4t si,i z - co.c x z and note 2 sin x cos x I- (sin x - cos x ) 1

( I Isin(x—u)l
.-- dx.
37. i,
J N L sin (x + 0)1

38. —
x sin x +- cos x )

ANSWERS

1. (i)logtan s.. (Ii) sin _isin h 2x +

x.(iii)x --sin 1 sin 4
(IV) 2c ' )." x -cosx (v)-.x L sin 4x
(vi) sin 4 r--sin I i ( vii) -j I 3x - sin 4x t sin 8
(viii) 5in -, x .sin
1 S , ( x) -.cos

i!1'r -nx .sinx (xii) toni - cotx.

(Iti,C.x sec -'x Sec'X

.-
76 INTEGRAL CALCULUS

2. (i) - cot I x -log sin x. (it) tan' x - tan x + x.

(iii) tan x (1 + tan I x + tan" x) . (iv) - cot x - cot X.

(v) -cotxcosecx - ! cot xcosecx + ! log tan .x.

(vi) tan 3 X(j + ! tan 'x) .3. 0) log tan Ix + 2 cos x.
(Ii) 2 sIn x - log ;ec x + ton x) (iii) . log ( sec x + tan x)
1 1 +'/2 sin x
(iv) -- log - '12x sin (v) tan x + - tn 7x.
(vi) log tan +x - x cosec x. 4. (1) -. 'Jsi (7 sin x - 3 sin 3x).
(ii) 2 cot I x( 4tan x + 2 tan 2 x - ). 5. (1) &
- + tan x
6. .log tan ( .}x _ftani-).
- 3(3 tan x +4)
I I + 42 cos x1 43 + tan x
S. (I) log '12 coax . (II) log '13 - tan x
9. (1) 1 log tan ( I n + x) (ii) I tan x +tan -'('12 tan x)
10. j logtan( . E + x). 11.jlogtan(x +4x).
12. (1) - - cos + 1 cos x. (ii) sec x + 2 cos x - .'cos3x.
13. (1) Sec r + log tan - r. (ii) Pan I x + log tan x
(iii) . log sin 3x - log sin Sx
(iv) . I cosec x - log ( sec x + tan x
14. (i)tanx -2cotx -cot'x.
(ii) j. (tan'x-cot 3 x) -. 3(tanx-cotx)
15. (I) 2 '/i. (ii) log (coax + sin x + 'Ixj.
16. (i) I- tog 2+
2 tan x I
(ii)- tan-' Itanx
--,-- -
L tan
17. (1) tan-' tan 2 x)(ii)-• tan-' ( -- tan Zx )
18. (I) 2 ' i sin x - 42 log tan (x +

(Ii)x + I
iY. 2 tan2 ! - x
l+tanx
 SPECIAL TRIGONOMETRIC FUNCTIONS Ti

20. (0 (4 r + log ( sn x + cos x))

(ii). 2 cosec a tan (tan 4a tan 4x)

21. (i) 4 (x+ log (sin r + cos x))

(Ii) 1
13
Z+j log (2 sin x + 3 cos x).
(ill) sin - ' (4 sec 1 4x

log tan
_____ + -- an
t-'- .

ar 8 log(acosx bsinx).
23 +
1-
a
+

2 , tan -ix + bi
24.--- 8j-3 tan t___1_b,)Ja>b;

log I atan4x + I, _ •I(b - a2)l if a < b.

a1)
J(b 2 - 8 +(b' 02
I4 2tan x ^ ?
25. (1) 4 tan -' 4 .' 5 tan 4 x + 4) (ii) - log
3 2tan4x+4
1 + 2 tanh + tanhx
(ill) 4 log . (lvi log - tanh x
4 - 2 tar.h - 12

(2 + anix)
26. (I) 4 tan ' (4 tan 4 x). (ii) 4 log 2 - tan 4 x
a)
log c s (
r -
27. (i sina o

I 52) tan x) if > 5;

28. -
82)tan (a _____________

I a tan x _42 - a2
ifa <
2) a1)'
•

2a7(b' - log a tan x +'I(b' -

52) bog ('[- b 1 cosx ^osr + b'sin2x),

29. 1(a' -
-1 b 22
if a 5;
__________
cosx)ifa <b.
> ,- (
30. - 25 2 log(a +bcosx) _2ab 2 (a i- bcosx)
78 INTEGRAL CkI'

31. 3 iog (2 COS X 4 5 sin X ) 2. *2. i) Lg (1 4 "'

(h) tan ' ( I + tan x ) 33. 'Ir 3 - 4 in x
34. 29ir.' ('I2sin+x).

35. (i) sin x -cosx — - log tim (x +-n)J.

(101og tan 1 x ti an Z fx . 36.'12 sin (sin x - cos x)

37. coccos(cosx-a) -i log (ut

sin - cosx
x sin x + cosx
 CHAPTER V
RATIONAL FRACTIONS

I Method of breaking up into partial fractions I

5.1. Integration of Rational Fractions.

When we have to integrate a rational fraction, say, 1(x)! 4,(x),
iff( x ) he not of a lower degree than 4, ( x) , we shall first express
f(x) / 4, ( x) by ordinary division in the form

C,xP+ C,xPi + ... + C0+-

V(X)
4, ( x)

where C P + ....... . C0is the quotient, and w(x)isthe

remainder and hence of a lower degree than 4, ( x).
r' (x)
Then i dx = C, x + i -- dx.
j4,(x) p +I

So we shall now consider how to integrate that rational fraction

' ( x ) / 4, ( x ) in which the numerator is of a lower degree than the
denominator. The best way of effecting the integration is first to
decompose the fraction into a number of partial fractions and then
to inkrate each term separately.
We shall not enter here into a detailed discussion of the theory
of partial fractions for which the student is referred to treaties on
Higher Algebra, but we shall briefly indicate the different methods
adopted in breaking up a fraction into partial fractions according
to the nature of the factors of the denominator of the fraction.
We know from the Theory of Equations that 0 (x ) can always
be broken up into real factors which may be linear or quadratic and
some of which may be repeated.

• When f( x ) and (x) are algebraic expressions, containing terms Involving

positive Integral powers of x only, of the form
80 INTEGRAL CALCULUS

Thus, the general form of $( x) is

A (x - - - y)F(r +
((x - 1)2 + m l ) ... ((x - l')Z +
rn'2)'
Case I. W hen the denominator contains factors real linear, but none
repeated.
To each non-repeated linear factor of the denominator, such as
x - a, there corresponds a partial fraction of the from A / (x - a ) ,
where A is a constant. The given fraction can be expressed as a
sum of fractions of this type and the unknown constant A's can be
determined easily as shown by the following examples.

Ex.1. integrate
5 +
+ x 1 - 6x dz.
r3+x26x=x(x2+x._6)x(x+3)(x2).
( P.P.78fl

Let x2+x-I A B
_____ C
-j'____
x+3 + -2
Multiplying both sides byx(x + 3)(x - 2)weget
+ x -1 = A (x + 3)(x - 2) + Bx(x -2) +Cx(x + 3)..
Putting x = 0, -3, 2 successively on both sides, we get
A
= -i-, B =,C
the given integral is
JJdX If dx If dx
6J x J x+31 2) x-2
= .}iogx + 4log(x + 3) + flog (x - 2)
Integrate
Ex. 2. dr.
( I - a)(x - b)(x - C)

Here the numerator Is of the same degree as the denominator and If the
numerator be divided by the denominator the fraction would be of the form
I +, where Q=(x-a)(x- b),z-c) and P is of a lower degree than Q.
 RATIONAL FRACTIONS 81

Hence, we can write

_______ A B C
(x-aXx--b)(r- c) x-a x-1, X-C (1)
(x-a)(x-)(x-c) +A (x-b)(x-c)
+ B(x-c)(x-a) +C(x-a)(x-b).... (2)
Putting x = a, b , c successively on both sides of the above identity
(2), we get

A - I,' - ____________
- (a-b)(a--c) ' - (b-cXb- g ) ' - (c-a)(c-b)
from (1), It follows that the given Integral
C a C. dx b' ( dx
Jdx + ( - b)(a - c)J + (b - c)(b _a)J x - b
( dx
(c - a)(c - b))
= X + (4
43
)(4 )
b3
Iog(x - a) + (bc)(b_a)i06_

C3
(c - a)( c - b ) log ( z - C)'
+

Case II. When the denominator contains factors, real, linear, but
some repeated.
To each p-fold linear factor, such as (x - a )', there will cor-
respond the sum of p partial fractions of the form
A,- A, A1
+ 1- .........+
(x - a)' (r- a) (x-- a)
where the constants A, , A, - .. . A1 can be evaluate
easily.

Ex. 3. Integrate

Let
X2
f( + 1)'(x
A
dx.

C
(x + I)-(x + 2) (x +I -+ (x-+1 ) Cr +
Multiplying both sides by (x + 1)2 (x + 2), we get
= A(x + 2) + 8(r + I)(x + 2) + C(x +

I nleq,a: Calculus (main) -8

$2 INThOXAL CALCULUS

Putting x - 1, - 2 successively, wegot A 1, C 4.

Again, equating the coefflcl.sits of x I t-n both side,,
;.B=-3, since C=4.
dx dx dfx
the given inferal
(x+1_)- jx+ +J
- I - 3 log (r + 1) + 4 log (x + 2)
+I
Not.. The partial fr&ctiovo In the above case can also be obtained in the
following way. Denote the first power of the repeated factor, i.e., x + I by z,
then the fraction Now, divide the Numerator by the Deno-
minator of the second fraction, after writing thorn In ascending powers
of a, till the highest power of the repeated factor, viz., z 3 , appears in the
1\ 1 4z' -1 _.I + - 4
remainder. Thus the fractionZ 3= - I - .z +- I
1+z/ z2 Z I+Z
Now replace; a by x + I, and the required partial fractions are obtained.
Case III. W hen the denominator contains factors, real, quadratic,
but none repeated.
To each non-repeated quadratic factor, such as x 1 + px +
(or, x2 + q, q * 0) there corresponds a partial fraction of the form
(A x + 8)1(x 5 + px + q), the method of Integration Of which is
explained in Art. 2.5.
x
EL 4. Integrate (x - j)(x .p 4) dx.

A Rx+C
Let (x- 1)(x' + 4)i x +4

A (x 2 + 4)+(øx + C)(x - 1).

Putting x 1 onboth sides, weget A = 3.
Equating the coefficients of a 2 and ron both sides, we get
A +BO and C-B . 1; hesce,B-,Cj.
 RATIONAL RACI1ONS 83
the given integral becomes
I
f dx I Jf ___
5) Ti
x4 dx
z+4 dx =-----'i- J J
Ir22dX 41d7
m

= - log(x- I) j lg(x2 - 4) + tan'

Ex. s Integrate
I
J dx
(x i a l )(x 1 + b)

(x + a 2 )(x' + I' 2 ) - b ['7f-;•-p-2 + UI]'

the given integral '1 2- b2

1
jf -J 2
dx
7*+47 1
[1 tUfl
x xlI.
I tan-'—
= '-j; -
dx
Ex 6 Integrate Jp-;-i.
Since x'+ I = (x + I)(x - x + I).
I = A + Bx+C
let us assume r.1 x+I
I A(x- x + I) + (8r+ C)(x + 1).
Putting x =- I, we A
Equating fle coefficients of x2 and the constant terms, we have
A + B = 0 and A C =I B - i ,C =+.
the given integral becomes
lf dx
— i-------
3jx +1
--If r-2
-i -----dx
'Ij x2-r+I
(2r- 1) -3
3J r+1 6J x2-x+l
it di itIt 2x -1
dx
3j x + 1 —-1
+I
dx+ 6 2 -
2j x x + Ix2 - x

=Iog(x+fl_,log(x_x4I)+J
dx
(x -p' (;3y
+
14 INTEGRAL CALCULUS

^ Iog(x +1) — .iog(rt -r + 1)+

_I tan1
vr)
/2x -

log(x +1)- 4- log(x 2 -x + 1) + tan-' /-

_2x,3 1
6 T3

Case IV. W hen the denominator contains factors, real, quadratic,

but some repeated.
In this case we shall require the use of Reduction FOrmula to
perform the integration, for the general discussion of which see
Chapter IX.
dx
Ex.? inttgrsc5
(1 +
Although lids case comes under Cm (IV}, it can be treated more simp-
ly as follow.: Put tan 0.
t ,ec2OdO
... I =J cos l OdO
.1 sec4O

- J+(1 + cos29)dO +( ++51n20)

I 2 tan o
= (e+ + tan*0} =f [tan lr+1

5.2. Two Special Cases.

(A) In many cases, if the numerator and the denominator of a
given fraction contain wen potters of x only, we can first write the
rractlon in a simpler form by putting z for x 2, and then break it up
nto partial fractions involving z, i.e., x 2 , and then integrate it.
Ex. 3. Jnesrat ej dx.
+ -:-2

Puttingx' =z, wehave

_________ - z z A B
x 4 +x 2 -2 - z+z-2 - (z+2)(z-I) • 7 +.
- I ' say
z-
z A (z- I) +B(z + 2).
Puling z - 2 and 1, we get respectively A = , B
 RATIONAL FRACTIONS $5
_______ 2 1 1 1
x4+x2_21xt+2x2i
25 dx + I f dx
+2 3 rt1
212 tan r11 x-1
+
32 1os—.x+ I
(13) II, in a fraction, the numerator contains only odd powers of x
and the denominator on,y even powers, then it is found more con-
venient to change the variable first by putting x 2 z and then
break it up into partial fractions as usual.
Ex 9 Integrate x dx
+ 3x' + 2
$
Put x 2 = z. 2x d. - x3dx =

1f zdz
2Jz2+3z+2
Now, x 2 +3zs2
Z = Z =
-••- . say
(z+U(z+2) z+I +z+2
We determine as usual, A = - I , B = 2
t dz •1
i = -} [2Jj_ dz -_J_-_j-j =2 [2 log (z+2)_
log (z+I)]

= log(x 1 + 2) _Iog(x2 +1).

5.3. Integral of the form
f dx
J (x - a)-(x-
where m and n are positive integers and a and b are unequal, positive
or negative, can be evaluated by putting x - a = z ( x - b).

Ex. 10. Integrate

j (x - I ) 2 (x - 2)
Put x-1=z(x-2).
1-2x 4x -
dz
I -z ( 1 -z)2
86 INTEGRAL CALCULUS Lx. V

Hence, the integral transforms Into

(1 - z)3 1 - 3z + 3z - z3
5 5
z - 3logz + 3z
I x-2 ) /x - l \
-3Iog(—J+3 -I
x_\ I (x
-IV
-
-k---- x-2, ( x-2/

EXAMPLES V
Integrate the following:
(x - I)dx
1.
S (x-2)(x-3)
rdx
2
$ (x-a)(x-b)
(x-1)dx
'
• S (x + 2)(x-3)

xdx ' 3xdx

4 GO
• (i)S x2_ j2x+35
x2dx
• S (x-I)(x-2)(x-3)
2x + 3 ) dx 1) dx
(x 2 ^
6
• 5 x3+x2.Zx 5 x(x2-1)
xdx (x - I )(x - 5)
7 (ii)J (x_2)(x_4)1
• (i)J x2+7x-+12
I-3x2 xdr
8 dx. (ii)
(i)S 3x - x' $ (3-x)(3+2x)
XIIX dx
9 (ii) f,
• (i) S (x-a) ( x-b)(x-c) (x- )(x -. b)
I C. P. '81 1
dx dx
_j)3 (iv)
V)
S (x - 2)(x ) (x + 1) 2 (x ^- 2)
Lx. V RATIONAL FRACTIONS 87

10. (i)S (x + 1)(x

xtdx
+2)2 (u)I (3- x) 'dx.
X , + x'

")S X -
dx dx
11. 00 )2
3 x' - x + I —71
J:x(x
dx (x + 1)dx
12.
(i)S (x 2 - 1)2
dx
(ft)J (x - 1)'(x + 2)2
(3x + 2)dx
13.
(j)J (x - IP(x +T-)
dx
(1I)J x(x + I)'
2+2
14.
(i)J (WJ 1-x'

15. (j)J x' X

dx. (li)J x(x' dx-1)

X2 dx
dx.
16.
(i)J I- (li)J -I
xdx x 2 dx
( x 1 + 4 2 )( x + b2 ) (11)5(22 + S I )(X 2 +
17.
S
xdx
18. S(2 2 +
m 2 )( 2 + fri)
GO
x + a 2 Xx 3 + b2)

dx
19.
J (x' + 4 2 )(x + b)

xdx
20.
(1) J (1 + x)(1
(x1I)
+Xt)
(Ii)j dx.

21.
4 +x+I
dx.
J X

xdx x dx
(it)J X 4 - x2 - 2
22.
f x- x 2 - 12
dx
23. (2 2 + 4x + 5)2
j
x'dx
I
- (x' +1)(2x' 7- 1)
88 INTEGRAL CALCULUS Lx.
dx
25
J x(1 +x+x2+x3)
Jdx
,-. x4+x2+j
+ x2 + I = (x 2 +x + I)(x 2 - x + I)]

27. (i)J
t
x4 +
dx.
i• (ii)J X +1
dx.

[x 4 + I = (x 2 +x'12 + 1)(x 2 -x'/2 + I)]

28
(i)J
dx
cosx(5 + 3cosx)
(ii)5 sin2x- sin
dx

dx e'dx
29.(i)J I + .. g2 Ie -3e-+2
I C. P. '851
dx
30. .(putcos x =
J sinx(3 + 2cosx)
dx
31. Show that j,()

'I
= i [iogx + £ (- I)'(-)
log( x + r)].
'-I

where •(x) = II (x + r).

F-0
a.
32. Show that dx
J 7T
N

log (x-a,),
f'( a,)

where f(x) = II (x - a,), [a 1 *akif i * k).

r-1
 RATIONAL FRACTIONS 89

ANSWERS
1. 2 log (x -3)- log (z -2).

2. __!- (alog(x - a)- blog(x -b)}.

3. i f3 log (x + 2) + 2 log (x - 3)).

4. (i) } (7 log (x -7) -5 log (x - 5)1(11) log ((x -2) 2 (x + 1))
5. .log(x -1)- 4 log (x -2) + ! log (x -3).
6. (1) . log(x-1)-logx _ 1 log(x+2).0) log (x2-1)-logx.
x-4
7. (i)x 27 x - log —.
-x- 27log(x+ 3) + 64 log (x+ 4).(ii)
2 x-2
8. (i) . log (x(x _3)4}. (II) _ ! log (3 - x) log (3 + 2x).

a 2b2 log (x - b)
9. (1) lo g (x - a)
(a-b)(a-c) (b-c)( b-a)
C2
) log(x - c).
+ (c

I I x
a)2 log -.
T& - a)(x - ) + (b -
x-1 x-2 x-2 1 ( x-2
-----3 log - + 3--- -,
x-2 x-I x-I 2 x-I

(iv)
x+I x+1 I (x+\Z
x+I x+2 x+2 2
4
10. (i) + . g ( x + I .(ii)-- 4 log x + 4 log (x +1).
x+2 l
1 1ri-I 1 x
fl. (I) - + - log— . (Ii) - + log--
2(x-l) 4 x-1 r+1
I ri-I I x
12. (I) -log --2x2 1
4 x-1
12 1. x-I
- - log --- )

{ x-2 1 x-11 x 41+3

13. (I) + - log --j . (II) 2 log - +
(x - 11 2 x+12(x+1 )2
90 INTEGRAL CALCULUS

1
14. (1) tan' 2x+1I
, - log,1 I - x
+

(ii) — log( I- x) +
T
tan-' 2x + I 7
N3-

1. (I) tlog (x 2 - 1)-log (x 2 + I )} (ii) j-l og (x 4 -. 1) - logx

16. (i) . (log(I + x) - log(]-x)) --tan'x.
I
(ii) —  log i-I--tan 'x
4 x+1
17. (I) (7' 2) log, x 2 +b 2 I ( atan- r_btair'_.}
2 .(ii)------2  a

18. Ci) 2( a2
I 1 log(x 2 + a 2 ) -b2 log (X2 + b2)j.
 b2) (a

a3 t'
(ii) x +
b 2 - a1
tan -.x—
a + at—b 2 tan b

19. a2 1
____
{log(,
x + b b x}
2) ^ —a a

20. (I) .- . log(1 + x) + . log(1 + x') + tan'x.

(ii) - 
I _________
+. tan l2X+I
log 'l( x 2 + X+
- 1) '13

21. 1)- log(x 2 + x + I)).

11 tlog(x 1 — x +

22. 0) —7I log—.-- '13

x-2 +— x .( h )i(log(x1_ 2)-logx1+1)).
x+2 7 tan'?

x+2
23. ( tan (x+2)+145). 24. tan lr-tanI(x'12).
25. logx - . log(] + x) - !log (] + 2) -!tan-lx.
1+x+z I
26 4 log 1-x+x 2 + tans
1-x
(r'13)
27; (i) I log I—+ x'12
x-12 +r'
+ r
I / x'12 \
an -' j— )
I x'12
tan' 1-x1
(ii) —f--
 RATIONAL FRACFIONS 91

28.
(u) -Iog (1 cos x ) 1 og ( - cosx ) - -Iog (1 - 2 cos x )
29. (i)r+Iog(1+e').-2lOg(1+2C'). Oil .'log ((e'-1)(e'+3)fl
30. _log (I + cos r) + j Iog(1 - cosx) +. log (3 + 2 cos r).
 HAFTER VI
DEFINITE INTEGRALS

6.1. Thus far we have defined integration as the inverse of dif-

ferentiation. Now, we shall define integration as a process of summa-
tion. In fact, the integral calculus was invented in the attempt to
calculate the area bounded by curves by supposing the given area
to be divided into an infinite number of infinitesimal parts called
elements, the sum of all these elements being the area required. His-
torically the integral sign is merely the elongated S used by early
writers to denote the sum.
This new definition, as explained in the next article, is of fun-
damental importance, because it is used in most of the applications
of the integral calculus to practical problems.
6.2. Integration as the limit of a sum.
The generalized definition is given in Note 2 below. We first
start with a special case of that definition which is advantageous
for application in most cases.
• Letf( x) be a bounded single-valued continuous function defined
in the interval (a, b ),a and b being both finite quantities and b >a;
and let the interval (a, b) be divided into n equal sub-intervals,
each of length h, by the points
a+h,a+T h,...,a+(n-flh,so that nhb _a;
then Lt hlf(a)+f(a+h)+f(a+2b)+...+ f(a+ih))

i.e., shortlyk Lt 0 h f ( a + rh), ( nh = b - a),

b a Ef
or,"-4
Lt-
n-(
(since n-.- when h-O)
'i.e., which doe, not become infinite at any point. See Author,' Differential
Calculus, Art. 1.6.
 DEFINITE INTEGkAj.

Is 4eflned as the definite integral of f ( x) with respect to x between

the limits a and b, and is denoted by the symbol

f(x)dx,
J 4

where 'a' is called the lower or inferior limit, and 'b' is called the
upper or superior limit.
Cor. Putting a . 0, we get

fx)ühZ f( r ), wh,renh b.

Note 1. f(x)dx is also sometimes defined .s

Ja
M

Li h 1 + rh), or U h E ((a + rh);

k-.-.
O h O
rO
these definitions differ from one another only In the Inclusion OT exclusion
of the terms hf( a) and hf( a + ih ) ,i.e. ,hf( b) which ultimately vanish.
It should be carefully noted that whichever of these slightly different
forms of the definition we u,., we always arrive at the same result. Some-
times, for the sake of simplicity, we use one or the other of these definitions.
Supposing the Interval (a, b) to be divided Into n equal parts each of
length & by the points x ( a ), x , r2 .......... x ( = b), the definite
Integral
a-I
f (x) dx may also be defined as LI f(x,)& .
JI r-o
Note 2. The above definition of a definite integral is a special case of
the more generalized definition as given below.
Let f(x) be a bounded function defined In the interval (a, b) ; indict
the Interval (a, 1') be divided in any manner into it sub-Intervals (equal or
unequal) of lengths & , 82 ........ . 6. In eachsub-interval choose a per-
fectly arbitrary point (which may be within or at either end-point of the in-
;.rvaI :and let these points hex C..
94 INTEGRAL CALCULUS

Let Sn Z 6,f(,).
r=I
Now, let n increase indefinitely in such a way that the greatest of the
lengths 6 , . ...... ,, tends to zero. If, in this case, Sri tends to a definite
limit which Is independent of the way in which he interval (a, b) is sub-divided
and the intermediate points C, , C, ....... are chosen, then this limit, when it
exists, Is called the definite integral of f( x ) from a to b.
It can be shown that, when f(x) is a continuous function, the above limit
always exists.
In the present volume, however, In Art. 6.4 we prove that if, in addition
to /(x) being continuous in the interval, there exists a function of which It
Is the differential coefficient then the above limit exists.
In the definition of the Article above, for the sake of simplicity, f ( x) is
taken to be a continuous function, the Intervals are taken to be of equal
lengths, and , ,.... ( are taken as the end-points of the successive
sub-intervals.
The method of unequal sub-divisions of the Interval Is illustrated
in Ex. 5 below.
6.3. Illustrative Examples.
'a
Ex. 1. Evaluate froøi first principle e'dx.
"a
From the definition,
- I
e'dx = kD h L c'5 , where iih
fn
a r-o
Li
u-s h[e' + ea . .. +
0
= It. I11 4a
=
= e(eba h
1) —r--j ,sincenhb - a,
=e-e', [5ince uoè
-CI }..
 DEHNFE RMORALS - 95

Ex. 2. Find froi,, the defin ition, She vth, of

f x dx .
Jo
From the definition,
.1
x'dx I, £ (rh) 2 .where nh = 1,
J C

+ 2 2 h t + . . + n'h')
=h :o h 11 'h'
= Lt 1h 2 ( 1 2 +2 2 + ...

I (,, + I )( Zn + I)
= 1-40
Lt It'
6

= -- Lt (2n3h + 3i 2 h 2 .h + nhIt2)

=-(2 + 3J1 + h').since th I,

I I
=-.2=-

Ex. 3. Proveab i,iitiodx = -

Here by the definition,

-,,dx= IJ I I
k_0[-+t------I (a+(n-.jh)) 2J
I
(where nIt b - a)
Denoting the right-hand .et-le by $ , since, obviously,
j 9 _____ and <
(a + rIt (a + h If a $ (r + I) It) F. + ( r - I) h J(a + rk)'

w ege.S >A [+ +
) (as (,i-I) h)(a+nh)]'

[Cs
 96 ITNITEGRAL_CALCULUS

>!
(. - a + nh)' i.e., [since nJ, b - a]

Also, 5< h i i
h) a(a+h) 5hija+(nj5h)'

/ 1
i.e, <(-_ 1 /1 1
- I , i.e., I---
a-h a+(n-1)k k s-h b -h
'1 1' '1 1
Hence,a (s---) <S<(___
b a - h b - li
and this being true for all values of Ii , proceeding to the limit when
h ..-. 0, (-
! clearly tends to (-I - and S by definition
r b gjx r i
becomesj -- , an d hence
j b-=---
For an alternative method, see Ex. 5; here m = - 2.

Er. 4. Prow by summation, sin r dx = cos a - cos b.

n-II
I sin x Li li 1 sin (a + rh ), where A = b - a,
J I dx h -.
O
r- o

= LI l ( sin a+jn(a +h)+ sin (a +2h)+...ton terms j

= Li h • sin (a +<n - 1)!} +nh

_______
sin!2 li
= 1L
_ 2
hO sin2 !h s in + (n - fl.}

+h [Cos
= h -.O sin±h cos t a +un_I
(a_.h )_ 4}j
= Lt [cos (e-4h)-cos(.a+nh-lifl .,since Li -.2--
h -sO
 DEF1NrFE INTEGRALS 97

a + nh b,
Icos(a+ h ) - cos(b--h)I, since =
= h
= cos a- cos b.

Es. S. Evaluate x' dx where in is any number, positive or iregal:vc.

.Ia
integer or fraction .but I - 1(0< a < bY
Let us divide the interval (a, b) into n parts by the points of division
a, ra, 2 ,,, ar - , ar",where ar= It i.e.,r
Evidently as u -^, r= (b /a) 1 ' a - 1, so that each oI the intervals
r - I ),ar(r - 1) .....ar -
- I) -40. Now, by the generalized
definition, as given in Note 2, Art. 6.2,

I '
Ja
dz Li (a'.a(r -1) + (ar)'.ar(r -1)

+(ar2).(a7)(r - 1)+ .. ton terms]

Ion terms)

= Lt 1' + I *0i

r-1 -11
= L I aint T _ • -i L(r) J

Li
,_.1 ? • a-I "1 a ) -.1-1}
{( !

- Li

• _ •)
,_.i --T  (- 
- b
M +I

Since :I__1 being of the form

[

0-L
1
- 7_h Is.-
# I ) in + I
Integral Calculus (main) -9
 UM
TWRAL CAM
Nc*t 1. bkln* xn'iou l (a, ' )is tt 'cable in a , 6), a urn-
.ue li teft if 'i.e ,nmatlon S a; giver in N ' i- 1, Art. 62, exists. S, it s
immaLerial in what node we calculate a. The same remark ho!dt or the next
example.
fb
Note.2. !nv
e val uating I
Jo
d xn^ - 1,6 > 0) wc may hrst

evaluPeJ zó IC<a<b) s. above, i.then n-iake a-3 0+.

a

EX . 6. The de[nitios

b
a log- (0 <a <6)
J-
As In Ex 5, divide the Interval (a . 6) into ,i parts by the points of
division a • • w' ............... Cr" Y. a,", where a," = 6
I.e.,, = (b/a)' /'. Evidently asn.-+., r (b/a) 1 ' -sLsothat
each o the intervals a - I ),ar(r -. 1) ....-, 0. Now, by the general-
ized definition,

= Lt (r - I) = Ii n(r-.- j)

= Li n b a P /1 I)

= Li
h-*0L
Ie .i
h loga
6 where h = ! log -.6
n a

=log 6-[slnceLtj_.=I
r

Ex. 7. Find a!, initi The DQIUC of

I
f
Jo
sec' x dx.

By definition, the required integral

a
I U it sec: 'rh,where nh =..
Ii-. 0
r-1
 F.rNrr lNrLGRLS

Now,scc(r- Hh see rhse rh<sQcrh.sec(r •

since "c increases with XiflO<x4t
I - sin(r+1)-'l
stnh cosrh. cos (r + l)h
- I
( tan (r + 1)1* -tan rh} .

Simdarly,sec(r- l)hsecr = --- ( tan rh-tan(r-- 1)h).

sin h

Thus, I lies between It ------ E (tsnrh -tan(r -

. O sin
r.1
'V

and hL.'O
L (tan (r +1)h - tan rhj,

i.e (tannh-tnO)and

Since nh = it, and LI ('t / sin l) 1 ash -. 0, both the above Un,tt
tei d to tan 1r i.e., 1.
Tence, I has the value 1.
6.4. kefinition of definite i utegral based on the notion of
boun
We h ye i wo methods of refining definite integrals: one based
on the notion of limits, the 1.er based on the notion of bounds.
The first method based he not-cm of limits is given in Note
2, Art. 6.2.
The second method based on tLe notion of bounds Is given
below.
Let the interval (a , b) be divided in any manner into a num-
ber ( say n ) of sub-intervals by takiig interme-diate points
a=X x1<x2...
Let M, and m be the upper anc2 lower bounds of f( x) in the
r-th sub-interval (x, -' ,x. ) and let 5, denote the length of this
sub-interval. The lower bound (denoted by J ) of the aggregate of
 INTEGRAL CA LCULUS

the sums S = EM, 6. (obtained by considering all possible modes

of sub-division) is called the Upper Integral and is denoted by

I f(x)dx,
and the upper bound (denoted by j ) of the aggregate of the sums
s = Em, 6, Is Zalled the Lower Integral and is denoted by
IF

5
When the lower and upper integrals are equal, i.e., when j =
then f( x) is said to be integrable and the common value is said to
be the integral of f( x) in (a , b ) and is denoted by

L bf(

It can be shown by what is known as Parboux's theorem that

both the definitions are equivalent when / (
x ) is integrable.
Note. The Integral defined above, when it exists, is called a Rlemann
integral, as it was first obtained by the great mathematician Rlernann.
6.5. Necessary and sufficient càndition for integrability.
We give below, without proof, the necessary and sufficient con
dWon fo. the integrability of a bounded function f( x
7

If there be at least one pair of sums S , s of f ( x ) for a sub-

division of the interval ( a, h ) su. that
S - 5 < £,
wherec is any arbitrarily small positive number, then f(x) is in-
tegrable.
Note. It can be easily shown that the sum or difference of two or more
functions integrable in (a, b) is also Integrable in (ES, a).
6.6. Integrable functions.
(i) Functions continuous in a closed interval ( a , b ) are in-
tegrable in that interval.
 DEFINITE INTEGRALS 101

(ii) Functions with only a finite number of finite disconti-

nuities in a closed interval (a. b ) are integrable in that interval.
(iii) Functinn - .i aa interval ( a , b
re irtegrah]e in that interval.
6.7 Important Theorems.
1. 1f ( x ) i integrable in the closed interval (a , h ) and if f ( x)
^O for all) x in (a,b ),then

a O(b > a).

Since f (x) ^, 0 in ( a b ), it follows that in the interval (x,. 2 ,

the lower bound In, >_ 0 and therefore
Inr5r ^! 0.
j, which is thi upper bound of the set of numbers s, ^ ! 0

Since f( x Y i intcgraHc. x ) dx
J R

and hencc f (x )dx ^ 0

5
Alternatively.
Since f(x)is integrable in(a..b),
b

f(x)dx LI f (c
5 n -4 -

Since f(x)? 0 in(a,b), :. f (, )2t 0 in(a,b).

Li 1J(, ), ? 0 in(a,b).

f(x)dx ^ 0 in
f a
 102 INTEGRAL CALCULUS

Note. It can be shown similarly that if f(x) :5 0 in ( a , b) then

f(x)dx 0.
J a

II. If f(x) and g (2) are integrablein (a, Li) and f(x) L
, g(x)in
( b ), :hen

S:f(x)dx 5(X) dx (h > a).

Consider the function w ( x ) f( r) gCz

Then Y (x) is integrable irt(a, b)ar,d w( x) ^! 0 in (a, b).

by (1),
$ y ( x ) d x -> 0 in ( a , b ) ,
gf ldx ^ ! 0 in (a,b),

i.e., f (x)dx ->Jg(x)dx.

UI. If Mand in are the upper and lower bounds of the integrable
function f(x)in(a,b),b > a,then

a) S
5 f (x)ix 7^ M (Li - a).

Since in 15f(x)5 M in (a,b),

( f ( x ) - m) ^ 0 in ( a , b ) ,

5(1 (x ) - m)dx -,:t 0

f(x)dx f dx, i.e., 2! ,n(b

f a
^ rn
a
a
 OEFD4E n(rEGRAI.s 103

Similarly, since M - f ( x ) ^! 0, we can show that

M(b a) 2!
s :f(x)ax
Hence the result.
This is known as the First Mean V alue Theorem of Integral Cal
Cu
Cor. The above theorem can be written In the form

f(x)dx = (b a)g,when in
J
and if further 1(x) is continuous in (a • ) then f(x) attains the value i
forsomovalue of x such that a S %b ,andso

i(x)d (b -
J a
IV. If f(x)and g(x) are integrable in(a,b) and ifg(x)
maintains the same sign throughout (a , b) , then

in and M being the lower and uppet bounds of f(x) in (a .b)

Let us assume, for the sake definiteness, that g ( x) is always
positive in (a b
Now, in S f(x) :5 M in (a,b).
Since g ( x) is positive,
mg(x) 15 f(x)g(x) 5. Mg(x),
f (x)g(x) —mg(x) ;! 0,
b
(f(x)g(x) - mg(x)ldx ;! 0
J
ex uthors' Dfferentiai CalCUlus.
 104 INTEGRAL CALCULUS

pb
f(x)g(x)dx 2!mJ
q J a a
and f(x)g(x) - Mg(x) 15 0,

x )g ( x )dx - Mg ( )) dx 0,

M fbg(x)dx.
f(x)g(x)dx

f ab f(X)g(X)dX=J ab S(x)dXwhere m ^ <M.

Cor. If further f(x) is continuous then f(x) attains the 'alue j for
some value of x, whore a 5 5b, i.e. , f ( ç ) =
when f(x) is continuous
rb
f(x)g(x)dx = f() J g(x)ds.
.11

Note. This is the generalized form of the First Mean Value Theorem. The
theorem III can be obtained from this by putting g (x) = 1
V. If (1) is bounded and integrable in the closed interval ( a, b
and if

F(x) x is any point in(a,b), t'en

=ff(t)dtwhere

(1) F( x) is a continuous function of x in ( a , b)

(2) If ( x ) is continuous throughout ( a , b ), then the derivative
ofF(x)exists at every point of(a.b),id = f ( x ) .
 DEFINITE INTEGRALS 105

(3) If f(x)is continuous throughout (a,b) and if •(x)be a

function of x such that '( x) = f( x)throughout ( a , b ) ,then

F(x) f (t)dt =(x ) -(a).

(1) Let us consider a point x * h in the neighbourhood of x

in ( a , b

Then, F ( x + h)
=J: 1tdt

F ( x + h) -F(x ) =Jf (t)dt

.J f (t)dt

.11
f (t)dt =
=5 z
by COT. of (Ill), where i lies between the upper and lower bounds
of f( t ) in the interval ( x , x + ,'i 'Sincef( 1) is integrable, m and
M are finite and so is i.
Lt (F ( x + h)- F(x)) = Lt pth= 0.

Lt F ( x + h) = F(x).
h -.0

Thus, F ( x) is a continuous function of x in (a, b ) -

(2) We have F(x+h)-F(x) =Jf ( I )dt

= hf(ç). where x 5ç 15x +

since 1(t) is continuous. I See Cor. of (III)
F( x + h)-F(x)
= f(ç). for h;t 0
h.
When h -* 0, —* x and f() ---.f( x ) , since f(t ) is continuous.
 ThTEGRAL CALCULUS

F(x + h) -F(x)
• LI
It exists, and f(x),
i.e., P(x) = f(x).
(3) Since f ( x ) is continuous throughout ( a , b ), as proved
above,

• F'(x) = f(x), i.e., F'(x) =

F'(x) - '(x) = 0.
Let i(x) F(x ) -x ).
= 0 everywhere in(n,b).

Hence, i(x) F() -(x)=aconstant cin(a,x)(1)

F See Differential Calculus A rt. 6.7, Ex. 1. I

When x = a,F(a) =f :f (t)dt = 0.

Since, from (1), F(a)- (a) = c, •. -$() = C.

Consequently, from (1),F(x) = (x) + c =(x) -(a),

= ( x) -a)

In particular,

1(t) d •(b)- •(a).

J a
Note. The relation given in (3) is known as the Fundamental theosen, of
Integral Calculus. F For an alternative proof, See A rt. 6.72. I

6.8. Change of variable in an integral.

To change the variable in the integral f( x )dx

by the substitution x = ( I) , it is necessary that

 DEFINITE INTEGRALS 107

(I) 0 ( I ) possesses a derivative at every point of the in-

terval a5 15 , where(a)= a and i( 0)= b, and ' (t);e 0
for any value I in ( a,
(ii) fL( t)l and '(t)are bounded and integrable in(u, ).
When the above conditions hold good, then and then only we have

I)) ' (I )dt

f a
x)dx
=J:ft (
Illustration

dx
Let 1
=f -I
1+

Putting x = tan o, we get I =f io =

Putting x / I, we get

de
I=- +t2
=

The reason for the discrepancy lies in the fact that I lt does not pos-
sess a derivative at 1 0, an interior point of ( - I , I); in fact, the func-
tion itself is undefined when I = 0.
6.9. Primitives and Integrals.
If ' ( x) f ( x ) , then 0 ( x) is the primitive of f(x). The in-
tegral of f ( r), on the other hand, is
LI ZJ ( C,)) S. ,or symbolically

x ) dx, i.e., the analytical substitute foran area

1:
in case f (x ) has a continuous graph.
The distinction between the two is that while integrals can be
calculated, primitives cannot be calculated.
The question as to whether a primitive exists and the question
 I Gs INTEGRAL CALCULUS

of the existence of an integral off ( x) in ( a , b ) are entirely inde

pendent questions. It is only in the case of continuous functions that
they are the same.

Indefinite integrals can properl y be dr'c,

of primitives.
The connection between primitives and integrals is represe.
by the Fundamental Theorem of Integral Calculus, viz.,

dx =

Illustration:
I I
(I) f ( x ) =- x . sin - - -
X

= 0 (x = 0).
Here,
-
%1 = f(x) for x ^t 0 and 0 for x = 0,

so that primitive exists, but f(x)

5 dx does not exist.

(ii) f(x) = 0(r 0), = I(x = 0);hercn(0I)f I(x)dx

exists and = 0, but no primitive exists. 0

6.10. Illustrative Examples.

Er. 1. Show that

<5 0
'(4 - x +fl <
We have 4>4.- (x 2 - 0)in(0,1),

or, 'ii'>I(4_z1 tx5).

I . I I

L dx

5'
. 0
dx, i.e.,
<f 0
x'
 DEFINITE INTEGRALS 109

Again,4 -x 2 <4- .x l * r l in(0,I).

I I
:.
1'4 -x )> '1(4 r 21)

I dx I dx
j •;(4 - x ) > J '1(4 -z 2 +
I I
I- •. I ,1.it - dr

L'jo j- r
Hence the resu l t.

Ex. 2. If f(x)dx ex i s ts,

show t hat f(x)dxl sj f () j dx.
a JI
Wc have l f(C )&. + f( )6, + . ... + IF I
!1 J(-ç )I I S1 1 +1 f(2 )11 I + ........ I f(ç)I I 81.
Lf (ç ), I -'5 £ I f(, )11 6, I
Li I Zf( )S, I !; £1.1 f(, )I I
f.bf(.)d"l r b
S f(x)I dx.
a
OIherwi

SinccJ f(r)dxexis ts , I 1(x)I dxedsts.

a .. J a

We have - f(x)J S f(x) 5 I f(x)I.

fb
I f(x)I dx5 f(x)dx^J I f(x)I dx
-f I
a

i.e., I f(x)dxl f(x)I dx

J a a
IC INTEGRA L CA LCULUS

6.11. Geometrical Interpretation of 1(x) dx.

Let the function f ( x ) ,y

which we suppose to be finite /0
and Continuous in the interval D
(a , b) [b > a 1, be represented
graphically and let =f (x ) be
the equation of the continuous C
curve PQ, and let A C, BD
be two ordinates correspond- X' 0 A A' B
ing to the points x = a, x b, '

meeting the curve at finite

points.
We have Oil = a, OB =b and .. A B = b - a.
Let A B be divided into n equal parts each of length h
..nh=b-a, Or, a+nh=b.
Let the ordinates be erected thro-ugh the points whose absci-
ssae area + h, a + 2h, ...., a + ( n - I ) h to meet the curve
at finite points.
Let us complete the set of inner rectangles A CC' A ' ..... . and
also the Set of outer rectangles.
Let S denote the area enclosed between the curve y = f (x),
two ordinates x = a, x = b, and the x-axis.
Let S1 denote the sum of the inner rectangles.
S1 / ( x) monotone increasing I
< S. I

Now,S 1 =hf(a) + hf(a + h) + .....+ hf(a i- n.-1h)

=h £ f(a+rh).
"0
Let S 2 denote the sum of the outer rectangles. .. S 2 > 5
 DEFINITh INTEGRALS III

Now, S hf(u + h) + hf(a i Zh) +...+ hf(a i- nh)

h Z f(a + rh) — hf(a) i- hf(b) I since a + nh bJ.

'-0
We hv€. S < S < S2
Now, let the numbr of sub-division increase indefinitely, and
consecjucntly the length of each of the sub-intervals diminishes in-
definitely.
This, a n •-* , ! - 0.
both hf( a ) and hf ( b) —, 0 , since 1(a) and f (b
are finite.

b
h Z f(a + rh)
S'-4
Al
Lt
-. 0 =5 f(x)dx,
a

r-1 b
and S1 --, h L f (a + rh) f(x)dx.
h-.0
LI
=5 a

Since we have always S 1 < S < S2,

b
S
=5a f(x)dx.
Thus,
5 f ( x) dx geometrically represts the area of the
space enclosed by the curve y = f ( x), thejordinates x = a , x =
and te x-axis.
Note. The arguments here postulate q'concave curve. Similar argu-
ments apply for a convex curve, or even for curve which alternately rises
and falls in the interval. I
 112 INTEGRAL CALCULUS

6.12. Fundamental Theorem of integral C4ICUIUS.

If f(x) is integrable in (a, b ) I a <b 1, and if t.ire exists a func-
tion • ( x) such that 4'(x ) = f(x)in(a,b), then
L.
f(x)dx = (b) -
Ja
Divide the interval (a, b ) into n parts by taking intcrrnedito
points
<x, b.
Then we have, by the Mean Value Theorem of L'iI(crcntiaj Cal-
culus,
(x) -(x, 1 ) =( x
N n
- 4'(ç,)&7 =E I(r,) -
r-I
[ where 8, = x, - I I
= 14 x )-
1 ( x )J+((x 2 )-Ø(x , ) j + .....

= 4'(x ) -(x 0 ) = •( b) -
Lt
0
E ,' (c,) 8, = ( b) - (a), where Ols the greatest of
the sub-intervals 8,. Since f ( x ), aa1 hence' ( x), is integrable
in (a,b),therefore
b b
L i Z 0'(c)8, = I •'(x)dx f(x)dx.
J =f

Note 1. The above theorem establishes & connection between the hit egrs.
Zion as a particular kind of summation, and the integration as an operation uivers,
 DEPDffS DA 113
to differentiation. This also establishes the existence of the limit of the sum
referred to In Art. 6.2, Note 2.
Note 2. From the above theorem it Is clear that the definite integral is a
function of its upier and low limits and not of the Independent variable x.
Note 3. It should be noted that U the upper limit is the Independent
variable, the lategra! I. not a definite Integral but Pimply another form of
the Indefinite Integral. Thus, suppose If(s) dx • (r); then

5f(x)dx =$(x) -( a ) (x) + aconjtant Jf(x)dx.

6.13. Evaluation of the Definite Integral.
By the help of the above theorem, the value of a definite integral
can be obtained much more easily than by the tedious process of
summation. The success in the evaluation of a definite integral by
this method minly depends upon the success In the evaluation of
the corresponding indefinite integral, as will be seen from me fol-
lowing illustrative examples. The application of the above theorem
In the evaluation of the definite Integra' is very simple.

Suppose we require to evaluate J f ( x ) dx.

First evaluate the indefinite integral Jf(x) dx by the usual methods,
and suppose the result is (x).
Next substitute for x in ( x ) first the up per limit and then the
lower limit, and subtract the last result from the first.

Thus, 5 1 t)dx = $(b) -

Now, 0 (b) -, (a) Is very often shortly written as 10 wl

It should be carefully noted that in a definite integral She arbitrary
constant of integration does not appear.
For, If we write if( x)Jx - •(x) + c - (z), say,
Iniegral GaICL11US (main) - io
114 INTEGRAl. CALCUL.US1

theJf( x)dx W (b) a) (0(b) +c) — (a) + C

= $(b) —
Thus, while evaluating a definite integral, arbitrary constant ,tad
not be added in the value of She cir?ur• lig indefinite integral.
6.14. Illustrative Examples.
'r. dx.
Ex. 1.
Evaluate

I x dx =
J

r x*l r -t
I x dx I I = -- I b' - •' I; * + 1*0.
I L fl+1i 1+IL .1
a a

Ex. 2. Evaluate Cos 2xdz.

J o

f cos'rdr=4f 2cordx+$(1+cos2x)dx
=.}x ++ sin 2x.
*12 x/2
c2x dx =[y x 1+7sIn2x
1 ]
5
I I
= 3t + T Si n T X.

(1

Er. 3. Evaluate dr.

fJo
dx
j 1+x J \I+X
= 2$-_!_-_ dx _Jdx = 2 log (I + x)- x.
 DEITNFTE IWflGRLS 115

i +r)- f 21og2-l-2iogl 21og2-1.

1• 4. EDaIuattJ__i

I. dx 1 x
J a+z a a

Note. Two paints should hc noted when cvJuatl.ig a definite ineg,-aI

for which the indefinite integral involvex an inverse trigonometrical func-
tion.
(I) The result must never be expreir.d In degrees for the ordinary
rules for the differentiation and integration of trigonometrical functions
hold only when the angles are measured in radians.
(Iii) in qtAbstituting the litufta in the Inverse functions, C&C should be
taken to choose the right values of the expressions obtained. UnIr* other-
wise mentioned, usually the principal values 3re used.
6.15. Substitution in a Definite Integral.
While ticrating art integral by the substitution of
a new variaV, it Is sometimes rather troublesometa transform the
result back Into the original variable. In all such cases, while inte-
grating the corresponding intergal between limits ( Le., corespond-
ing definite Integral), we can avoid the tedious process of restoring
the original variable, by changing the limits of the definite integral to
correspond with the change in the z'erlable.
Therefore in a definite integral the substitution should be ef-
fected In three places .l) in the Integrand, (U) in the differential and
(iii) in the limits.
 hiS ThITEGRAL CALCULUS

The following illustrative examples show the procedure to be

employed.
6.16. Illustrative Examples.

f
1 si,i1x
E.1. Evaluate

Put pin - 'x = 0. :. dO 1 dx

x1)

0 and I are the limits of x; the corr spdtng limits of 0 where

0 sin -l x are found as follows:
When x = 0,0 = sin 1 0 0.
When x= I,8= sit) '1 =tc.
*/2 1/2
849=
1=50
Note. Of course this example can be VIed out by first finding the
indefinite Integral in terms of r and then subUtuttng the limits.
F' I________
Ex. 2. Evaluate a1 - x 2 dx.
Jo

Put r.asln0. :.dx=acos3d.

Also, when x-O, 0=0,and.ienr=&*,9=f*.

I a coi' OdO.. coe 040.

NowJo
J (I +cos 20)J0 0 uin2e].

i as [e ;+.1n29j

EX. 3. EIlslust.f -
a
 [DEFINITE INTBGRALS Ill

Put x a Cog 2O+ gin 'O. .. dx 2( - a)ain eco.9d9;

also, x -a = .tn 2 9 a(1 - co. 28) • ( - a)M819,
X = ( I- .1n 2 $) -acos'O

when x a,( - a) sin 2 O 0.

sin OO since *a. ..9=0.

Similarly, when I = PAP - a) c05 0 = 0.

cos $=Q. ..

I = 2 ( - a) 1 Mn 2 8 coø'$iO.
J a

Also, f(1 - cos 48) dO • 9 - sin 49.

/2
c9)d9 =(-a) 2 [e_.sln4o}

i( -a)2 (l- ! sin 2id - l a: P - a)1.

E. 4. uate
Evaluate
I ' i( (
a
x -
a
dx
)(0 - )
(B a). 11. E. E. '791

AsinEx.3,ptltr = aco'O+ sin1G.

i
= 2dO=2.-*11.
0

.1.
4I3)..
0

Put x sin 9. Then dx cos ode; also when x = 0,9 = 0, and

when x = 9
It
' coo Od9
cos2OcosO = JO
I see 20d$
0
 III ) INTEGRAL CALCULUS

*16
4.log too (* + 8)]
- [
=}( log tan -4t - log tan 4.%l log (2 + 43).

EN, S. MOR that = 2

0

Lt sins .. cos 0d dx,

also when 8 = (, x, 0 and when • = r, x = 1
It I

=J r' (I-
I ,ln' 0(1 - O). Cos 8i8
0 0

[7,]l•[]t
f 'x' dx- fx'dr 1 2

6.17. Series represented by Definite Integrals.

The definition of a definite integral as the limit of a sum enables
us to evaluate easily the limits of the sums of certain series, when
the number of terms tends to infinit y by identifying them with
some definite integrals. This is illustrated in the following ex-
amples.
In identifying a series with a definite integral, it should be
noted thit the definite integral

1 )dx hta + r,when nh = b a,

f J( =h0
may be expressed as

Lt b a j (a + = f
I-4 N N
J
In the special case when a 0, b I , we have h = I/n
 DIWINITh UfTEGRALS 119

Hence, in this case, we have

Lf !Ef(!) J!(x)dxr

(As if we writer for rn and dx for I / n. I

or, puttting h = I / n,L t h £f( rh) f( x ) dx.

= 5
I As if we write x for rh and dx for h . I
6.18. Illustrative Examples.
Ex. I. Evaluate Li _
Lt n+2
(n +I
++. . .. + I . (H. S. '88J
Dividing the numerator and denominator of each term of the above
series by. it, the given series becomes
(1 •l
= Lt S -+ + .....+

+.it 1+.
it it

S n
= Li - £ = L i )i L '--- [putting It = -
'I-.- it 7•
'I

= 1 ---dx
I -tx
=[log(1 log 2.

Ex. 2. Evaluate
2 4 6 2ii
2 \T
((+. )(i^.)(l+.) i.I +--iJ

Let A denote the given expression then

log -Lr +-4)

 120 • INThGRAL CALCULUS
Ex. VI(A)

I,
Li log A = 1.1 L 2 - log
7-1
1
2xlog(1 +x2)dr
=J 0
r2
log zdz, [putting I + = zJ
=j
2
= [zlogz - z] = 2 log 2 -1 = log
Since log Li A = Li log = log 4-C
Li A, i.e., the limit = -C
Ex. 3. Pro that +
Li
1 +2" +3 +

m+I
Left side
=
fl-+-.fl [( -
l) (2) 1-
N a
-) I
I, a
Li r)
N fl
z (
U
h ( rh T [where h
i_
Ir
1
-• ____
= x"dx-
J o -

EXAMPLES VI(A)
I. Find by the method of summation the values of:-
b
5
b
6) e dx. (ii) 5 e h dx.
5 X3 dx (iv)
5 (ax + b)dx.
Ex. VJA) IDEPINrFE INThGRALS • 121
b
(v)
f sin x dx. (vi)
f Cosa do .

(vii)$ x dx. (viii) $+ dx.

(ix) I sin nxdx. (x) I cosec'xdx

J o

Evaluate the following integrals (Ex. 2 to Lx. 12)

,1 _______
2 (I) x 3'I I + 3x 4 dx . (h t 'J 2ax - x' dx.
J0
dx ' dx
(iii)
J x(1 + Iogx)
I
I H. S. '851 (Ic)
I 0
(x 2 +1)'

3.
Jof (H. S. '801

4 (i)J sin - ' xdx. (ii)f tan xdx.

(iii)f( Cos -I x)2dx. (iv) f log ( I + 2x ) dr.

vfxtan -t X) 2 dx (vi) x2 I(4 - x 2 )dx.

sin mx sin nx dx. I J. E E/82 J

0

•
(ii) J cos mx cos nx dx (m, n being integers )
0
 122 INTEGRA L CA LCUluS Ex. V I(A )

(jIl) sin x sin 2x dx.

50
6. (i)5 sjn 2 nxdx. (ii)Jcosanxdx.
° (n being an integer)
I I
.0 xdx .. dx
7. (1)
J (l +x2 ) 5 (a 2 + X ) 3/2

(l$J (ax — x A(x - 1X 5 — x))

2

lx t 4z
8. () r x sin xdx. (ii) sec x dx.
.1 0 Jo

(iii)J (secO - tan G) dO

0
I
9. (i) Jo tan x dx. (ii) tan 2 x dx.
Jo

I P
I
10. (i)J cos 2x cos 3xdx. (ii)J sin 2 xcoOxdx. (H. S. '821
o 0
.1. 1
(iii)J x cos x cos 3x dx. (iv) J sec'O A.
a
11.(i) 5 x log r dx. (ii) x2 sin x dx. (Ii. S. '81 1

52
(iii) sin cos ( a 2 sin' 0 + b' cos' )do .
V1(A) 1 D€PlNtE INThGRALS
2
• dx __________
dx
12. ) ( Ii)
a2 + ,
J x(1+ 2x)'

(ill) Jodx (a>b>O).

a + b cos x

dx
i - a cos x + 7 (0 <
a <I )
(Iv) S
0

Show that ( Ex. 13 to Ex. 28 (ii) I:-

lug 2
-dx o
13,.-
$0 1+e'
.5
14 i9.9a =log( . ) log tab).
x
a
S
IS. I sin'
I +t
at = 2atan-' a— log (1 + a2).
(H.S.'851
2
16. (05 T(2 - x)dx =.

dx
(ii) $ x3TV?T+) = . log f H. S. '85 I

17 2
j(a2+xl)2 dx=-
j
sin xdx '
18. i+052=l+tan v
J,
C ___
19. cos'x.l,l sin xdx =*••
0
 124 TNTEGRAL ('AI.CLJIJIS I Lx. V1(A)

dx '1
20.
a
(1)5
2 cosx +b 2 slnx L
. ( a, b > OJ.

' sin 2 xcos 2 X1

I (sin'x + coax)" g
Jo

dx_
log 2.
zi ' I0 4+

- dx
tan '
5 f2 + 4 sin x

- dx
2. (l
+ 3 co g 4 tan-' 4.
50 5 x

dx
(II) =-'- log 3.
3 + 5 coo x
0

dx
(111)5
1 + 4cot2x
0

dx 0
23. 50 + cos 0 cos x

cos xdx
24. 1 0 (1 + sinx)(2 + sin x)logf
4* sin2.r
dx
25. 5 +

dx x a2 +
26. (1)5 (2 cos 2 x +b 2 sir. 'x)'
0
E. E. '88j
(Multiply numerator and denominator by ec 4 x; then put b tan x a £5210J
Er. V1(A) DBPINflE JNThGRALS J2

is
2 x sin x cos x __ _____
(g+b) .ta,b>OI
f
.0
(a' coB'x+b 2 eln'x)'

27
t'
f: 1 1 2
dx = e - -
J (i (Iogx )2) log
3
28.
2

1 (1 + x)(1+ 2x— x') =

29. Evaluate the following

1 2 11
(a) Lt -+ + I.
I-4 N+2M ,I+NINJ

ft
+ ......+
n 1
N-*-
[ I
2+12+nl+22 112 +
H. S. '86)

(Ut) LI- ) ( n2_22) •+(n2(n1)2}]•

II -•

1 1 11
jv) Lt F +..+ - . I.
11j
p -4 LRf l(4n2')
I Write n 'I( 2m 2 - 11 2 ) in the last term.

12 22
(v) Li
[ 11 P '2'
IC. P.'84 I
ri nl _____ 11
(vi) $-.- + ....+ — I .
LI
V n + (n + 1)3+ (n+2)'
.fl2.(ft_fl2
tvii) LI
ft . - ft2

(viii) Li £
II
r-t
 16 C1JU)S I Lx. VI(A )

1 2 !l '/
('x ) L i((1 + ;)(
1+ • (' +-.)} .

) i_((i +.)(i +-) ••e

ft

xt) n+r
Lt Z -
0

(xiI)
n !_

(xnt) U
n .....
ri n+l
- +I-
n
I
+2
+

[,H. S. '83, '88)

(xiv) Lt — .- [ i ++ .....+
3-] .IJ.E.F.'86)
(xv) L I i-
r d( + 1)s 4(m + 2) + ... +
1

r ,1-iirn
(xvt) LiI—
N- I. fl"
a
dx 4
eg
sintl
-
J (x +a)+'Ix J cos'e
0 _f0
find the value of a.
31. 11 a be positive and the positive value of the square root
is taken, show that
•1
Jdx
2 if a < 1;
'( I - 2ax + a' )
-1

-2 ifa>I.
a
32. If ?n and it are positive integers, show that

(i) sin mx sin lx

5 =f
Er VI(A ) I DEFINITE INGRALS 127

(ii) f sin mx cos ,ixdx = 0.

-x
a if m * n
(iii)J cos ,nx cos ,gxdx=
{ ifm=n.
- I

ANSWERS
1. Ci) (e -• e ) . (ii) ( - e )1k. (iii)
(iv) a $ I, (v) I . (vi) sin b -sin a. (vii) . (viii) 2.
(ix) (1 - cos,ia)/n. (x) I.
2. (U -. (ii) . xa . (iii) (iv) ( x + 2fl

3. 1. 4. 6)5 - I. (ii) . x - . 1og2. (in) x- 2. (iv) . log 3.

(v) I it ( 3 t - I) + . Iog2 (vi)-,r -,-43.
in(m -n)x sin( m + n),t
2(m — ii) 2(m + n) (ii) 0. (iii)!

(. (i) . x. 00-1 7t . 7. (i) 42 - 1 <" a2 '/2

(w) - X. 8. (i) I . (ii) log ('/2 + 1) . (iii) log 2.
9. (O . log2. (ii) I -r. 10. 6) 1 . (iU-. (iii)-(n-3).
a2 +ab+b2
(iv) 1. T (ii) n - 2. 3 a+b
x
12. (U
4a (ii) 1og-..
5 (iii) '/(a -b2) COS

GO 29. (I) log( 1 + rn) . (H) I x. (iiD--x. (iv) + x.

m
(v) jlog 2. (vi) . (vii) j. (viii) -}z + I . (ix) 4/e
(x) 2e 1/2 ( - i)• (X i) .-+
4l
o g2. (xii) ij x. (xii ) log 3
(xiv) 2. (X v) 1 42 -!3. (xvi) 1 30.
 123 DJFEG*AL CALCULUS.

6.19. General Properties of Definite Integrals.

b
(1) 5a
f(x)dx
=5a
Let f(x)dx =(x); ..Jf(x)dx =(b) -(a);

then, Jf(z)dz =$(z);

..5 f(z)dz =(b) -(a).
(ii)5a f(x)dx = - 5b f(x)dx.
Let =$(x);
5 (b)

and 4(b)J -$(b) -$(a)

-5b f(x)dx =— I(a)-

Thus, an interchange of She limits changes She sign of the integral.

(iii)Ja f(x)dx =Ja f(x)dx c

f(x)dx. (a<c<b).

Let
5 f(x)dx $(x);
..5 f(x)dx (b) -$(a).

RIghtide ($(c)- $(a)) + ($(b) — (c)) • •() -

Generalization.
b CA
f(x)dx f(x)Jr f(r)x +
Ja f "' +J
p C.
• f(x)ix +
Jell
f(x)dx,
when a < c1 < c2 < .....< c < b.
 DEFINITE INTEGRALS 129

(iv)f Je
f(x)dx
=J
f(a - x)dx.

Proof. Puta—xz,..dx—dz;
also when r 0, z a, and when x = a, a = 0.

nghtside = 0 z)dz =Jf(x)dx.

_S:I(z)dz 5
*12 *12 /2
Illustration : 5 sin x dx sin ( - x) dx
=f cos x
=5
na a
(v)f f(x)dx = n f f(x)dx, iff(x)=f(a + x)
J o Jo
I C. P. '86 1
Proof.
,2a
f(x)dx f(x)dx+ J f(x)dx+..+J. f
J0 =J 0 a

Put z + a = x , then dx = dz,

also when x = a, z = 0, and when x 2a, z = a;
• 24 ,a
f(x)dx =f
(z + a)dz f(a +
J 4 0 =J 0
a

=dof f(x)dz.
Similarly, it can be shown that

f(z)dx 2' f(x)dx f(x)dx;

J =J a 0

Inlegral Calculus (main) -11

 130 INTEGRAL CALCULUS

and so on. Thus • each of the integrals on the right side can be

shown to be equal to f ( x ) dx . Hence the result.

Illustration:

Sincesjn s z n'(,t + x) .•. sin'xdx =

Jo Jo
2a a a
NO I
Jo
f (x)dx
=5 f(x)dx +5 Ma - x)dx.
0 0

21
Proof.J f(x)dx
Proof. /(x)dx f(x)dx.
0 0 a
I By (iii)
Put x F 2a -z in the 2nd integral ; then dx - dz,
also when x = a,z = a; and when 2a, z =0.
the scccd tegral on the right side, viz.,
2.
f f (x )dx f f (2a - z)dz
=5 f(2a - z)dz.
BY GO I

- x)dx. tBy(i)j

Hence the result.

(vii)
5 f(x)dx = 2 f(x) dx, iff(2a -x) = f(x),

and
5a f(x)dx = 0, iff(2a -x) = -f(x).

4 These two results follow immediately from (Vi).

 DEFDE INThGRALS 131
Illustration
Srncesin(, - x) = sin x, and cog (n -x) - - Cos j,
ft*12
sin x dx = 2 sin x dx; and cos a dx 0
Jo Jo Jo

b '-- 2 Jf(inx)dx.
.10 0

and f( co a) dx = 0, If f (cc.s a) is an odd function of cos a.

Jo
r+1 a
(viii) j f(x)dx = f (1(x) + f(- x))dx.
-a

Proof. ff(x)dx=ffx)dx+ff(x)dx
Now, putting a = -

ofxdx i:- z)dz= f(-z)dz

=-

f0
Hence, the result follows.
Cor. If f(x)isanoddfuitctionofx, i.e., f(- a) -
p +1
f(it)dz 0,
J -a
andUf(r)i5cflfuflcsiotofx,j.e,f(.x)f(x),
a
f(x)dx - 2J
J -a 0
*32 iNTEGRA L CA LCU[i;

f
I p .11/2
9In'xdx =0. alf

I .11/2
sin'xdx =2 1
p/i

Jo
.SlflSx1Jx.

6.20. Ill ustrative Examples.

By the help of the above propertios of definite integrals we can
evaluate many definite integrals without evaluating the correspond.
Ing indefinite integrals, as shown in the following examples.
, 11/2
Lx. 1. SI,o, that log tin r dx 0. 1 H. S. '851
J0

'J log 11/2

p11/2
tan x)dr (ay(iv,Ar,.68J

f
11/2
0 log cot: dx =
g -• lo tan r dx
;
21 :

Ex. 2. Show that

f 0
/2
co x) dx (C. P.188

11/2 in

fsin (Tx,j +f cos())

dx

. 11/2
I '4C011Z
J0 cos x)

r '2 f!7.
+ o
 DEFINITE INThGRALS 133

1/2
rr-x + 1/2
1dx
1/2

=5 'I( sin x) + q( coo dx . lIz] 4 .

=)t.
1/2 1/2
*
Ex. 3. Show that log sin xd* -J log coo xd; log
-
- 0 0
C. II. '86

iogsinxdr log sin x)dx

J =J
1/2
= Jo
I logCos xdx.

By Art. 6.19, (iv))

1/2
21 =I log sin x dx +f log cos x dx
0

1/2 1/2
= ( log sin x + log cos x)dx =I log(sin x cos x)dx
5O JO

1/2 1/2
log
og 0_)dx (log sin 2x - log 2) dx
L 2 5

fn /2
log sin 2x dx - log 2
=
Put 2x =z dx=.'dz.
1/? 1
log sin 2x dx log sin zdz
J

= .flog sin xdx =f 0

/2
log sin zdx = I [By(z), Art. 6.l9,I

21 = 1- - log2; .. 1 = 2 log =log

 134 INTEGRAL CALCULUS

Ex. 4. Show that 101( l x)

= -Iog 2. .( C. II '61 & C. P. 'BSJ

Put x=WO; ..dx=seOOdO;alsowhnx...og.0.

and when x = I, e
) I
I log (I + ,ane)de log (I + tan(., - 0))dQ.
=J =J
F By A rt. 6.79. (iv) I
/E
Now, I + tan u -
V4
- 0 \a = 1 + 1- tan = 2
/ l+ tan 0 I +tanO

I
= 5 2
log + tan = f (log 2 - log (1 + tan 0)1 dO

21 =
5 log2dO -

t. log 2;
5
.. I = -
t:1 +
log 2.
tanO)dO it. log 2 -

*4
Ex. 5. Show that C
— dx = 0.
J -a I+ x

i =5 dx +
Xe
dx = Ii 4 '2 say.

Pu t t ing x =- z in the first integral,

a 2
C =1
11=
xc xc C xc
r1

pa 1+z2 dz-1p I+z 2 dz =-i --

p 1 +x
0 a

Hence the result.

6.21. Logarithmic and Exponential Functions.
The fundamental concepts of Calculus furnish a more adequate
theory of logarithmic and exponential functions than the met-hods
adopted in elementary books. There an exponential function is first
introduced, and then logarithm is defined as the inverse function;
 DINrFETNTFtRAIc 135

but in the treatment of these functions by the principles of Calculus,

logarithm is first defined by means of a definite integral, and then
exponential function is introduced as the inverse of logari-thm.
From the stand-point of these new definitions, certain impor-
tant inequalities and limits can be obtained more easily and
satisfactorily.
A. Logarithmic Function.
The natural lQgarithm log x is defined as

log x= (1)

where x is any positive number, i.e., x > 0

Thus, log x denotes the area under the curve y = I / t from
t = 1 tot = X.
From the definition it follows that log 1 = 0, and I since lit
is continuous for t > 0 1 from the fundamental theorem of Integral
Calculus it follows that log x is a continuous function and has a
derivative given by
d 1 . .. (2)
(logx) = .
Since the derivative is always positive, log x increases steadi-
ly with x (i.e., log x is a monotone increasing function).
Putting I = I / u in the integral for x, we get

di du -log 1
-=
log x=J z--=
j _
J
Putting t = yu I y = a fixed number > 0 I in the integral for
log( xy ), we get

TJ -Lm ;-f
XYdt du du
log (xy)=f
t/y I I

= logx_log(I/y)IOgX+l0gY ....(4)
 INTEGRAL CALCULUS

i this way, other well-known properties of logarithms can be

)ped.
ice log x is a continuous monotone function of x, having
ue 0 for x = 1, and tending to infinity as x increases, there
,e some number greater than I such that for this value of x
y e log x = I ,and this number is called e Thus e is defined
equation

loge = I, i.e. .r: = . - ... (5)

ponentia unction.
- y = log x en we write x = e . (6)
n this w ay exponential e Y is defined for all real values
In particuh' e I = 1 , since log I = 0. As y is a continuous
ion of x, x a continuous function of y.
x =e Y so that y = log x, and so

dx x dy
=1 I—
dx
=x=eY

i.e. , (e' ) = e . . . . (7)

More generally, (e -Y = ae Y .
dy
> 0) is defined as ell-$, , so that log a' = x log a
hus, 10' = e108I0
he inverse function of a 3' is called the logarithm to the base a.
h us, if x = a s', y = log. x.
Some Inequalities and Limits.
i) To prove 2 <e < 3.

log 2=5I<:<2; .. <l/t<1.

 uNITE INTEGRALS 137

5 a, i.e. • < I • i.e., <5 .. . .. 2 < e.

ç2
( dt di fdi l du du
J T = J TJ2T=J0TJ0r

(by putting t = 2 - is and = 2 + u)

du fdu . . ('di
= > i. e., > i. i.e., >j -r
4 4 J -T

3>e, i.e., e<.3

x
- 'clog (1 +x) < x(x >0).
(ii) To prove
l+x

From definition, log( I + x)

= 51 -
:.<t<1+x. :.1/(l+x)<1/t<L

I4X .i+X
iIXdjI
-I j di,
•'I

i.e. -----< log (1 1)<x.

1x
( iii) To prove Li- log (1 s x ) =1.

I IogO+z )
From (u), - -_________ -. < 1, and since I /( I x) and] both
1+x x
tend to I as x -+0, the required limit = 1.
a - I = log a..
iv) To prove x LI

................ log a, and that for X = 0 is log a, it

A .5 , i., .J'

follows, from the definition of the derivative for x 0, that

gS_g0
Li = log a
h

Putting x for h, the required result follows.

 138 LNTEGRAL CALCULUS

e' - I
When a = weget . Lt X
=
-40

xl
(u) To prove LI - I +- I =
-4i:
Since A log (1 + xl) = it follows that the derivative of log
I+ xl)for t = 0 is x. Hence, from the definition of the derivative for
r = 0 we get
Lt log(l +xh)
- x
h -
Putting h I /.we see that
I r\ I
ç _log t, J+-ç), t.e., ç LI log l+) =x.

Since the exponential (unction is continuous, it follows

LI (t + =

If we suppose ç —* through positive integral values only, the re-

quired result follows.

Putting x = I we get Lt ( = e.

(vi) To prove n( j - 1) = log

Lt
Since the derivative of e = e Y ,and that t,r y = U is I • we have, from
the definition of the derivative for y = 0
C C C -1
L h -.O h
Putting z / n for h where z is any arbitra number, and n ranges
over the sequence of positive integers, we get
( el/ll
Lt n = I , i.e., LI n( (e)-I)=z
Z j

Putting z = leg x, so that e x, the required result follows.

log x
vii) To prove LI = 0, when rz > 0
If 1>1 and >0, then t<t 0 1
 DF1N1TE INTEGRALS 139

('dl (
logx=J<J ,D_1
I dt.ie., < ,,.e,, < - forx >1.
Suppose a>

-Dforx>
--- I.
But (1 /x) —i O,as r--,since a>
Hence the result.
Note. Replacing x by n, where n is a positive integer,
Lt 2.8! = 0. when a> 0( n - through positive integral values).
viii) To prove Lt -- = 0 for all values of pi, however great.
From (vii) , x - $ log .- 0, when x - , 0
Putting a = I / 3 in the left side and raising it to the power a, we get
x' (Iogx)" —* 0, as x-e '. Now putting x = e ,so that log x =y,
the required result follows.
6.23. Two Important Definite Integrals.
A . If n be apositive integer,'

iz
cosx dx I C. P. '82)
Jo
n-ln-3 n-S
- n n-2 n-4 3[C P '84)
422'
or, n-1 n-3 n-S 42
is n-2 n-4 53
according as n is even or odd.
Proof. Jsinxdx =Jsjn-Ixsinxdx
= sin 'X.(- co s x) + ( n - 1)f sin - 2 x cos xdx
(integrating by parts)
* For other forms of these integrals see S 9.3.
140 INTEGRAt CALCULUS

= - sin" 'x cos x + (ii - 1)J Sin .2x(1 -Sin tx)dx

= - sin ,hxcosx+(n_1)J Sill , 2 xdX_(n_1)J sin" dx.
transposing - (n 1) Jsin" x dx to the left side and divid-
ing by n, we have
I' sin"-'x--
cos x (n- 1)1 2 x dx .()
I sin "x dx = --
1
+
fl
J J
52 L_)_ l 2,
"s in "xdx
n nJ
0 0

n-it
= --- sin" 2x di

}-icnce, denoting in "x dx by 1, we have

(2)

Changing n into it - 2. n 4, etc. successively, we have,

from (2),
fl- 3
fl - i
I 4; I_ 4
n.-5
- - 1,etc.

n ---I n - 3 71 - 5 3
-.-- 1 1
'°'

L! ^
n n- 2 n77 4 '3
according as n is even or odd.
42x
But i0 =J dx

Jw
cos 1 1.
and Ii
=
sin xdx
= [- J0
 I, p l (ND E IN 1EGRA LS 141

Thus, we gt the rquked value of

f 2 sin x dx.
U

Exactly in the same way it can be shown that

f20
cos x x dx
has precisely the same value as the above integral in either case, n
being even or od( 1.
Otherwise, it can be shown thus

f cos x Li r
=f cos n - x) dx
= 5 sin x dx.

Note. The student can easily detect the law of formation of the factors
in the above f.rrnul, noting that when the index is even, an additional fac-
tor t is written at the enI but when the index is odd, no factor involving
IS introduced. The formula (I) and (2) above are called Reduction For-

mula. (See Chapter IX. I

B.
50
s!n x cos ' x dx, rn , n being positive integers.
(C.P.'BsJ

5 sinxcos'xdx cos
n-- i lcos-xsinxsrn
t
= cos''x -------
ri + I
+ --- ..- i 'dx
in P I

[integrating by parts and noting sin x cos x dx =sinm

-+x
5 +iI
n—i1. COS 2x) COS . -2 xdx
,n+I 7n+iJi sinx(1
n'" 1 xcos - 'x
= - + U - I tI sinxcos
m+1 m+I j 2 xdx

- U
I Isinxcosxdx.
nt+l j
'ee Chapter IX, Art. 9.15.
 142 INTEGRAL CALCULUS

Hence, transposing and dividing by we have

I Sinxcosxdx
SI fl I XCOS IX n—IC
= 4---IS1flXCOs'2xdx
m+nJ (I)

(2
sin xcos ' xdx = rsln"'.'xcos.ix
72
I
j 0 '.
0

+ I
m +n j
0

n—I a
M + nji sin 'xcos' 2 xdx. . . . (2)
0

Again, writing f sin'xcCXX Jin. . x(cosxsinx)dx

and integrating by parts and proceeding as above, we get

I. . sin_-ix cos , .•Ix

I sin 'x cos -x dx
j rn+n

+ rn — It.
in + njI $In2xcosxdx
and hence taking it between the limits 0 and -in, we get

I siri'x cos 'xdx in — II sin 2 xcosxdx ....(3)

0
rn+nj
0

Thus, denoting sin x cos ' x dx by I., , we have from

J0
(2) and (3)
n—I
Jm,n = in +
rn — I . ... (4)
rn+
 DEFI1]TE 1N7EGRAL 143

r..i
Again, since SiflXCOs'xdx
Jo

rihc
= sin' ( . n - x)cos' (- - x)dx
Jo

rIt
=J 0
Sifl'xcos"xdx,

=1n (5)
By means of the formulae (2) and (3), either index can be
reduced hy 2, and by repetitions of this process we can, since m
and n are positive integers, make the original integral. viz.,
I,
depend upon one in which the indices are I or 0. The result, there-
fore, finally involves one or other of the following integrals:

5 sin x cos xdx -iJ dx=

} .. (6)
f sinxdr = 1;
J oJo
cos xdx = If
Thus, finally we have

sin mx cos x dx
J0 5 0
cos ' x sin ' x dx

- l.3.S ... (m - 1).1.3.5 (n — 1) n

2'
when both m and n are even integers; and
- 2.4.6...(m —1)
(n + I)(n + 3)...(n + in)
when one of the two indices, say m, is an odd integer.
By (iv) of Art 6.19,
 144 [1'4TEGRAL CALCULUS

jilt

J sin" (4n - x)cos ( .Ix - x)d

0

Ft

=1 0
(I)

From (2) of Art. 6.10(B), we get

n-I
= -- . I_, -
M
m+n
.1
1 (2) ...
I,._,,,, by (1)
= m +
If n is an even integer we can deduce from the (.r t result of
(2)'5y integration

n - I n-3
t•
rn+nmi-n-2""
- (ii - 1)(n - 3)...3.I
- (rn + n)(rn + n - 2)...(m + 2)
1.3.5...(n- I)
= - i sirtxdx.
(rn + n)(rn + n-2) ... (m+
The result now follows from (A) of § 6.23.
Note 1. The above definite Integrals are of great use in the application
of Integral Calculus to practical problems e.g., In the determination of
centre of gravity, in the calculation of area, etc. and also many elementary
definite integrals on suitable substitution reduce to one or other of the above
forms, as shown in the following examples.

6.24. Illustrative Examples.

Ex. 1. Evwluatef x' 4(1 )dx.

Put x = sin 0; .. dx = cos edO and I =COs

-X' 0;

also when x =0, 0 = 0, and when = 1, 0

Er VI(B) DEFINITE INGl 45

The given integral then reduces to

1.3.5.1 x 5r
I sin'Ocos t OdO = 14.6.8 1 =
J 0

Ex. 2. Evaluate I x 1 (1- x)1dx.

Jo
Put x =ain 1 0; .. dx = 2 sin 0 cos 8 dO
and when x =0,1, we have 0 0, fit respectively.

I=2 f 4 16
sin t $cos 4 OdO = 2 2. = -
0 o

Ex. 3. Evaluate Cos 'xdx.

0
Since cos ' x = -cos-(x-x) when it is odd.
and = cos (x - x) when it is ever,
by Art. 6.8 (vii), it follows that I = 0 when it is odd,

and I = 2 P x cos"x dx, when it iseven

n-i
=2
it n-2 n-4 -2-1--.(By
422
A rt. 623(n)1

EXAMPLES VI(B)
Show that

1. + b - x)dx f(x)dx.
=f
(ii)J::f(x+ c)dx
=f!x.
mb
Gii) f(nx)dx =.;. ff(x)dx.
f
Inlegral Calculus (main) -12
 146 flBoR4L CA LCULUS Ex. VI(B)

2 sin x
2. dx= . C. P.'861

r 2 CO5X—jfl
3. . dx ..
J I + srnxcosx

4.5 U
(acosx +bsln'x)dx 3w(a + b).[C.P.'851

S. sin 2x log tan x dx 0.

J 0

6. xf(sinx)dx f (sin x)dx.

J o 0

. j x log sin x dx = . log-.. I C. P. '75

r .
8. I xsrn•xcos 2 xdx =
Jo

f x sinx dx n2
9.

•
f sin4x dx
. o sin x
0

II. 5x.1a2 -x 2 dx= 0.

2*
12. sin' xcos . xdx = 0.
1
Jo
Lx. VI(B) DEFINTE D11BGRALS 147

•1
13. J log sin (4 * 0 ) do = log". tPulFio
0
I
logx W I
14.
5 I -x ) 2 dx
= log -2
(Pu g x sin 0)

15. 1
so
log (1 + tan 0)dO . log C. P. '76 , '83 1

16. fxcos 4 xdx =

Ix
17.(i)5 cos'xdx =n. (ii) 5 in'xdx 128

(C.P. '82)
52 sin 4
(iii) Ocos'O dO
- 2048

(iv)f sln
4 x cos 'xdx=
315
J o

'C
(v)j (1 Cos x)'dx--
0
+ 57

p.'
(vi)J sin 3 xcosxdx = 0. [H. S.'80J
0

,.II/2
(vii) J cos 1 0 dO = 0. (viii)
o J -2/2 sjn'x dx = 0.
I
18.(i) I x 3 (1- x)dx
Jo
 14 INTEGRAL CALCULUS Ex. Vf(8)

(ii)J x 3 (1-' x1 )dx =

0
a
x1 3i
(iii)I 0
i(a 2 _2)
dx

(iv) j .i - x 'd'
=
5

dx ----= . ( Put x=i I an O1

19. $0X2

r* x sin x I 1
20. (i) I _t•- L II. '75, J . E. E. 89 I
J0 I +cos 4

sin
(ii) + 1).

I -------
(iii) J dx ! ,t(x-2). IC.H.19641
sec tan
0

(iv) f!---.- 2N
jsin x + co

(v) (1 + 1og(42 - 1)1.

, (I-x+ r')dx=ln-log2.
(vi)Jcut
0

(vii) J 0
a 2 sin
xdx
+b 2 cosx - 2ab
a,b >0).
 DEFINiTE INTEGRALS 149
Ex. V1(l)

xdx n
log 2.
(viii) $ 1 + cos 2x + sin 2x =
0
_2 -
dx = log + 1).
(ix)J 0
(2x' - a1 )2

vd 7T1(al +b1)
os+bn2x)l = 4a
.', j 0 (1.E.E.'88)

I
=5
21. If I tan "0 dO • show that I,. = .- - I •

Hence find the value of tan 'xdx.

22. Show that, if m and n are positive and m is an integer,

1 1

5 i" -1
0
(1 - x)"'' dx
=5 0
x"' (1 - x)'' dx

1.2.3 .....(m - 1)
=

A NSW ERS

21.
 CHAPTER VII
INFINITE (OR IMPROPER) INTEGRALS AND INTEGRATION
OF INFINITE SERIES

7.1. Infinite integrals.

In discussing definite integrals we have hitherto supposed that
the range of integration is finite and the intcgrand is continuous in
the range. If in an integral either the range is infinite or the in-
tegrand has an infinite discontinuty in the range (i.e., the intcgrand

ly
tends to infinity at some points of the range), the integral is usual-
called an Infinite Integral, and by some Writers an
Improper Integ
ral. Simple cases of infinite integrals occur in elementary problems;
for example, in the problem of finding the area between a plane
curve and its asymptote. We give below the definitions of infinite
integrals in different cases.
(A) Infinite range.

(I) ft (x ) dx is defined as U f f (x) dx,

provided f( x) is integrable in (a, c), and this limit exists.

(ii)5 f(x)dxis defined as £—

i
f(x)dx,
providedf( x) is integrable in (c, b), and this limit exists.

(iii) If the infinite integrals

1:! x ) dx and
5 f( x ) dx

both exist, we say that

5 .f( x) dx exists, and

f(x)dx=ff(x)ax+ff(x)dx
i::
 INFINITE INTEGRALS ETC.

Not,. In the ab4v cases, when the limit tends to a finite number, ie
integral is said to be convergent, when It tends to infinity with a fixed s
it * said to be divergent, and when It does not tend to any fixed limit. Ii.
or infinite, it is said to be oscillatory. When an integral is divergent or
dilatory, some writers say that the integral does not exist or the integral h
meaning. (See Ex. 2. § 7.2 I
(B) Integrand infinitely discontinuous at a point.
(I) If f( x) is infinitely discontinuous only at the end point
i.e.,if f(x)-9aS x —a,then

LI d,c > 0,
ja,cf

provided f( x) be integrable in (a + c , b) and this limit exists.

(ii) i f ( x ) is infinitely discontinuous only at the end point
i.e.,iff(x)—"as x -+b,then
b b- E
f(x)dxisdefinedas Lt f(x)dx,E > 0,
f 5
provided f( x) be integrable in (a. b - c) and this limit exists
(iii) If f( x ) is infinitely discontinuous only at an internal point
(a < c< L'),i.e.,iff(x)—*oas x —,c,thcn
fb

—* JI
c-e

I a f(x)dx =, Lt I
fJJ a
f(x)dx + £ ,Lt
O
c*C
f(

when E --+0 and c' - 0 independently.

Note. It Sometimes happens that no definite limit exists when c
' tend to zero independently, but that a limit exists when £ = '-' (See E 7.
Art. 7.2. 1 When c c' , the value of the limit on the right side, when it ex
ists, is called the principal value of the improper integral and is very often

denoted by PJ f(x)dx.
a
 152 LVIIGRAL CALCULUS

(iv) If a and b are both points of infinite discontinuity,

then $ f(x)dx is defined as f(x)dx f(x)dx

these two i ntegrals exist, as defined above,
+5 when

c being a point between

a and b.

7.2. Illustrative Examples.

Ex. 1. Evaluate t dx.

'C
I I e- dx = Li (1C C ) =
Li-, J
= C .-. I
0

Ex. 2. Evaluate Jcosixdr.

0

I= Lt- j cos ix dx = Lt
(-4 (-3 1 but this limit does not exist.

I lence the iniea1 does not exisl.

dx
Ex. 3. Evaluate

a
L+
dx dx
- a
a
x2
f I+

dx
Ja I + x 2
= Li
c —.-
dx
1 +x 2 =5 €
Li (tan-'a -tan-'t)

dx
= Li
I ' ____
dx
= tan - 'a +,t;

1 +x2 €' -. - Li ''

(tan-tan-' a)
Ja= _____ a = C

=n - tan 'a -
Although this TigraI does not exist in the manner defined above, it if,
expressed in terms of Dirac's delta function [8(i)] in modern mathematics
Detailed discussion is outside the scope of this book.
 INFINITh INTEGRALS ETC. 153

I = (tan'a +it) + ( -I n-tan -l a) It.

Ex. 4. Evaluate f dx
Jo •
Here tends to as x tends to + 0.

•1 dx
dx Lt .11 O 3(1 -c"3 ) = 3.
I x2'3t -sO X u/ 3 c
£

dx
Ex S. Evaluate
5 -1 TT

Here —* as x -sO. an interior point of the interval (-L1).

dx dx
-j+Jj.
1J -i 0

-i dx = U ( - I )
Jt p
Now, 5LI
0

this limit does not exist. soj does not exist

Similarly, .Lx does not exist.

Note. In examples of this type usually a mistake is committed in this

way

Since 5.dx=_1 =-2,

[-n r:
which is wrong.
In this connection, it should be carefully noted that the relation

f ( x ) dx = F (b) — F (a ) cannot be used without special examination

J a
154 INTEGRAL CALCULUS
unless F' (x) = f for all values of x from a to b, both Inclusive.
Here, since the relation A (- ) fails to have any meaning
dx\ X X2

when x = 0, and I) is a value between - I and + i, we cannot direct-

ly apply the Fundamental Theorem of Integral Calculus to evaluate this
definite integral.

Ex. 6. showiltatf e-" cos bxdx= b1 a>O.

f oe_1 cos bx dx
= - ( - b sin bx)]C I Art. 3.3j

= 2 (e -l c ( — acosb+bsink)—(—g)).

I e" cos bxdr = Lt I e" cos bxdx

J0 J0

= r —.. [a2-
+b2 - lc (—acosbt + bslnbc)+ ai j
Now,£ Lt e(— acosbt + sinbc) = 0.
Since e - ' —+ 0 andcos bE and sin bc arc bounded. I
I
I e'cosbxdx = a
i0 a2+b2

Er. 7. Evaluate j• d
x
The itegrand here is undefined for r = 0.
1 i
V
-c
-- dx = LI I - + , Lt I . dx
£0
—*
J X £0J
-9 , I

= Lt [ log( _')1_ 1 + LI[ log x]

= Lt log e- Lt loge' = Lt log -

as c and e' tend to zero independently.
 IWO

INFINITE INTEGRALS ETC. 155

But this limit is not definite, since it depends upon the ratio a : c'.
which may be anything, £ and a' being both arbitrary positive numbers.

+ dx
But if we put a = a', we get
f 1 - =
log I = 0.

Thus, although the general value of the integral does not exist, its prin-
cipal value exists.

Note. .!. , when the range of integration is such that is neg-

5
dz
ative throughout, may be written, by putting z = - x, as
f ie
= [iogz} ' = [Iog(_ x)} ,forlog x is imaginary here.

Ex. S. Evaluate
5 (1 +x2)2
dx.

Put x = tan 0. .. dx = sec 20 dO ; as x increases from 0 to , 0 in-

creases from 0 tow.

a
'=1
tan 2 0 sec 2 OdO
sec4O 'f sin' OdO...in

Note. Thus, sometimes an infinite integral can be transformed into an or-

dinary definite integal by a suitable substitution. But whenever a substitution
is used to evaluate an infinite integral, we must see that the transformation
Is legitimate.

Er. 9. Show that

5 e x"
-' dx = it !, it being a positive integer.

Let I, denote the given integral.

'C
I, = Lt I exdx
C-.- J
0
156 INTEGRAL CALCULUS

= Lt ii- e' xI + ii e'x"' dx

C — .— LL .J 0 J 0
I Integrating by parts)
PC
= it LI I ex " 1 dx , since Lt e.&' -0.
C—.—
0

I See Das & Mukherjees Differential Calculus,

UT. 1i(,i flsE A u?7ITh, Mffl flu. Z liii). A

= nI =n (n - 1 ) I,-2 (as before )

= ,i(n -1)(n - 2)Ia-3 , = etc.

= n(n -1 )(n - 2) ... 2.1 f e' dx

Jo

= it ! ,since f e -, dx = 1. See Ex. I above)

Jo

7.3. The integral $ e2dx.

Since e- 1 2 ( = Ile ,' ) is positive and < I +xt (for x > 0)

x increases monotonically with X
it follows that
J e = 2dx

x 2 1
x dx
and i e- • i.e., < tan-'X,
dx < i
J j° 1 +12
o f See , 6.7.1
This being true for all positive values of X, however large,
and as tan t X increases with X and -4 as X -4 o , it follows
x monotonically increases with X and is boun-
that e - ' 2 dx
5
ded above.
 INFINITE INTEGRALS EIC. 157

Thus, the infinite integral j e' dx is convergent.

Denote it by I.
Now, a being any positive number, replace x by ax.

Then, 1 ae"dx.

ie2 =Jof ae 2 (1 - - 2)dx.

Since ae - ' " > is a continuous function for all positive

values of x and a (which are independent), assuming the validity
of integration under an integral sign in this case

if e2da
=J
(Jae2(1+ ') da) dx. .. (1)

Also for any particular value of x ae 2(1 *da

,f
F_i21 I C1(1 +,2)l 1
2(1+x2)L[
_e2(12)
L J0
I
2(1 + x t ) as
1 2.dx
Hence from (I), 12
=
I it
=-1--,or, I

i.e., e ' dx = -}
For an alternative proof see Chapter IX, Art. 8.21.
 158 INTEGRAL CALCULUS

7.4. The integral- sin bx dx.

0

— e sin bx
• dx,a>O.
50
x
Assuming the validity of differentiation under the integral
sign, we have

=5e-i cos brdx

b2
a > 0. 1 See Ex. 6 Art. 7.2. J

Now, integrating with respect to b

f dli 1 b b
u=a
a 2 ^b la=a—tan-'-4C=tan-'—+Cfl)
a a
where C is the constant of Integration.
From the given integral, we see that when b 0, u = 0.
from (1), we deduce C = 0.

t e-' sin bx Li
I dx = tan-' - . ... (2)
j 0 x a

Assuming u a continuous function of a, we deduce from (2),

when a — ,0,

( slnbx x
dx = , or - ... (3)
0
according as b > or < 0.
Cor. When b = I we have

(4)
 INFINITE INTEORAL.S ETC. 159

Note. There are other method, of obtaining the result. Student, may
consult any text-book on Mathematical Analysis.
7.5. Integration of Infinite Series.
We have proved in Art. 1.4 that the integral of the sum of a
finite number of terms Is equal to the sum of the Integrals of these
terms. Now, the question arises whether this principle can be ex-
tended to the case when the number of terms is not finite. In other
words is it always permissible to integrate an infinite series term
by term ? It Is beyond the scope of an elementary treatise like this
to investigate the conditions under which an infinite series can
properly be Integrated term by term. We should merely state the
theorem that applies to most of the series that are ordinarily met
with in elementary mathematics. For a fuller discussion, students
may consult any text-book on Mathematical Analysis.
Theorem A power series can be integrated term by term throughout
any interval of convergence, but not necessarily extending to the end-
points of the interval.
Thus, if f ( x ) can be expanded in a convergent infinite power
series for all values of x in a certain continuous range, viz.,
f(x) = + a 1 x +n 2 x + .... . to —,
S
then ff(x)ix.J(ao+ MIX +a2 x 2 . ..... ) dx
a a
S
zf a,x'dx,
I

or,
5 X
f(x)dx s:(ao + M i x +a1 x 2 + ..... ) dx

a,x' dx,
a
provided the intervals (a, b ) and (a, x) lie within the interval o
convergence of the power series.
160 INTEGRAL CALCULUS Lx V I

E. Find by integration the series for tan x

=1_x2+X4_X+ ..... tO.,ifx2<1.
I 1+
integrating both sides between the limits 0 and x,
JX
dx
- + x 1 - r4 +
J:T2
tan x=x_x 3 +4x 5_ +r'+ .....-I<x<1.

EXAMPLES VII
Evaluate, when possible, th2 following integrals

(i) dx xdx
1 f I+x2 x +4
Jo J

dx f
2. (I) f (ii) " dx.
-
2

r dx
3. (i)f-- (i i)J

sin xdx I dx
4. (ii)
frI cos 2x a
+ Cos X

2 dx
dx
S. (ii)
WI' x(1+
x) 2-x

'1 + x dx dx
•;
(i)f 1-x (ii) J0 (1 + x2 )4

4*
xdx xdx
7
(i)1 (1+x2)2 .(II)S 74 I
Ex. VII INPINITh INTEGRALS M. 161

dx dx
S. (i) j
(I - X)' 01) f
j (X + l)(x + 2)

Show that (Ex.9 to Ex.22)

dx x
9
J: (x'+ a')(x' +b2) = 2ab(a +7 ) .(a,b>OJ
xix.
10
f 0

II x'dx
ta,b>O J
f: (x' +a )(x' + ti' ) 2(a + b)

12.
L
e- (cos x - sin x )dx 0. (ii)J : dx 0

(ii) Divide the range (0, -) into two parts (0, 1) and (I

13 + dx

14. I
x log xdx = - (1),(fl>-1).
5
15 - - sin bxdx b
• Jo e
a' + b'
(a > 0).

16 dx dx
• (i)J o x1+2x cos e+1 25
0
x2+2xcosO+1

(ii)
+(1 +x' )} n' — I
where n is an integer greater than one.

inlegral Célculus (main) -13

 162 INM rRAL CLCt!Lt:3 Ex. VII

xdx 11
(th)J rr x)(l^z) 4

( . )J naxco..x 0, or -1
17. dx

accordirg as a >, < 0! = .• (a and b bc:g supposed

OsitVe ).

sin 2z cos J._x(sinx +Cosx)ld

C.
(u)5 +

18. sLflXd
f0 x 4

f
1'.. i !±-!j3
J0 x 16

( sinmx
20. dx.m,or_Tm
f
according as m>, or < 0.

21.

r 'nx'3 3n
22. dx—
JI0(s-)
x S

23. Find by integration the power series for the following:.

Ci> log (1 + x ) ; (ii) log( 1 – x ) ; (iii) sin 'I
Show that :• (Ex.24 t. Ex26)
dx
• C -------, ----r i sinX 1.3 sin4x
24. 0) j 2 'sin x •i-
+ 1-4 - +

• f dx x I x 51.3 x' ;(x2 <1).

"1>Ji-i
 Lx, VII D'9NTfl INTEGRA E11. 163

J -"i-
xsinx
dx x x
x-
(iii) +

b
442 +
(iv)
5 -dx log + (b— a)+

(v) 52 sin 2 d where e 2 < 1,

- f /\21 /)3\,4
2 11 2) 2.4) 3

dx
(vi) -j---------._-. , where k 2 < I

{ i. +(f ).k 4
2 )2 4 ,... I

(vii) + I.
J 1 ' x 3 5

dx
(viii) •o = 1 1
+ 2.4.5.2' +

-
2. ------dx = -
(— 1)
(2n +
0

26. (i)I
J i+x dx =
[ WeE
n'1. !! 1
6J

dx= -
 .INTEGRAL CALCULUS Ex. V II
164

27. (I) Show that if, a > 0,

•1
I I I
a --a+I
-- dx =
J i +x
- +4+2a+3+

Hence deduce the value of the series 1 3 + - - - +

(ii) Show-that, if a > 0, b 0,
- _____
I+x b
dx a a+b+ a+2ta+3b
J o
28. Show that

x 2 Plog (I+x)dx -21;711

2 [h+4••
J
( integrate by perU. I
ANSWERS
(II) does not exist 2. (i) !log
3. (11)3
1. (1) 3 X.
3. (I) principal value is . (ii) principal value is 0
4. (i) does not exist. (ii) does not exist.
6. (I) X . ( ii).fr-
5. (1) log 2. (ii) does not exist
S. (i) does not exist. (ii) log 2.
7. (I) 3 .(ii) 0
23. (1) z-3x 2 •3x 3 x4 + ........ . x<I

log + ...... r 2
--•i +--..... < 1

x5+ .... 27. (i) log 2.

I x 11.3 --
(ill)' + 2
 CHAPTER VIII
IRRATIONAL FUNCTIONS

8.1. In the previous chapters we have discussed simple cases

of integrals of irrational functions. We shall now consider here
some harder types of such integrals.
8.2. If the integrand contains only fractional powers of x,
i.e., if the integrand be of the form

Fx)
where F( z) is a rational function of z,
the substitution is x =
where n is the least common multiple of the denominators of the
fractional exponents of x. I See Ex. I of Examples V III. I
8.3. If the integrand contains only fractional powers
of (a + bx), i.e., if the intcgrand be of the form

F((a + bx)),
where F( z) is a rational function of z,
the substitution is a + bx = z'
where n is the least common multiple of the denominators of the
fractional exponents of ( a + bx) .L See Lx. 2 & 3 of Examples V III. I

I
8.4. Let the integral be of the form
x ( . a + bx • )P dx,
where m, n, p are rational numbers.
(A ) If p be a positive integer, expand (a + bx ) P by the
Binomial Theorem and integrate term by term.
I See Lx. 4(1) of Examples V III.)
(B) If p be a fraction, say, equal to r/ s, where r and a are in-
tegers and s is positive,
166 INTEGRAL CALCULUS

m + 1 = an Integer or zero,
Case I. If
the substitution Is a + bx =
m + I * an integer or zero, we apply the following
If
I
Case IT.
If m + 1 + • an integer or zero,
Case II.
the e-,al substitution Is a + bx' = z'x
If, however, the integer is positive or zero,
alternative subslitutiott is a + bx A =V
If the integer is negative,
the alternative substitution is ax- 2 + b =
which is practically the same as (1) of Case II, sometimes facilitates
the calculation. I See Lx. 2 of A rt. 8 .8. 1
dx
8.3. The integral of the form I 1 +d
+ b)I(cx_

Here the substitution is cx' + d = x 2 z'

Sometimes trigonometrical substitutions like
x = k tan 0, x = k sin 0 ,.x = k sec 9, etc. facilitate
integration.
I See Er. 28 (ii) of Examples 11 (A ) and
Lx. 80) and Er. 8(ii) of Ex-
amples V 111 .1
8.6. The integral of the form
dx
f
J (px 2 + qx + r)'I( SK I + bx + c)

Here we shall consider two cases only.

breaks up into two linear factors of the
Caje I. If px + qx + r
forms (mx + n) and (,n'x + n'), then we resolve
I/( mx t n )( inx + n)) into two partial fractions and the integral
then transforms into the sum (or difference) of two integrals of the
 IRRA TIONA L FLJNCIIONS 161

type (B) of Art. 2.8. (See Ex. 13 of Examples VIII. I

Case IL !Ipx' + qx-+ nsa perfect square, say, (Ix + rn) 2 , then
the substitution is lx + m = liz
In some cases trigonometrical substitutions, as in Art. 8.5, are ef-
fective.
If q = 0 b = 0 the integral reduces to the form given in the
Art. 8.5.
In all these cases, the general substitution is
I iax + bx + c
.I I I=z.
'4 \px2+qx+r/

Briefly, we have considered integrals of the type

dx

where. P and Q are both linear functions of x and P linear, Q

quadratic I See Art. 2.8(A) and 2.8(B). I and P quadratic, Q quad-
ratic. I See Art. P.5 and 8.6. 1
If P be quad -'it and Q linear, put Q = z
Also we have considered integrals of the type
C 1(x)
pQdX .

where f ( x ) is a polynomial, P, Q being linear or quadratic. I See Ex.

11 to 15 of Examples VIII,
8.7. The integral o f the form

f 1(x) _d
X
+ 2bx + cx +2bx +
where f ( x ) is a rational function of x
The denominator can be written as

.j((x +_).)+v(x+)+c}
nd hence the substitution is
 16$ DTBORAL CALCULUS

x + - z or x - - - a
a a
according as f ( x) is expressible In the form
I 1 I 1 I l /
or

If b - 0, the substitution

a' + = z or x' - =a
x la'
is sometimes useful. I See Lx. 19 of Examples VIII. J
S.S. Illustrative Examples.
C
Ex. i. IntcgrsleJ x34(I
dx +x' )
Comparing it with the form of Art. 8.4, we find here
m .-3,ii .3,,=-i,s =3.
si+1
Now, # an integer, but

i,,+I r
- I, (an lnteger). ... (I)
+
by Art .8.4, Case It, we put I + x 3 zx3

X3 (z' - 1) 1 • X ... (2)

(z' I )/3
dx 1)l/3 dz . ... (3)
(3

denominator =x4:

C * (1 +x' )3/3
ziz - - 1z , -T
• -- j X2

Alternatively, since (1) is a negative Integer, we can put

+I= a1

dx
Thus, I ,, .jx 4 (x' + l) 1 /3 dx.
'
( i( +-))
 IRRATIONAL FUNCflONS 169

Since x' + 1 = z .. - x' dx z2 dz.

i = _Jz a 2 dz= etc.

dx
Ex 2 2x + 1)4(x 2 - I
InteRrdIeJ (x2-

It is of the form Case II of Art. 8.6.

j
dx
=J (x- 1) 2 ((X -1)' + 2)

dz 2) ,putting Z — X
5 z 2 '(z +
It is of the form of Art. 8.5.
• S
'J2sec2Od8
2tan' O.J2secO ,
putting a='2tanO,

coseccot 0 d = -.cosce.
=-5 '(z +2),
I z, cosec 0
Since tan O =
I 'J(z + 2)_ —1- 4-2x
W + 3)
2 a 2 x-1
Ex. 3. Integrate the following
X - I dx. J _____ dx.
X +
)J (iTi) 4 '
id
J
x4+1
x• (•• x4+I

•1
(j) J - X
dx (dividing the numerator and denominator by at
= J a 1 +-a
I 1
(, l +)dx
at =j2?2 [onputting x_ a)
= 1, 1\l
J (\ x- — +2
a,1
I xl - I
= . tan y=tafl I x '/2 )

(Ii) It is similar to (I)

170
INTEGRAL CALCULUS

1_1

x 2
dz
=ir On Putting

I
= 2V log z-'12
log
------ I
=m x2 + 1-x'12
2 + +

(Iii) I - I f ( __ + I)- (r 2 - I)
X. +
- dx

i .:' + dx --If x 2 - I
x4 + I
_____
= tan -' (x 2 - I
_____ - + I- xV2
log

by (1) and (ii)

Ex. 4. intesraeef I X dx
xx2+x.

_1
(I __)dx

(1-_)dr
J (x++!)2

_•__,•_Z:__
f [putting x+Jzi

f( cosec 0 cot 0
= • cot 0 (putting z = cosec 01

= 540 9 COS4 = cosec - ( 2_!

)sin
-1 ri )
 IRRATIONAL FUNCTIONS 171
Lx. VIII
EXAMPLES VIII
Integrate the following :-
1+ c
dx. (Put xz4.I
T(1 +x)
J
dx
2 Ir
.1
^ 2TT4TX+2)
dx
(Put x+2=Z.I

3. jYTTT2+x)I3

(i)$x(1+X)2dx. 60J(2+X)dX.
4.
p_____________
dx
5. (i)J -j- 1-53-2dx. x A (2 +X2)Vl

f(l .+x ' dx

6. (i) j dx. (ii)
J
(iii) ('1(1 dx.
J X

P 1( xdx.
- x2 'Ixl( I - 2x) dx.
7.(I)
J -- J
dx dx
S TixTTi-)
8.
W TTx2+4)
x 2 tan 0; x = 3 sec 9 1
I Put (I) = (U)

P xdx dx
___________

J (x-I ) ( r + 2) J.(x2 )3/2 (x -

x' )) dx
10. (')
f '1(1 x 2 ) dx.
+x +- r (x +1(a 2 +
TT
j

C dx
+1) (iI+1)
11. j (
dx
12.

dx
13. 5 T9)(31+2
 172 INTEGRAL CALCULUS Lx. VIII

x+3
14. J(x2+5x+7).J(x+2) dx.
dx (xt+4x.s-4)dx
15. 1 J (x ' +5x+7 ) .J( x +2 y(ii)J
(x2+5x+7)J(x+2)
12+1
16 . (i)41
+
dx. (ii) f
x2 1 dx.
X, + 12 + I

f + dx.
17. ,

18. $ X , + xt + 1dx.

19.
f 1i-:(1_x2)dx
x )'I( x + 1)
(1 + x2)dx
20. J (1 - 12 )'I(l -3x +x4
(x2
21. jr xt(x 2 +x- 1 - I) X,
x2-x-2
22. f x(x-' -x 2 )3/2 dx.
f 1+x-2
23. J'J(12 +x.1_flI1

24 Integrate
5
by the substitution z = x
dx
r(x2_x+2)'
+ 'I( x 2- x + 2)
and show that the value is -y
I Jog I(x2_x+2)+x_42
I(x - x + 2) + x

25. Integrate dx
$ +2X- 1) ' by the substitution
Zx + 4W t 2x - I) and show that the value is
2 tan -' (X + 1(x2 + 2x - I)).
 IRRATIONAL FUNCrIONS 173

ANSWERS
1. 41tan (,lx)+Iog(1 + 4x)l.
2.2(x+ 2)1/2 — 4(r + 2)" + 4 log (1 +(x+ 2)1/4).
-- x 1' +
3. 2 tan' (2 + x) 212 . 4. (I) 3 x" + 11

(ii)
^ (2+x)"

________
+ 2
(2 +4x)2/2.
1 (2 +x )3/2 I.
—
1 (2+ x2 )I/l
. 12 4 x
+ J)
xn )(n- 1)/n
1 (1 +z3 )4/3 1 (1 +
6 (I) — 4 (Ii)

[ Iog sxl 41 .x 4 ) ]
(iii) - .4(1 +x 4 ) X2

)3/2
2 (X -x2 )3/1 2 (1- 4 (1 -2x)211
7.
- T 3 xs12 -

/ x43 4(x -9))

S. (t) - tan -' (..( x (ii) tan - (
________

+ 4))
4(x+ 2)-43
(x+ 2) 112 - 2(r + 2)" + 3 log 1(x+ 2 _) __+_73

I (-;).
1O.(i) 4(1 +x +X ) sinh •' ( 2x 3 )-
+ I __________

(31x))

(ii) Ix + 4(1 2 + x 2 )J
N

11. 4(x • 1)+ ,ih1x_.Sjflh

4(42 +4z + 5)
12
8 2x+1
cosh (5 +35
13. -sec' (2x + 3) + -3 1
x + 3)

tan
14.-j 1 4'1(z +
174 INTtGRAL CAICUI US

15.(i) tan -i( zil) - I

+ 2) -log
2 2)
(ii) 2 'I( x + 2) - ( x+i
tan -'

X - I (jj)1 log '

tan-' -- '• +
- Z

17. ta -' ( xi3 ) I

- log _

18 an ( 2 -x + I
2-1 log x
x3 )

Is /x'12 \
20. sin-1 (-•) 21.
) x
22. . 23. qlnh -'
2 )•
 MISCELLANEOUS EXAMPLES I
. Integrate the following functions with respect to x
x 2 +c03 sinx
(I) -.2 eorec
x 2 X. (ii) ( H.S. '84, '87 I
+1 (

1
(W) cog 8x -cos " x .
+ 2 cos 5x
3
tan - tan
(iv) tan a + tan x

(v) sec 7 x cow '• (vi) x 3 ( log x)2

Ii) sec x log ( sec x + -tan x) (viii) x cos x
(ix) sec x tan x + tan2x) (x) x Cos x.
(XI) (log x ) . (xii) tan -, 04). (xiii) log (1 + x 2 )
(xlv) x 2 g ln -'x (xv) 2' cos x. (xvi) e' x4
Integrate the fôflowtAg
x - 1V X2
di
2• ('
(i)JItX + 1 ,' (ii)J (1 +
log (1 + x) dx. sin ( log x) dx.
5 00 yS

4
• (1)5 (e
dx
+e')1
- (ii)5 dx
(1 + e' )(1 + e')

5. (i)f (a + x)'Ja'+xdx (11)5 (a 2 +x t )'a

dx _______ dx
6
•(i)J (1 +x2)I(i_x2) (ii)5 (1 - x'l(1 + x1)

7
•(j)I
dx
(x2 - 4)'I(x-1)
- (ii)J x
dx
( 2 +1) .t(x 2 + 4)
- dx
(ii) 5 dx -
S
•(i)J x2(1 + x2)'
dx
+ 1)
dx
9
(i)J (ii)
5
176 INTEGRA... CALCULUS Misc. I

10. (1) J'X3 4(x dx fix

a - 1) x(x + 1)2

11.
(i)f x+ 1
x 1 (x - )dX f
+ 1) dx.
X3

iI
12 ( ) 1+ sin x
dx.
(ji)fr2+ sin 2x dx,
1 + cos 2x
eltlfl-tx
(i)J
13
(1
xtan-'x
+x2)312
dx
(li)J (1 + l ) 3/2 ix.

14. (I) I dx (ii) f (a sin x dx+ b cos )2

% j I + cos2x ,

15. (05 ;;-;dx (11)5 sin (2 tan -' 'q' 1 -x ,Ix

+)
16 (i)J Ixdx dx
(x + I)(x +2)
(11)5
• x (x - 1)(x + 1)
x2dx dx
17. (i)J(x _I )2( + 1)
• (11)5 x'I(x
+ x - 6)
18 G) sin ax
• sinx + tanx .60 3 cossin
x-dx
5
x + 2 sin x
19
(i)S e - "
(x - I )(x
dx, (ii)
5 x 'I( 5x 2
3x2-2x-3
dx
- 4x

20.(•) f (x -
- 2)(x -3)
dx
'f(x-1)(x-2)(x)
d

21.
(i)J 4 dx
+ 18x + 81
- (ii)f (x dx
1 + 2x + 5)2

dx
22. (1)5 (1
—+ X
)312
+ (1 +
I ii) 54x J(2 + 2)dx.

Evaluate the following

3 1
23(i)J x1(I () I x' sin3xdx.
0 Jo
 Misc.! MISCELLANEOUS EXAMPLES 1 177

-t
24. (1)1 xlog(1 +4x)dx. (U) I Iog(1 + cosx)dx.
J o. Jo

25 (i)
f o
- dx
(1+xt)2
0
x dx.
1+x

26
•
(i)J 1
dx
x(1 +.T2)
0
xdx
(1 + x)(1 + x1)

_____dx
27
•
(i)fo 1—x+x2
dx
__________
(Ii)5(12),J(12).
0

28
•
f(i) x2 - 1
(x2 + 2
1) dx . (if) f
x
x4 +
t + 1
I
dx.
1 '1
Show that

dx
29 = 0.288 (nearly).
Jt I + x)(2 +x)
•1
dx
30.
fJ x+'I(a I
0
x2)4*
.1
dx
31. L2 _ 1)(3 - x))

32 • J 0
a
a 2 - x'
•3 + x2
- dx= (1 I A )a2 .1 Put x 2 a 2 cos 20.

I
33
f 0
dx
3+2sln x+ cosxl

34
Ia f-
lb dx
1ei b2 e--
. tan
a
I-iI paral Calculus (main) -14
11$ INTEGRAL CALCULUS misc. I

1)
+log?.
dx
33
J -log( X+

M. If C., C , C 2. C denote the coefficients In the ex-

pansion of (1 + x) where n is a positive integer, show that
C0 C1 C . C. 2 ^ 1- I
..........
1 2 3 n+1

ANSWERS
1. cotx i. tan - 'x). (II)r cos a+ sin a log sin (x-a).
Vii) isin 3x - . sin 2x tv) sin 2a log sin (x a) - x cos 2a.
lv) cot 1/1 (vi) . x' ((logx)' -jIogx +..J.
(vll)(Iog(secx + tan x)) 2. (vili) ( x 3 -6x)sinx +3(x' -2 )cos..
OX)jsec x'I(l + sec 2 1) + -} log ( sec x*

(x) -r sin 3x + uLcos 3x .+ ir sin X + 1 cos x.

(xi) x(1 3 .3l2 +61- 6), when I = log r.
(xil) (x + 1) tan -' ('/x) -'/x.
(xiii) zkg(I +x 2 ) - 2x + 2 tan x.
sinx s j'J( 1 - r ) ..( I - z2 )2.
2' I
(xv)
4fl + (log2)21 cos I x - cot (log?))
-

(xvi) e ' ( z 4 - ix 3 + 12r 2 - 24x + 24).

I I+
2. (i) tan 'x + +A 2 (ii) - 4(1 ,,2)2

3. (i)logr - ( + .) log (14x).(ii)_.22(cos log r+2 sin log x)

4 (j) 12 ( I + e 2' )' . (u) (I +e'

. 1)(2x 2 +3a+2a 2 )4(7 +x 2 )+a 3 log (x+ 4(2 +x' )J.

(j) (x + a) 2 ( 15x 2 - 12ax +43o2 ).

 MISCELLANEOUS EXAMPLES 1 179

'1(1 +r) $ z'12

6. (I) v2 tan' (ii)
log '1(1 7 z 1 -
) - x '12

7. (I)
1 2 'l(x 1) - x43 - (ii)
I
tan ( x'13
log '1( x 2 - 1) + 43 x + 4) )
2+3x 2 3 I
_______
8. U) - z')2 tanx. (it).5. (I) log
2x(1 + +
2x2 --1 , l(x2 I)
(II) (x 2 + 1). 10. (j) sec 1 +
2r1
x X
(ii) log--
• l'x
- J+x

+x 2 )
+6x 2x-1 .(li)logUr+'1(x' +i))'1________
L+ 3L + 2 log -

12. (i) x ( t i x - sec x) + log( 1 + sin x > (it)e' .in x

x - tan x (a +
3 > VT + x 2) (ii) ( I + a I M 1 + X 2 _)

14. () 11 sec x + log ( sec x . tan x (ii)

- Cosx
(a sin x + b Co s x)

1 1 + sin
15.(i) I log 1I -.+'12
'12 sIn x
• - log
j----
(iI){x'1(1_x2)_Co5X}.

16. (I) 242 tan-' - 2 tan -' 4r.

x I I I
(1) log-1 + ^ .tanx.

17. (i) ! log (x -1) 4 log(x 1 • 1).

ri 1)

(ii)\j. cog' q 2(x+3)

5x

IL (0logtan-x - tan' jX.

(ii) -(2x - 3log(3cox + 2sinx)}.
19. (I) e - ( . + 1) log ( i $ 1) . (it) sinh
(- )
ISO INT1GRAL CALCUf (IS

20. Ci) x + 2 (log (x -2) - Iog(x -3)).

log (;c
(II) 9 log (x- 3)-.- SIog(r - 2)- 1).
—
21. (1) ( tan' x +

r+1 ri-I
- tan 2-+ 8 + 2z + 5)
2 2 i- xI(2 + x ) — 2
22. (i) 2 tan '(1 + x )1 . ( ü) 3(x + '1(2 + x2))

23. Ci) j-t (ii)1

27 - U
. 24. (1) - — 2 log f) . ( i n log 1-
25. (i). (ii)i/v2. 26. (i)- log 2. (ii) - ii.
2. cot 2.
27. (1) 13. 00 . 28. (j) --L(ii)
 CHAPTER IX

INTEGRATION BY SUCCESSIVE REDUCTION

AND BETA AND GAMMA FUNCTIONS

9.1. Reduction Formulae.

It has been mentioned in Art. 1.6 that, in some cases of integra-
tion, we take recourse to the method of successive reduction of the
integrand which mostly depends on the repeated application of in-
tegration by parts. This is specially the case when the Integrands
are complicated in nature and depend on certain parameter or
paramtturs These parameters may be positive, negative, negative
+42)./2,
or fractional indices, as for example, x "C", tan" x, (x 2
sin x cos x • etc. To obtain a complete integral of these
trigonometric or algebraic functions, we first of all define these in-
tegrals by the letters 1, 1. U , etc., introducing the parameter or
parameters as suffixes, and connect them with certain similar other
integral or integrals whose suffixes are lower than that of the
original integral. Then by repeatedly changing the value of the suf-
fixes, the original integral can be made to rest on much simpler in-
tegrals. This last integral can be easily evaluated and knowing the
value of this last integral, by the process of repeated substitution,
the value of the original integral can be found out. The formula in
which a certain integral involving some parameters is connected
with some integrals of lower order is called a Reduction Formula. In
most of the cases the reduction formula is obtained by the process
of integration by parts. Of course, in some cases the method of dif-
ferentiation (See § 9.19 below) or other special devices are adopted
(See § 9.20). In the next few pages methods of finding the reduction
formula of certain integrals are dlscl!sscd
Case I. Integrals involving one parameter.

9.2. Obtain a reduction formula for x' e" dx

5
Let 1=5x_ e'dx. ... (1)
 182 INTEGRAL CALCULUS
Integrating by parts.

I e
x e'5 dx= x— n
X . e" dx, (2)

or, .1 = xe" n (3)

a -
Note 1. It may be observed that the integral on the right-hand side of
(2) is of tfe same form as the Integral in (I) except for the power of x, which
is,' - I ,and which can be obtained from (I) replacing iby i - I on both
aides. If n be a positive irtng' ,: cevcly"" ""
finally depend upon to =Ie'-' dx e'/a, and Is thus known.
Note 2. In evaluating (3) from (U we could Integrate r first, but L- that
case I, would have been connected with 1, ., i.e, with an Integral whose
suffix is greater than that of the original one, which is not usually desirable.
A little practice will enable the students to choose the right function.
9.3. Obtain reduction formula for

(L)
5 sin xdx;5 sin xdx.
(II)
5 co." x dx;f cosx dx.

(i) As in Article 6.23A(I) of the book,

I =fain'.xdx

- sin' - ' x . COS x

n + ,t - ' fsin-2 xdx.

• I - sin 'XCOSX n–I

II
+ ... ( 1)
is the required reduction formula.
Also by (I), taking limits of integration from 0 to -L ,
 I]4TtGRA71ON BY SUCCESSIVE REDUCTION 183

J.JsInhxdx=iJa_z(hI>1) ...(2)

Similarly,
COS :,mn x + It - I '_••
(lj ) i. = Jcosxix n.' (C.P.'861

and j. = cos 'x dx = I. - (,s > I). ... (4)

Note. If the integrand be sinh ux or cosh I x a similar process may be
adopted.
9.4. Obtain reduction formula for

(i)Jtan 'x dx ; ( ii) 5 tan x dx. C. P. '89

( n , a positive integer)
x. tan 1 x dx
5 tan x dx = Jtan
2
Here, 1.
= 5tan , 2 x.(sec'x - 1 d
= 5tan" 2 x. sec 2xdX _J tan ' 2xdx

= tan— x
-
Thus,
=
tan,-Ix —i2. ... (1)
n—I
Also, taking limits from 0 to ii,

_l4 W4 #14
r x 2 xdx by (I)
- fJo tanxdx = tan--' I
a -Ja tan
I ... ( 2)
1.-a .
=
Note 1. If n be a positive integer.
 184 UnW RA L CA LCULUS

tan"xdx - tanx tan s-3 x tan'-5x

J - ti — I - +
If n be odd, the last term js ( - 1) )/
'J tan x
- I) (I)/2log,ec x.

If n be even, the last term is( - I)(- - 2 )/ 2f tan 2 x dx

-J tanx - x).
Note 2. If the thtegrands be cot- x, tanh ' x, coth 'x the same process
may be adopted.
9.5. Obtain a reduction formula for

I. .fsecxdx =fsec'-2 x. sec'xdx.

5 sec" x dx.

Integrating by parts,
I. = sec-2x. taflx_ 5(n_2)sec . - 3 xsec x.tan x tan
xdx
= Sec -2 x tan x—(n - 2)5 see - x( see 1 x_ 1)dx

= S ec -2 xtanx_(n_2) [f see . xdx _J See .-2xd1J

Transposing and simplifying,

15=
sec'2X tan x n - 2
n—I + n_I 12. •.. (1)
Note. U the Integrands are cosec r. sech -x, cosech x then proceed-
ing as above we can get the reduction formula for each of them.
9.6. Obtain a reduction formula for e" cos ' x dx.
Let 1 = e cosxdx. 5

Integrating by parts,
ecosx nt e_
a + aj
-I COS'X.Slflxdx
 INTEGRATION BY SUCCESSIVE REDUCTION lBS

e' cosx P1
rS
= + - I - cos-' x. sin x - - Ie"
a a La

x n - 1 ) cos" - 2 x (- sin x) .sin x+ cos - x.cos x} dx]

((it

ne"
= - COS ' x +- cos'' x. sin x
a a'

_$e{ (n_1)cos'2x(cos2x_1)+ COS x)dx

- e" cos''x(acosx + n sin x)

a2

-- [nf c'- cos"xdx - (n_1)5e cos2xdx].

Transposing,

e" COS , -'x(a

COS
( n1) x+n sin x)
1+— I,.= +
a' a' a2

e"cos 1 x(acosx +n sin x) n(n-1)

+ —I_
n'+a'

9.7. Obtain a reduction formula for j ( x' + a2 )" dx.

Let I (x' +a' )" dx.

f
Integrating by parts (taking I as the second factor),

= x(x ' + a 2 ) -5 n(x ' + a 2 )-I .2x.xdx

x(x' +a' )_2nJ (x' +a' )- -1 (x ? +a' -a )dx

186 INTFGRM. CALCULUS

= x(x 2 + a )' - 2n (x 2 + 0) dx

+ 2na2J(x i + a')' dx.

Transposing,

( + 2n) I, = x( z' + a ) + 2na2L-i

- x(x 1 +
2n + I
a) ^ 2n
2na
+
1 lr.i

Note. It may be noted that here ,i need not be an integer.

Put n = and compare with § 3.5 (C).
9.5. Obtain a reduction formula for ax 2 + bx + c ) dx
5(
Let 1=5(axz+bx+C)1x

If a be positive,

1. = a' (z 2 ± k ) dx, where z =x +

5
=ac4 - b (1)
and k2 4a3.. .
and if a be negative, say, = -a',

(a')' (k 2 -z 2 )' dx,

5 b4a'c+b'
z x - , and k ... (2)
where = = 4a'

But (1) and (2) are similar to that of § 9.7 above, and can be
evaluated by the same process.

9.9. Obtain a reduction formula for

dx
dx
Let I ,then I,-i J(X2 +a')'
 INTEGRATION BY SUCCESSIVE REDUCUON 187

Integrating by parts.
I
- ( n - 1).2x.x dx
(1 2 + at )..-1 -J (x' +a 2 )
'x2+a2-a'
+ 2(n - dx
(1 2 +a2 l)J (x + a2) 

x +2(n-I)l_ - 2(n-1)4 2 1
= X? 2).

Traisposing.
I
2(n - I )a 1 2n - 3)I-i
 (1 2 +a' ).-1 + (
=

I x 2n -3
i.e., I. = 2(n - I)a 2 (x2 + a 2( n - I)a'
dx
9.10. Obtain a reduction formula for
5 (ax z + bx + c )
dx (I)
Let I.
bx + c)
i ( ax 2 +

If a be positive,
I f dz
1. ;;j (z' ±k 1 )n

4ac - b 2 ..
where z=x+ b , k2
- 4a'
. ( 2)

and if a be negative, say, = - a'

dz
- r-J

4a'c + (3)
where z x- and kl 4a'
Both (2) and (3) can be integrated by the same process as in
§ 9.9 above.
188 INTEGRAL CALCULUS

Note. In Article 5.1, Case IV of the book, we have remarked that when
the iritegrand is a rational fraction in which the denominator contains fac-
tors real, quadratic but some repeated, in general a reduction formula is req-
uired. Thus, to integrate such functions, separate repeated and non-repeated
quadratic factors and for repeated quadratic factors, use the result of the
above Article.
x dx
9.11. Obtain a reduction formula for J( ax I + bx + c)
j
where is is any positive integer.
xdx
Let
1'J1( +bx+c)
- 2ax + b - b
Noting that x - 2a

b ________ dx.
1.xaJ (2ax +b)
'-t dx
2a$\T(ax2+bx+c)
—

N0wJ Jax(2ax +b)X jdx

2 +bx + c)

= 2a +bx +c . x _J2(n_1)x 2 4Y T1 4bx+cdx

(ax2 +bx+c) dx
= 2x I\x__________
2+x +c_2(f l_1)J \I(aX 2+bx+C)

= 2x al.
-.Jaxi+bx+c_2(n-fl1 + CI,, ? 1.
n-I r +cI,,2
I,, =_.Tax2+bx+c__-_ [aIm + cI,,i
a a
b
---I,,

-1
=— + bx + c- (is .- 1)!,,
a
(2n - flb (is- 1)c
a-? I,,
- --
 INTIiC,RATON BY SUCCESSIVE REDUCFION 189
Transposing and simplifying.

I=— 'Jax+bx+c (2n

-
-lan1)b 1 ._ I - (n -an1)c In-i
an

Case II. Reduction formula involving two parameters.

9.12. Obtain a reduction formula for $ x ( log x ) dx
n, a positive integer).
Here, since two parameters m, n are involved, we shall define
the integral by the symbol In,
=Jx ( log r)n dx.
Integrating by parts,
X
I, m+1 .x " dx
logx)-_--Jn (logx)11
x

-
t.logx) - " $x"( log x'dx
m-
(logx)* n
m+1
i.e., l . '.=m- - (logx ) - 211n+ I
Note 1. Here we have connected 1...,. with I. , - i and by succes-
sive change the power of log r can be reduced to zero, i.e.. after n opera-
tion, we shall get a term in, • i.e., Jr - dx, which is easily Integrable.
Thus, by step by step substitution, I.. . can be evaluated. It may be noted
that when two parameters are involved this is the usual practice.
Note 2. Students must be cautious in defining these integrals. Here, as
for illustration, In,.. * I.
9.13. Obtain reduction formulz for

(1)
J
(a + bz)M
____
dx; (ll)5 x m (a +
dx
bx)'
 190 1NTEG(AL CALCULUS

t(a + bx) dx. (n # 11

0) Lot i x
J

Integrating by parts.

= (a+ bx)' + —
Mb f (a+ bx)'' dx.
It - I X.
— (71 - I )x'

(a + bx)m mb
(1)
= - (i* - 1)x"1 + n-

dx
(ii) Let I. , . (a + Fr

Integrating by parts,
I nb f dx
(m_1)x 1 (a+bx) ni-I) x' (a +bx)"'

I
= - (rn - I)x (a +

- n (a + bx) -a dx ... (2)

1) x' (a +
MX-

I
- Thx'T+bxi
- 71 an
+
rn-I rn-I

an 1 rn+n- I
(rn - 1)x (a + bx)a+ rn

Changing n to 71 -1 on both sides.

I in+n-2
Im a(n_1)xm (a+bx)' +

Note. Formula (2) or (3) can be taken as the reduction formula for (ii).
(3) Is more rapidly converging. The other ways in which these integrals can
be expressed are left to the students. s See also S 2.2, Er. 9 .1
 1?EGRA7I0N BY SUCCESSIVE REDUCTION 191

9.14. Obtain reduction formula for

(Of xM(1 -x) ix; (ii) 5 a x" (1 O N dx.

(I) Let!.,. =Jxm (1 -x) dx

+ .(1- x)-' dx
m+1 m+ 1j
r
m,1 mi-lj
- x)" i r
m+i
Transposing and simplifying,
x' •1 (1- x)' n
ni+n+I +Im1. m+n+1

(ii) I1J, x-( 1 x) dx, by above, this

5

L in+n+1 J m+n+1'''
0

+ 1J_...-t.
Note. In Integral Calculus J.. Is usually denoted as P., the first
tulerwn integral. It is also referred to as the Beta-function.
[See § 911 below.
It is interesting to note that J =
Le., a..... = A.'. although 1 , . *
9.15. Obtain reduction form ukr for
(i) 1,=5.inxcoaxdx;
I
(ii) j_ sin " x cos 'x dx
J
(m , n being positive integers.
192. ThITEGRAL CALCULUS

iii itria.ieu. •

and obtained
S jfl ,N•I XCOS ,II X n-I
1_,. = + m +n
?n+fl
sin"xcos'x m-I in a ;nilar
or = - +
m+n
J,,... 2
rn + t
way, and when m and n are positive integers.
n—i rn—I
m+n 2 m+
Using § 6.19 (i"), we also see that J... = J, .

9.16. Obtain a reduction formula for dx. I n ^ t

5
Lt i... fsin"xcos-"xax. .. (fl

Consider 1'..q fsinPxcosxdx

p+q p+q
(byS9.lS above l
Changing q to q + 2,
5j1F4IO$*IX g+I
.*•pq2.rP.
p+q+2

Transposing,
- SXC X p sq + 2
r r pq.2.'.. ( 2)
-- q+I
Iq 1* 01
Now replace p by in and q by - is in (2) and use the defini
tion (I).
Then,. (2) becomes
I sin m—n+2
I_, . =fl - I cos'x - n —1
 INTECKAflON U Y SUCCESSIVE REDUCTION

f x
9.17. Obtain a y cdu1ior formula for cos . [n^11
J
r. dx --
Let I_
= j -
Consider, as before,

1'pç
=J sin hx cog xdX

- -
pi-g + 2
+ !p.q.2
q+I q+1
I as in § 916 (2) above I
Replacing p by - m and q by - n and using the def. of !.,
1 1 +
m+n-2
Im , = n - 1 stn"xco gx n-I
9.18. Obtain a reduction formula for

i=Jcos"xcosnxdx
connecting with (I) I_ ,. (ii) 1 ,, (m n
(i) Let
= J Cos "'x Cos nxdx

cos-x. sin nx m
sinx) sin nxdx
(I)
Since sin nx sin x = co g (n - flx - cog nx Cos x,
COSXSIflflX m
= Cos (n — l)x
+ - •cos" x.(
n nj
- cos flX COS x ) dx
Cos 'x sin nx 7fl F -
=M + -;

Inleqral Calculus (main) -15

94 INTEGRAL CALCULUS

Simplifying,
Cos x sin nxm
+
=m+i +
(ii) From (1),
COS'X sin flX In
+— (cc "'x sin x). sin ,xdx.
Again integrating by parts.

Jr.,.. COS x sin nx r

COS' sin x cos ux
+m
It n..l --
fl

+ if ((m-1 )COS -2x(-sjnx)siiix

• Cos -tx. Cos x) Cos nxdx]

COS 'x sin nx m( Cos ' -1 x Cos nx sin x)

-
f((m_i)cos-2x(cos2x -1)

• cosMx)cosnxdx
- cos 1 x(nsjnnxcosx - mcosnxsjfl.x)
nz

+ mf ((m -1 + 1) cos'x cos nx

-(m- 1)cos'2xcosnx)dx
cos x ( n sin nx cos x - m cos nx sin x)
n
+ - (m - 1)1..-2,r. ]
Transposing and dividing,
fl Sin I1X COS X - 11% Cos flX Sin X
= COS IX

mm-1)
- n 2 - m2
 [NTtGRA TIOJ'I BY SUCCESSIVE REDUCTION 195

There arc three other integrals of a similar type.

(i)fcos . x sin nxdx, (ii)J sin xcos nxdx
and (Iii)Jsin x sin nx dx,
which can be treated in a similar manner, and connected by a red uc-
ion formula cithi with 1.. - or with 1,., 2 . In each case.
Foi instance,
(?n + P1 )Jco5"'x sin nxdx = - cos 'x cos nx +m
(n 1 -. rn 2 )$ sin "x cos i*dx

= (nsinnxsinx + mcosnxCosx)sjfl-'x
- m(m - ; etc.
Case III. Special devices.
9.19. Obtain a reduction formula for dx
dx
5 (a + b cos x)
Let I
5
(a + bcosx)
Consider P = sin I(1)
(a +b cos x)"-1
dP
dx

- ((a + bcosx)-112
- cosx(a+ bcosx) + (n -1) b(I— cosx)
(a + bcosx)'
(n - 1)b + acosx— (,i - 2)bcosx
(a + b Cos x)"
- A + 8(a .bcosx) + C(a + bcosx)3
...(2) (say).
- ('
a + bcosx )
 96 \ ITGRAL c,\:.( L

Then comparin g thc corIhcjcnts

A+Ha 4- Ca - I ) I' U 'Call a c! ' 01 2 )b
Solving

A -(n-1)-, (2n - J),C - (3)

substituting those val uec of ABC in 2 we git

d - ( - 1)1 a - ) I
dx b a4b _ cos t
(2 pi )an 2
-.

1, tc,rating both sde with rcs 'cct to x , and using the de in!-
tioi4 of 11

( n 1)a /,2 )
( 2 - .3) a ii - 2

g ill x
- ( - I t( a - b (a b cos x)

(2n -3a ( n--2)

I., 2
+ (h (n-1)(a-b)1"

A !ter,iati'Jr ,nct.z -!

I..vt I' = ---------------and

---I V -. a -'co-
(a. )

V-a
COS I
=

.Jf' dIx=
' f in I COS 1 si ( ( - b sin .v
- 1)

V- a (n -f lb f -•a
= WnT + v
(n - 1)(u 2 ..j,2) a 2 -3) (n -2)
- -
by" -4- -
 t41 t(,J Tft). H'' st?('( '.vr REDt(IJ')N

Intrting both sides w r t. 1 and ung tht' definition

C dx
I }t result lollows

Note. When n , apositivc itog pc.$ d ap1hcation l the

abovc tJito,ri IOrniiIj, I. will tj!tirnaIcv dcl,tid or j ehta ho. Ov
intrabIe '-r P 6 4 2

9.20. Otauz rCduL lionmulj' ar x a + bx ) P dx.

5
In this integral, usually denoted as b ine y n ial diJt rrnli215, t rer
,nctQrs ate rivo!ved and this intcgtil, written a

I'x li can be conraa'a ted iIt at N , ON

5
of h-' lntgr.i

LI) ' dx

(Iv) 17 , , • r (a4

i - )P d

(Si) t., . J - a- h

Ii) I., r )- h i - t:.atiig b:

- --

ía) 'Ia" )1 _i.;

I -
Il - a .n' I
198 INTEGRAL CALCULUS

Again, as above,

= + )P
fll+ I (a bx"
nbptx
bx' + ?JX )p-1 dx
- m + I jr (a +

[writingx ..s =-x .'(a +bx -a)].

Transposing and simplifying,

x 1 (a + bx" )r anp
+ .(2)
,ip +m -+I np+ ,n +
Changing p to p + i in (2) and transposing, we get a connec-
tion with the integral (iii). VZ
x "' (a + bx' )p.1
= an(p-s. I)
, (p + 1) + m + I
+ an (p + I)
Also changing m to m- n and p top + i in (I) and transpos-
ing, we get
x' (a + bx 4 )p-.l
nb(p + I)
M- n + I
- nb(p + 1) LP.l ... (4)

To get a connection with and I. ,,, write

X. = _!_ nbx
nb )

= _i. 1 X 111 .( fbx ).nb.xhIi dx.

,*b J

Integrating by parts and simplifying.

 INTEGRATION BY SUCCESSIVE REDUCTION 19c

- XI (a +bx' )F.I
- b( "p +m + 1)
a(m - ,i + 1)
- b(np + ,n + 1)
Changing m to m + n in (5) and transposing.
x' (a + bx )p*I
- a(m + 1)
b(np + m + fl +1),
... (6)
a ( m + 1)
These six formulae of L,., can be obtained by another method
Write P =x"' (a +bx' )U
where )and p are the smaller indices of x and (a + bx ) respoc
tively in the two expressions whose integrals are to be connected.
Findand express it as a linear combination of the two in-
tegrands. On integration the result can be obtained.
To illustrate the above statements we shall find a connection of
with
Ikvidcntly X =m, pp. .. P,:X* I (a+bx' )p.1
dP
- =(m+1)x(a+bx)P+I+(p+1)x- *.nbxI(a +bx

=(m+1)x(a+bx)P.(a+bx')
fl
+ n b ( p + x" (a +bx")F
=(m+I)ax(a +bx)'
+ b(np +n+m+1)x"(a +bx)F
Integrating with respect to x,
P = (m+ 1 1 4 +b(np + n +ivi +
- x(a+bx")P' b(np+n+m+1)
- - a( m + I ) - a ( m + I)
which is the same as (6).
 20) LN1EGRAL CALcUI US

Similarly, the other five results can be obtained.

For another illustration see sum no. 7, § 922.
.21. Beta and Gamma functions.
In many problems in the applications of Integral Calculus, the
USC of the Beta and Gamma functions often facilitates calculations.
So we give below an account of those functions —their definitions
and important properties, some of which are, however, men-
tioned without any proof.
Definitions

(A)5 x' -1 (1– x)"-' dx denctcdhyB(m,n)

0 j?n>0,n>O)
is called the First Eulerian integral or Beta function.

(B) e- x -1 dx denoted b y r ( n)
n > 0]
1C.P.'84,'81
is called the Second Eulcnan integra l or Gamma function.
Here m and n are positive but they need not be integers.
Properties
(i) By property (iv) of Art. 6.19, we get

(1 - i)-I dx 1xi (1 - )- 1 dx.

fO .10
B(m,n) B(n,m).

(ii) r (1) e • dx 1. See Ex.1 of A rt. 7.2.

* Results (v), (vi) and (vii) are given without any proof here The procfs are
based on "double integration" which i s treated in chapter 20 ot the
present book. Nevertheless, the results are extremely important in
applications and are to be carefully rcniembercd.
 INFE(;RATlo BY SU('CESSIVE RE1)UCTION .')I

(iii) As in Ex. 9, Illustrative Examples Art. 7.2, it can be shown

that even when n is not a posit;'r ir.rr,

dx = nJc lC.P.'SO

r(n+1)=nF(i).

When ii is a positive iptr,

r( n + 1)= n! IC.P.85,'BSl

(iv) Writing kx for . in (U) iy get

kxn ldL_.L!) .lk>O,n>OIICP631

(v) B(m,ri) IC.II.'86)

I ( m +

(v:) in)r1.-ni).-(O<in< 1).

sin mit

('ii) PIh.; r '--. (vii I C. II. '86

V ( ) I () = = it.
sIn 11

r() C.P'821
Alternatively, we can dcdt' thc v3ltI' of I in the follow-
ing way.

Putting in -= it = - in (v)

2 f1' )2dx
1(1)

I on puttli

Il : i
20 1)7EGRAL CALCULUS

Hence the result.

(viii) B(m,n) (1 x'-'dx - xi'-' dx

=f + x )"' -
0 -f (1 + x)'
9.22. Standard Integrals.
• L, r(2-J) r(.59-!)
(1)5 sin PO.co 3 q J =
o
21
g+2 [q>-il

Left side
=5 0
(sinO )Il (1 - sin O 2 )q12 do

4 • I

f' (1-x)1dx
1

= -

Ion putting sin 0

=- +1 'g +
(p
2 1) = Right side by (v).
B
2
(Compare § 6.23 B. I
r(P)
(2) 5 0
sin0O fo cosOdO= - ________
r(P2)

The proof is similar to (1). 1 Compare § 6.23 A. I

(3)5 e-'dx =.'/,i. C. P. '83J

0

f C -4 - 1
Left side = -1 dz I on putting x 3 = zI

=.}r(-)by(I3)., by(vii),I
Compare Art. 7.3J
 INTEGRA11ON BY SUCCESSIVE REDUCTION 203

9.23. illustrative Examples.

Ex. I. Obtain a reduction formula for tan x dx and hence or
5
otherwise find the values of (1) tan x dx; (ii) Jtan* x dx.
5
From § 9.4 formula (1), 1.. =J tan xdx -
- _tx 1_2

(I) Is =Jtan s xax =tanx -Is,

= tan I x - where I 5 tan x dx = log sec x.

=
1=.tanhr_tan1x+Iogsecx.
(ii) 14 = tan 'x 14 ; 14 = tan ) x -12
tan 5x
I = —.--- - lo, where Jo = 5 dx = x.

14
tan'x Ian 3 rtaflX
I
[Compare § 9.4., Note I in these two cases

Ex. 2. Obtain a reduction formula for sec' x dx.

5
Hence find the values of (i)5secx dx. (ii) 5 sec' x dx.

From § 9.5,I, =f sec xax sec 2 xtanx + n -2

-
n-I

sec 4 ztanx 4
(I) .. 1 =5 se xdx = + --I

s'xtanx
ec3 + 2 1312 _J sec Zxdx.LInx.
14

14
sccx tan x 4 sec 1 xt anx +fltanx.
2.4
5 +1 3
 204 LNTEGRItL CALCULUS

(ii) Also, 1 7 =Jsec 7 xdx = 0tx + -Is

scc-' x tan x 3 sc. x tan x I

4

=f sec x Jx = log ( sec x + tan

1, sec x tail sec x tan x + 3.5 sec x tan x

= .— - -
6 6 4 4 6 2
11 5
+ 6 log ( see x * tan x >.

Ex. 3. Ol'tain a reduction formula for e- co., "x dx, (a 0

Jo

and hence find the value

of , COS I x dx

From § 9.6. rpladng a -- a

I,, = e I d
Jo

" cs" 1x(-oos

& *

=
n .4'
C

fl +Q
n(n--1)
In 2 I since 1: e ;- -
is the rc'qiIi red red LJCtI on formula.
4 54 4 20
I

4 3.2 4
13 I, ••i
32 t43.4

4 4 7UI
1-1 2+4

E. 4. Obtain a reduction for

J • )PI77

Hence find the value of J --j---------------

dx
 LNTE(,RAiON BY SL'CC}-SSIVE REDUCTION 205

Let F,
= 5 kscgriting by parts

K
=
(x 2 + a1
'
I (r-a 2
.1
S
______________
)(/ii. 2xdx

i-a 2 gj7
--
4 ( ,aZ)i)i dx

= 1i + iil, ni21,

Changing n to n - 2 on hoti sides,

I _____ n-I
= T T TY' ( x 2 * TTfl^ + 1 2.

The result can he obtained from q 7 by 5 tibstituti ti v (n/2)

p'ace of n and changing the definition of 1,,

• 17 - f d
I 41
i-Sa2
I x 2 1
15
--j --r---• •. + 13 = 1iTTi7F
1 r 4 x 2.4 x
Z(T7i5s ii +3.5.(22)3/•T;_T
. 5. With the help of reduction formula, find the value of

Scos 'x

From 9-1 6, we ;c't he gin c ri! form of the red uc tlnii form ula as

Sc
sin"x I sill "' 1 x ni—n + 2
ix = ______
- !, • 2lfltIJ.
n-
o^lx- lcos" 1 x i-I
- - 3
= 1 5,4 1 5 ' -
'i -
S
2 - - - 1;
c(lSr I
206 INTEGRAL CALCULUS

Also 15 , 0 = f sin s xdx

sin 'x Cos x 4 sin'x cosx 42
ii.coSx
—
5 -- 3
(from § 9.3 (1) 1.
I sin'x I six
n' I sn'x
I35 -
- W cos'x
I . 4 sin'xcosx 4 2
Slfl X COS X + + K T COS I.
+ .

Ex. 6. From the reduction formula for

5 cos " x CO5 nx dx obtain the value of

5 cos 3 X COS Sr dx.

From § 9.18 (I), 1.

= 5 cos - x cos ,ix dx
- Cos _x sin nx ,,t
,+,I + m+n
Here in = 3,n = 5;

13 , 5 =J ' COS s x Cos 5xdx -- COS'

r sin x 3
8
5 +••

cos 2 Sin 4x 2 cosxsln3x 1

124
= x6 + + 1O3
= 4

10.2 =
f
COS 2x dx =
-i--
c09 3 x sin 5xcos 2 x sin 4x cos x sin 3x sin 2x
8 + 16 + 32

Ex. 7. With the help of the different reduction formula jbr

m (s+ bx I )F dx, find the values of

f
(i)5x 3 (a + bx 2 )' dx. (ii)J ( b2 )4 dx.

(i) Here m = 3, ii = 2, p = 4, and since p = 4 is positive,

(I) can be connected with § 9.20(1) or (2).
 INTEGRATION BY SUCCESSIVE REDUCTION 207

Using (1),
= 14 (a + bx2 )4 2b.4
2.4
4 - •j
x' (a + bx 2 )' 2.b.3
6 - 6b.2.2.

x' (a + W)2 21.2

172.2 '9.'.
8
x'° (a + bx 2 ) 2.b
10

-
J dx -
- 12
x 4 (a + W)4 bx' (a + ) + b 2 x' (a )2
'3.2.4 = - -+b- 2
4
- (..+
- bz 2 ) b 4 x'2
10
Using § 9.20 (2) the result can be obtained in a different form,

(II) For this, the suitable formulae are § 9.20 (3) or (4).
Using (3), replacing p by - 4,
I 11 2(-3)+3+1
1324
- 2(— 3) (a + bx)" -2a ( - 3)
I _________ I
6a (a + bx t )3 + ' 3 13.2.3

13.2.3 I 2(-2)+3+1
_-
2a( -2 ) (a-
+b, I )2 2a(-. 2) 13.2.2
X4
- 4a(a +bx')2
I x' I
6a (a + bx' )3+ iY. (a + bx2 )2

Ex. 8. Find the reduction formula for f x dx

I),
j (a + 2bx + cx')'" x

J
and hence obtain the value of 2 x dx
(r' - 4x + 5)
208 [NTEGRAL CALCULUS

=f Ix
- (m+2br+cxi)

Consider 1.. 2
x 2 dx
Integrating by parts.
= -+ 2'x +
(a cr

n ( x!(2cr+?i)
(rn_I)(a+2bx.cx 2 )rn_1J (a+2t,x+cx2)hiX

n( C x'dx
(rn1)(a+2bx+cx2I

C
j (a + 2bx+ cx2)'
Changing ri to ( n - 1) on both sides,

n-I
(,n_1)(a+2bx+cxZ)-1 .(2cI,, , , +2b1).
rn-I

Dividing and transposing,

i__N
2c(n - l)(a + 2bx +crhjT
rn - I
+ 2c(n -1) 1. - 2,. -1 _! '- - L'
c (I)

x 2 dx (a+2bx+cx2)
Also,
fbx.cxi)"IJ (a + Zbx +cx2)N dx

= + 21, I_. , , +

Substituting and simplifying,

• 2b(rn- n)
I__N c(2n-rn-l)(a+2bx+cx1)'-T + c(2n_rn1)'N

a(m -1)
+ c(2i- 1) I2N - •.. ( 2)
rn -

Either of (I) or (2) may be regarded as a reduction formula.

ilence, using (2),(a = 5, b = - 2, c = ],here).
 I1 EAT!ON BY SIJCCESS1'E REDUCTION 209

i•4
- 4 (. 1) 1 2,4 +
[ - - )

5)3Jf 4 (-2)
12. 4 1.4
I -, SIx 4x 5 4T

r 1 1' -4(-3)
I - oT' 4. - 5) j —6

2 2
jI dr
( 4x)4J ((i-2+I}

dz
T. I z =2- xI

ar 2 1 using § 9.9 successively = X (say)

Then 11, r - _!! .2),

2 124x 4 21
'2 4 - - +

3$ 9 t. 4
I):
3.5 15
4443 46
354 3,51•$ '2
=- T 2
+
.
Yf.
J. &sin zdz(n >0), thell prove that

=
ii,+ 't't - lu. 2
lntcgritin

I.. = [ - -} f
Inlegral Calculus (main) -lb
10 DrrEGRAL CALCULUS

{[ ' $^P.
J0 - X-2 sin zdx

'I ( -. U
'I ( P - 1)
1
LL il$ -
fJ
jam
____ ----L-dx, V = I
J
(LJi.)dx,
SIfl X
beinp ex i g ,t shot' fj
= =

*/ 2
in (2* + 1)a •- Mr(7M - )x
Jo

=
J 0
' 2 --.
2.coi 2nxsi
-;;
--- :x .2
JC
cos 2*x dx

,. in2nx
£ I £.fl I - 0 k'raIt egral values of *.
L
- 5.1 511-1 .

Now, S2 - 5 .!. Sin, x

*/2
h

S.i .i =511 *.

Also, V ,1 , 1 -
+ I)x — sin 2*x
- dx

.ls(2,i + 1)x.Mnx
dx

fX/2
m(2* + x
sin x
30
 TEGRAflON BY SUCCESSIVE REDUCTION
211
v,-v,,, 1 . s,, =4 * , v,. -v .4 * ...., V 3 - V 1 =4*.
adding, V, -V 1 =(n- 1)rcI2.

Since V 1 =f dx =4 * , :. V. =4**.
a
EL 11. Show that

cor () =-x ; (0 r(4)r( 1 ) .2 ;

ciu,f sii 4 eti'ede .1 4* 5in'ecc3eda = fr*.
0 0
wr(f)r(!+i) .4r()
[since r( * + 1).x r(R),A..9.2J (Eli))

r( - .13' .-53 r( - ) -2-r (-i)

.-'/*. I by Art. 9.21 ( v ii)

(ii) Left ilde.r(4)r(1-4)i7_.

(iii) By Art. 9.21 (A)(1),

First Integra,. r()r(i-) !.&L1*L!1 3

i a aa * a
i r (6) 5!
- = 31-2 x
By Art -6 -19 (lv), Second Integral - First Integral.
show +
Ex. 12. met r '•4) IC.fi.'asj
2r( +T

____ 2iu - I
- 2 +i)
r(++)_r(2*+3)(
2* -I ___
2 r (2a2_ i) by Art. 9.21 (1W )
- 1 _____
2 r(2*23.i)

- 2* - 1 2* - (2* - 3
2 2
r 2
212 INTEGRAL CALCULUS

2n-12n-3 2n -S S31 f
- 2 2 2
I By repeated apptiat ion of the re.z.lt o f the above A rtiek

(Zn- 1\(2n - 3)(2n - c31

It (1)
= 2'
Now multiply the nt:rneIator and Jenorniflator of (I) by
21t 2 )( Zn - 4 ) 4.2

2m 2n1X2' -:c2i- 3 5.4i2)

• r(n +) =
( 1. I Nr
-2-,-T;
J, , • i I
=
Note. The .ih,vc . 1,e written %O the t'rii
1(u)! .n
It is an irnpoTta:o j Ut Vzn u. I .1hei1R4ti45
Note. 2. The ng' ide of (1) ...n be wnttr as ",here
C) the
notation ( a ), den (a 1 I • 2 ) (a n- I
• .r ( '), 1j).
Ex. 1i.Shou. ), lI 8f m n)L (m • ,1)=ti.l)g(n *l,m I

r(,ir(n 1(m n)1(l' rlrm)r(n)

Left side - _______ 1(1 ,, rti +,n+n)
r(M+n) + n)

rc:r(lC' )
Sitiarlv, right de -------• -.

Hence the result

E*. U. ftafio1e

f
Jo
ad Md its t'.2111C when Ca -

rs. ia tip. When .0. Y 0.x -1 /

 iNtEGRAtiON BY SUCCESSIVE REtWCTIOM 213

.1
i._j I y&I (I - y)ki dy
a

- .2k I r(a + k)1( ..k

r(u . 2k)

w:1cii 1

F( 2k • )

114 _LLLLLL2
1(2k + i ,r I & -. 1)
Ibvix U and \cte 2 of Art 9.23

2 k

FXAMI'LI.S t\

I. Ot'Liin a i..-dusti'r t'rrr.il.i t.r 'e ' ( n* - ) and

hen,v find kh4 v.duc of j i'. dz

2. that I 'e" 1 (a'' ' - a2x1 . 6at - 6).

3. Find the rdunort to, ,nula for

(1)1 cot " td.i. 4i0 1 cowc

4. It I,. s:.h O .1k), then ch,w i)tt

nL .inh ' 9 0 - ( ii - I ) I

5. ("bi.un .'r.- icti'n t)n..ut.e tot

(i) I tan OJt$ (Ill) J cch0dO

6. 'Show li.t if 1. -' f ' sn i- dt then

tx
-r
•4
- - ----------------
ill r b I,',2
' ,
Fit + -
+ 00
I.. - 2.
a' s r 1 b 1 a2
214 INTEGRAL CALCULUS E. IX

7. If I = Ix • cos bx dx and J = 1 xN sin bx dx, then show that

(I) bI x sin bx - nj,, -
(ii) bJ = - x" cosbx -,- nI,..-i
(iii) b21 .x- l (bxsinbx + ncosbx)- n(n- 1)I.2.
(iv) b'J =x"-' (n sjnbx -bxcosbx)-n(n-1 )J,-.
8. Find the values of the integrals:

(i)J (x 2 -6x + 7)S dx. (ii) J dx

(x 1 +
dx r x dx
(iii) f (x 2 + x + I) (iv) - 2x + 2)
J
9. Show that
2 +x )12 na2
i (a2 + x 2 )fl dx= x(a n+1 4-

find also 13

'40. If I deduce that

1(1 + x 3edx(n >1),
) - 2nx
I (1 +x 2 —e(1+
42 x2)'1
a
2n(2n -1) 4n(n -
+ 1_i -
a 2at

11. Show that ifu =1 x x 'Ia -x'dx,then

- x' (a 2 -x2 )32 n -1
2 U-2
- rz+2
12. Find the reduction formulae for
f xdx • ( dx
)J x"q ( x 2 - 1)
( Ii
J'I( 2ax - x 2 )

13., If 1,. = Ix 'Ia - x dx, then prove that

(2n + 3)1k = 2an1 1 - 2"(a -
 Lx. IX TNI'EGRATION BY SUCCSS1VE REDUCI1ON

Hence, evaluateJ X 3 47X - x2dx.

0

x dx
14. 11 u I , then show that
JI(ax + bx + C)

(n+1)au, +(2n+1)bu, +ncu,.. 1 .x Iax 2 +bx

15. If 1 = I (sin x+ COS x)"dx, then show that

COS
= -(sin + COS x)2. 2x + 2(i -
16. Show that

dx 2n- 3
(I) 2n_2I_t
- J0 (1 +X 2) r

- dx 1.3.5.7 ,
f (1 + x ) = 2.4.6.8

17. Show that, if I, cos" x dx and J, sin

= j J
(I) I =J. (ii)I I (n >2).

18. With a suitable substitution, using the previous :cmp-

find the values of
I -
_ dx
dx (ii)
- (jS 0
- x ) 5
0
(1 2

(Ii being a positive integer)

19. Prove that if u x tan - x dx th;a

j
in +)u + ( ii - 1)U,..2 y-
216 INTEGRA L CA LCULUS £x.LX

20. If n a 2and 1. (1 - x3). cos ,n--cdx. then

show that ,, 1 I, = 2n(2, - 1 )I, - 4n(n - 1)1, 2

/2

21. If U. 0 sin O dO and n> I • then orov that

= $0
i— I
u, = --u..2
Pt 7I

dx
22. 0) Obtain a .eduction formula for ' •
j

and (ii) evaluate f (1

dx
)
Fat
0

23. If 0(n) = x - log r dx, then show that

Jo
+ 2)— (2n +1)o(n 1)+ i 2 '(n) U.
"/4
24. If I. I tan "0 dO, then prove that
Jo
+ =

25. Show that x t (log 0 . Ix =

J0

26. If ., a
fx (1 - x) - I dx, then show th
Jo
a
( m - 1) ( n 1)!
=
(m + n- I)!
m and n being integers, each 1
 LNThGRA1iO) fY sUcCL.'sYIVE Rr,UcflON Ho
Lx. IX

27. It m • i are positive integers. then show that

L.a,-x)dx
rn±fl

Hence prove th3t I,. , = m r

2$. 1-iid he v.iIt", of

ut)

I• 44SX r
(r ) --------

. If I.., -- 4 .;nt .1% bji :-Ic

(7n -- ,, in n-- 2

-i ) "j I I I -

30. Obtain . rdti ' i I rrru,i Ijr

co5'r sjfl ndx, und i * Ic tbt , vj'tit

c,sx si -1 32 dx

31. 1, 1., J 51.1 2 (( 4 S nr .1 4'iI

'72 coc 1O

i_rn

32. If I.,. -J' Sin -.% j.I)% P!

218 INTEGRAL CALCULUS Lx. IX

,*/2
JN &1flX sin nxdx, then show that
0
(m + n)IM = 3fl . fl* -mJ, • ,. 1 (in>
*/2
33. If f(m. it) Cos x Cos nxdx,then show that
=J 0
m
f(m,n)
=m +
f(m - 1fl 1) m(m-I)
m2 82f(m-2,n)
'it
M- - Lit + 1),

____
and henceshowthat f(m,m) -
- 2*1

34. Obtain areducUon formula iorf dx

+ b sin x ),
35. Find the values of

dx
x dx
(i)J (1 + Cos a Cos x) 3 .(il)f (1+k sin x)k<l)
36. Using the integral I x(a + bx ), dx, find the values of

xs (1 ^ x' )7/2 dx.

f
1(1 + 2x' )fl dx.
I Use § 9.20 (5). I I Use § 9.20 (4). 1
•1
dx
x' '1(1 -x2 Use § 9.20(6).l

37. Find the reduction formula for 5 x' T2-ax- x 2 dx.

•2a _______
_

Hence show thatJ x 42 ax - x' dx

6m*2
= ,i -
(2m+1)!
__________

0 2 (m+2)!
 Lx. IX I1rFEGRATION BY SUCCESSIVE REDUCTION 219

38. If I xe-
10

- xe' sin xdx,

=10
then prove that (m being an integer > 1)
(i) I = . j m(1.._ i -J.,..i ).(ii)J,,, =m(1 1 +J_ 1 ).
(iii) I.. -MI. +m(m - 1)1,2 =0.

39. Show that sin 2nx cot xdx =

0

40. (1) ! u, = J cos nO cosec 0 dO, then show that

2 cos ( n - 1,) 8
- U.•_2 =
n-i
P. sin (2n - 1) x s'n2nx
(ii)if dx , = d x,
5 sin x J sinBx
then show that n(P, 1 - P. ) sin 2nx
and Q, i - Q = P. .
41. Prove that, if

1-o_-!X
I =
Jo I - cosx
dx where n is a positive integer

or zero, then J, 2 +J =21,..i

sin it
Hence prove that A =-
J0 s,n 1 0 2

sinnO =
42. (i) Prove that J A0 or it according as
sin 0
0
n is an even or odd positive integer.
220 1NI IGRAI. CAL Ct1US Ex IX

(ti BY I a reduction f(),mulj or o t herwise, prove that

.irnO
a po;!ve Integer.

43. it 'I I' .1 p&icitive u ltv gc r , then

I .0' t P? i
di
I- t

and dJue that j.: fT- ) di

J eo ' x ,i ii -- axthen show that

41. if 1.,, ,.

1, , I I _ ?I 2
---.- 2"'
J :' •.- --2 I
1 :3 ri

45. SF.:,w that t' ' ;in x dx

n(n- U'n-2).. 1.2 1

n')a' .(,1.2)2) ( a 1 3' )a?

if n is
n(n 1 i 2). 2 .1 1
n I.i2 - 2a
it P7 7' CVCfl

If 1, 'c 0 ' sin ) JO . then prove that

J
•tt.

a' ) -- I )(a h 2 )f,

 IX iN lcK.\1i>N BY S I1CFVF kJ'DU(I tON ?2l

47.1(1, Jta 2h sinco ..x -.'on'x) 'ixthp

prove that

'(fl,- 1)(ab.hl1,.? 22'i i- 1,(u +t')1..,4 4i1,

h ns 2 r Sal I k) (' a) '.,n i Cos x

-
a cos I x * 2h sin T . b sin a x ) -
I Ai'y the a !tri nat iw inn hoi of ' 19
48 Show that

tiJ •A )'M r)'ñ. 2r-q

Ftp-.-q.2

-, -1 •q > - ii
Put I .- 2y J
I-On• I )F(ii • 1)
UiIJIA

j'n .1.'i>-H

I - b l7I I

49 So ' tbct

J0
e'- d -I (
2 )
>
Put I 2 y
5) c•,-th

r •
rIdtw d
J0 f 1.
0

I rut I z-
SI. S'wtht
( .111)U(IrI 'fl = ( n,I)8( i
= in 13 - rnn).
222 INThGRAL CALCULUS Ex. IX

52. Show that

........ ..... .F()
0
=4
1679
Combine 1st and last factor, 2nd and last but one etc. and apply for-
ula (vi), § 9.21.1
53. Show that

$ dx
(1 —x')" - . (Putx' = z.j
0

54.- Show that the sum of the series

I I m(m+1)
n+I n+2 2
+ m ( m +I)(m + 2)
3! n+4
- r(n +I)r(I — m)
- Fn—m+2) , where n>_I and m<I

R. S. n + 1,1 - m) = I' (1 z) x, etc.]

55. Show that
i sin 21 r(m)r(n)
f (asjn2O+bco32e)_.dOa.b.r(3
I A pply A rt. 9.21 (viii).

ANSWERS
1. 1.a

14 = -
f
e ,:
1x444 + 4x0 + 121 2 . 2 + 24xg + 241.

3. () In - cot*' I..2 .(tl)!,. eotxeosec - x i.-2

- I -i
 INTEGRATION BY SUCCESSIVE REDUCTION 223

1 9ecb
tanh_j +I._z 2 9tanhQ -2
.(U)I.
it-I

[ (r2-6x+7) 20 (x'-6x+7) + j.j

5-jj- .
2016
8. (i) (x-3) _,(xZ -6xi7)'
_____
2OI612(X2
11.9.75
6x7)1
20.16.12.8
+ 209753 (x -

5x
_______
6x

5x
-
2016.12.8.4 1

5
11.9.7.5.3 I
(Ii') T^I)424 (x i + 1) 2
fl+jt*flX
16(x2
2x+1 2x+1 4 tan
T x+1) + 3(x 2 •x + 1) + 3T3 '(2x+1t)
2x + 5x + 7 fiT_2x+2_j.a1nh1.(x_1).
(iv

9. j x + r )1l1 82X (g2 + x2 )1I2

3
4 -F 8 4 log(41 t + xy).

12. (1) iiI, = - x" t 42ix -x + (2t - I )al_

- I it- 2 13.
(it - 1)x"' it -I
IS. (, ) I_hui.3 ..!,lfnl.odd
it n-2 3
n 3 1 it
it is even.
and
....
(ft) 2R_32N_$ Ix
2it-22n-4...
and If it I.
22. (I) 15 I x 2it-2
2it -1 (1+ x') '1(1 +zt) 2it
2ii-2 2n-4 2
(ii) 2i,-1 2iu-3

28 - (Ii)
8 cos4z 4cos2x 8 *in x
40% 3 un 3 x + 3 uln x + 3
(lv) 2 11 tan"' x + 2 too t/2x_cot5flz).
224 LNTECtM. CALCULUS

- x cis nr P1 1
.,-o. i ,, = -- I - -
3

34. (n .-I)(a b2)!,1 +(2n3)a1, i-( -2)I

35. (i) ( 2 + cos (1 ) COSC sit

I k 2 ( Li n . . k
(ii) (an - )
I-.- k 2 -
I + k. n z - k )

36. IL 1j. 991 + s ]

(ii) (1 • t' ) (x - I ) (iii) 2 '3

I (2.' - * 2 )3fl C2jn * 1 la

37. 1.. = - _______- - -- ______ - I.
PU -
- 2 m * 2
 CHAPTER X

AREAS OF PLANE CURVES

I Quadrature J
10.1. Area, in Cartesian Co-ordinates.
Suppose we want to determine the area A 1 bounded by tht
curve y = f ( x ), the x-axis and two fixed ordinates x a and x
b
The function f( x ) is supposed to be single-valued, finite and con-
tinuous in the interval ( a , b)

Y
J F

X
Fig.]
Consider the variable area QLNP = A, say, bounded b)
the Curve y f ( x ) , the x-axis, the fixed ordinate
QL wherc
OL = a and a variable ordinate PN where ON = x. Clearly.
A has a definite value for each value of x and is thus a function of
x. When x is increased by an amount Ax ( NN' ). A assumes ar.
increment E4 = the area PNN'P'. Now, if f ( x
1 ) and f( x )-be
the greatest and the least ordinates in the interval Ax,
such that x :5 x1 :5 x + tx ,x x , x
Clearly the area AA lies between the inscribed and circumscribed
rectangles UN' and FN'
i.e., f(x )x <z.4
<f( x 1 )x.

(1)
The process of finding the area bounded by any defined contour line I!
called Quadrature,
the term meaning 'the investigation of the size of a
square which shall have the same area as that of the region under con
derat ion'
inkgrai Calculus (main) -17
226 INTEGRAL CALCULUS

Now, as Ax approaches zero, by the continuity of the

functionf ( x) at x f ( x 1 and f ( x 1 ) both approach f ( x), and
dA
-AA
tends to - . Hence, relation
as the relaon (1) is always true
Ax dx
we get in the limit
dA
dx
by definition, A = Jf( x ) dx + C u F( x) + C, where C
is an arbitrary constant, and F ( x) an indefinite integral of f ( x)
Now, when x = a, PN coincides with QL ,and the area becomes
zero. Also, when x = b, the area A becomes the required area A .
0 F ( a ) + C and A = 1(b) + C.

F(b)— F(a) f(x)dx.

a
The definite integral

x ) dx, i.e., y dx
1:1 (
-:herefore, represents the area bounded by the curve y = f ( x) , the
and the two fixed ordinates x = a and x = b.
Note. Aim alternative method of proof of the above result, depending
)fl the deflruom oi a definite Integral 6s a summation, has been given in
Vt. 6.11.
Cor. 1. In thesan%e way, It can be shown that the area bounded by any
:urve, two given abs*iss (y = c , y d ) and the y-axls is

xdy.
J
Co t . e. If the axes be oblique, o being the angle between them, the cor-
esponding formula for the areas would be

gin 0 y dx and sin to x dy respectively.

Ja J C
 AREAS OF PLANE CURVES 227

10.2. Illustrative Examples.

Ex. 1. Find she area of the quadrant of the ellipse + =
between the major and minor axes a

Fig.2
Clearly, the area being bounded by the curve, the x-axis and the or-
dinates x = 0 and x = a. the required area
fI..
= ydx
Jo

= a
Ja2 - x 2 dx [since -s -- I
for the curve
0

COS
= a O.a
COS OdO( putting z = a sin O)
ab-

.X/2
=
ab
(1 +cos2O)dO
ab
=T 8 + sin 20
---- j.
óz I
T T1
Cot. 1. The area of the whole ellipse is clearly four times the above,
i.e., = nab.
COT. 2. Putting b = a and proceeding exactly as before, the area of a
quadrant of the circle x 2 s y 2 =a 2 is.*a2 ,and the area of the whole
circle =
 228 INTEGRAL CALCULUS

E,. 2. Determine the area bounded by She parabola y 2 = 4ax and any double
e,dinate of it, say, r x1.
The area OPN is bounded by
the curve y' = tax, thex-axis and
the two ordinates r = 0 and r

(5- L area O!'N

=
X

f0o
x,
y dx =
r1

JO
'I4axdx

I The P os i tive v a l ue of y is taken

since we aro considering the positive
hg.3 side of the y-axis.

•i
1r
3/2
= xi y 1 (wherey =PN ='Tiar 1 .
The parabola being symmetrical about the raxi, the required area
POQ
y1 =e y
= the area of the rectangle contained by PQ and ON,
the area of the circumscribed rectangle.
Cor. The area bounded by the parabola and its latus rectum =
Ex. 3. Find the whole area of the cvc!oid x = a (0 sin 0), y = a (1 - cos O,
bounded by its base,
The area of half the cycloid,
viz., area AOC, is evidently
bounded by the curve, they-axis
and the abscissae y = Oand y = 2a.
Hence, this area is given by

fxdy
Jo
Fig.4

a(t3 + sin0),asjnod9 r since y = a(1 - cosfl)

=j 0 I x =a(6 + sing)
 AREAS OF PlANE CURVES 229

= -OcosO +sinO
a [ + + (O _.sin20)1=.7.
Hence, the whole area of the cycloid is 3ita
Note. It should be noted here that if AM be drawn perpendicular from
OM
A on ZTX4 , the expression y dx represents the area
J
0
and not the area OAC.
Ex. 4. Find She area of the loop of She curve
xy z + (x + a) 1 (x + 2a) =0.
i-lore lotus first of all trace the
curve. Th e equation can be put in
the formy 2 = - (Cr + a) 2 (x + 2a)}/. 'I'
We notice that y = 0 at the
points B and A where x = - a
and x = - 2a, and y -. ±
when x -, 0. For positive values
of x, as also for negative values of
x less than - 2a, y 2 is negative and
so y is imaginary. There is thus no
I
part of the curve beyond C) to the
right, or beyond A (.z = - 2a ) to
the left. From A to B , for each FigS
value of z , y has two equal and opposite finite values and a loop is thus
formed within this range, symmetrical about the x-ais From B to 0, each
value of x gives two equal and opposite values of y which gradually in-
crease in magnitude to as x aproaches 0. The curve, therefore, is as shown
In the figure.
The required area of the loop = 2 . area APB

JPT3a)d
= 2.Jdx = 2f
and substitut i ng z for x + 2a. this reduces to

21(a - z)
Z
dz
J o
 INTEGRAL CALCULUS
230

Elije .2astn- cos-; ae

=21 a Cos $
Jo
2 81
[putting z 2a

ER
7,2 I cos8(l - cose)d9 2a 2 (i 1)
=
Jo

=a 2 (4- it).
and two given or-
10.3. Area between two given curves
InIPS.
Let the area required
Y P2 be bounded by two
given curves y = f ( x
and y = fz ( x ) and two
given ordinates x =a and
Q2 x = b, indicated by Q Q2
P 2 P 1 Qi in the (figure - 6),
O)t M]
! PNQ2 x where OM =a and ONb.

Clearly, area Qi Q2 P2 P I Q area P1 MP 2 -. area Q1 MNQ2

b
b
f, ( x)dx J-8 12
=S a

=Ja (fl x- ji (x))dx

b

=1 (yt _y2)dx,

0f the two curves P . P2 and

where yt and Y2 denote the ordinate s
Q Qz corresponding to the same abscisS'
 AREAS OF PLANE CURVES 231

10.4. Illustrative Examples.

Ex. 1. Find the area above the x-axis. •,tcluded between the parabola y 1 = ax
and the circle x 2 +y 1 = 2ax. 11. E. E. 891
The abscissa of the common
points of the curves y 1 = ax y
and x 2 + y t = 2ax aregiven
by r + ax = 2ax, i.e., x () and
x = a.
We are thus to find out the .2
area between the curves and the
ordinates x = 0 and x a
above the x-axis (i.e., for positive
values only of the ordinates).
The required area is there-
fore Fig-7

(y1 - y a ) dx I where yi 2 = 2ax - x 2 and y2 = ax]

J 0

=J(2ax_x2 -.)dx.
Now, putting x = 2a sin 2O,

'I2ax - x 1 Za sin
J odx = J o
I/ 4. X/4
= (1— cos49)dO =a2 [9stn40] =
a25
Also, f SrdX = 'Ia [. x" 2= a2

Hence, the required area is

a2 -4 a 2 = a2 ( -
Ex. 2. Find, by integration, the area of the ellipse
ax + 2hxy + by
:32 INTEGRAL CALCULUS

The equation can be put in the form

by + 2hxy +(ax 1 - 1) = 0.
V

X X

Fig. 8
If y y2
be the values of y corresponding to any value of x, we
have
22
=4I,2x 2 t'(ax l _U'4b-(ab-h2)X 2,
y1 -Y2

ab - h 2 being positive here, since the.conlc is an ellipse.

The extreme values of x , where thp ordinates touch the ellipse, are
given by
fb
- = 0, i.e., X _.±.s1j ab - h2
v2

The required area can be treated as bounded by two curves, MP,L, LP1M
respectively, both satisfying the given equation, but one having a single
value y for y coresponding to any value of x, and the other also having a
single value ys for the same value of x. -
Hence, the area required

sb-h 2 _________
I Tiiab -h 2 )x 2 dx
=I (yi —y2 )dx
bj 1 b
ab
44 ab-112
and putting '( ab - h 2 ) x = lb sin 9, this becomes
sf2
2
1
_____
cos2OdO
ab- 2 )
 233
AREAS OF PLANE CURVES

Note. The area of the ellipse can also be obtained as follow.:

Assuming the equation of the ellipse referred to its major and minor

+ = I by the theory of Invar-

axes as axes of co-ordinates to be
-- = at' - h'.
lanta as given in Conic Sections, we know that --j .
Now I from Ex. 1, COT. 1, Art. 102 1 the area of the ellipse Is
it
na= 1b— hr)
x )3 /(a + x )) and the
Ex. 3. Find the area between the curve y' ((a -
asymptote.
To trace the curve, we notice that y is imaginary for values of x greater
than a or less than - a. At x = a, Y
y = 0, and froma to_a,foreaCh_
value of x,y has twoequal and
op,.osite values, tending to ±
as x apcoaches - a. At x = a. the
x-axis touches both the branches. x' o A x
The figure is, therefore, as
sh:wn, symmetrical about the

The required area between

the curve and its asymptote i qFi.9
therefore

21a
dr
21 ydx a^X

and substituting z for a + x this reduces to

.25
21
Jo

Cos 0 29
=21 2acos19j-. 4a sin 9 cos 940 I where z 2a sin
Jo
1.3 5
= 160 cos'OdO =165 2- =355'
Jo
 234 INTEGRAL CALCULUS

10.5. Areas in Polar co-ordinates.

Let r = f(0)bea
curve APB, where 1(0)
is supposed to be a finite,
continuous and single-
valued function in the in-
tervala < 0< P. The area
bounded by the curve, and
the radii vectors 8=a and
0 is given by the
definite integral
Fig. 10

I
5 r dO,
L
i.e., } ( f ( 0 )) 2 dO.
Let A denote the area P0,1 , bounded by the curve, the given
radius vector 04, i.e., 0 =,a, and the variable radius vector OP
at vectorial angle 0 a < 0 < Then for each value of 0, A has
a definite value and so A is a function of 0 If Q be the neighbour-
ing point r + Ar, 0 + AO on the curve, we have
EA = the infinitesimal change in A due to a change
AO in 8
= the elementary area POQ
and this clearly lies between the circular sectorial areas OPN and
OQM, where PN and QM are arcs of circles with centre
0.
Thus, -I r2 AO <A .z i2 (r +

i.e., !(f(9))2 << . ( f(O +0))2

AO 2
Now, proceeding to the limit, and remembering that 8
being continuous, 1(8 + O)-3 f (0) as z0 -i 0, we get f (
dA
= i (f(0)) , i.e., .}T2
Thus, A =} f r 2 dO +C = F(0) + C, say.
 235
AREAS OF PLANE CURVES

Now, taking P coincident with A and 8 respectively and

denoting the required area A OB by A 1 , we get
+ C.
0 = F(a) + C and A = F( )
U
F(a)= -L o r1dO
where A 1 = F() - J

is here assumed as concave towards 0. A

Note. I. The curve A PB
similar proof with corresponding modifications holds even if the curve be
convex, or partly Concave and partly convex or wavy, in fact of any form.
As in the case of area in Cartesian coordinates, the above
Note. 2.
result can also be deduced directly from the definition of a definite integral
as a summation.
t r1 and(8) r2 =f2 (0)
Cor. The area bounded by the wo curves is = f
and two given radii vectors 0 = a and B

(r2 1 - r, 1 )dO.

A lternative proof
OB be the radii vectors cor-
Let A B be the curve, OA and
responding to 0 = a and 0 =
Divide 5 - a into n parts each equal to h, and draw the cor-
be the points on the curve
responding radii vector s. Let P and Q
=a + (r + l)h and let Us
corresponding to B = a i-rh and 8 to 3 . With centre 0 and
suppose 0goes on increasing from a
PN , QM as in the figure.
radii OP , OQ respectively draw arcs
Then the area OPQ lies in magnitude between
4 0P 2 .h and !QQI.h,
and 4tf ta + (r + It
i.e., between 4(f (° +rh)j'h
it is clear that the area
adding up all the areas like OPQ,
A OB lies between
n-I
Il—a
4 E jf (a +rh)) t h and! £
7-0
 236 INTEGRAL CA LCULUS

Now, let it .-. , so that h -, 0 ; then as the limit of each of

the above two sums is

ifa (f(0))'dO,
it follows that the area AOB is also equal to the definite integral.
10.6. Illustrative Examples.
Ex. 1.
Find the area bounded by the cardioide r = a (I - cos 0)
The curve is symmetri-
cal about the Initial line,
since replacing 8 by-8,
does not alter. Beginning
from 0 = 0 and gradually in-
creasing 0 to X , the cor-
responding values of r are
noticed, and the curve is easi-
ly traced in the figure - 11.
Fig. II
Now, the required area is
evidently, from the above article,

2.fr2 dO = a25 (1- cosO)' dO = a - =na'

Note. It should be noted that the area bounded by the cardioide whose
equation is r= a( I + cos 0) is also
-
Ex. 2. Find the area of a loop of the curve r = a cos 20.

Fig. 12
 AREAS OF PLANE CURVES 237

In tracing the curve, we notice that, as 0 increases from 0 to . n, r

diminishes from a to 0, the portion A P, 0 being thus traced. As 0 increases
from to tr, r is negative throughout, and the corresponding portion of
the curve which is traced is OP, 14I13 0. Then as 0 increases from 1 it to . it,
r remains positive and the portion OP4 A '5 0 of the curve is traced. As
8 increases from i it to in, r is again negative and we get the portion
OP4 B'P7 0 of the curve. Finally, when 0 increases from I it to 2it, r is posi-
tive and the portion OP. A O of the curve is described. The curve thus con-
sists of four equal loops as shown in the figure.
It t5 now clear from the figure that, area of one loop
= 2. area A PO
f7It
r2 dO = a2 cos 2 29d0 = -ira.
= 0 J0
Cor. Hence, the entire area of the curve, i.e., the sum of the areas of the
4 loops =.} itai.
Note. All curves of the type r = a sin nO, or r a cos nO may be
similarly traced, by dividing each quadrant into n equal parts, and increas-
ing 0 successively through each division. If r be found positive, the traced
portion of the curve will be in the same division; if r be negative, the traced
part will be in the diametrically opposite division. Anyway, when the curve
is completely traced, it will be found to consist of it equal loops if n be odd,
and 2n equal loops if it be even.
Ex. 3. (i) Find the area of the loop of the fohum of Descartes,

x + y' =3azy.
(ii) Find also the area included
between the fo!ium and its asymptote
and show that ills equal to the areof
the loop. X
(I) Transforming to cor-
responding polar co-ordinates by
putting x = r cos 8. y r sin 9,
the polar equation to the curve be- Fig. 13
comes

cos 0 sin 0 - (1)

cos'O + sin 39
 238 INTEGRAL CALCULUS

As 0 increases from 0 to +2,r at just increases from 0 to (3a I '2),

reaching the maximum at 0 = .. 2, and then diminishes to 0 again, thus
forming a loop in the first quadrant.
The required area of this loop is
I 2 3
- -X
1 sin OcosO
r2dO=—iI
[2 9a2
= — i 3d8
2j 2 j (sin
.3 Otcos 30)

t1dI
= lputtmg I = tanOl
9a2J 0 (1 +t)

9a 2
LI 2a I - 1
2 r - - j (1 +i) -.--
2 Lt 1+
0
£-9 .. L. +

(ii) The equation of the asymptote of the folium is

+y+a=0.
X ... (2)
Its polar equation is r = -a
sin 0 + cos 9 .
Now, r-. If (sIn 0 + cos 0) - 0, i.e., If tan 0 -4 -1
i.e.. if0-s.2.
the direction of the asymptote is e -.
The asymptote intersects the two axes at A and B • where
OA =a and OB = a, i.e., OA = OB.

Hence, the area of A OA B = +a2 . 4) ... (

Area between the folium and Its asymptote = the triangular area
OA B + the limiting value of twice the area between the curve and the
asymptote in the second quadrant (from symmetry)
= !gZ + the limiting value of twice the curvilinear area OKPQA O
+2o (say).
Draw a radius vector rr,lonit an angle 0 with the r-axls, such that
< 8 c w. Suppose It cuts the curve and the asymptote at P and Q
respectively.
 AREAS OF PLANE CURVES 239

Let us denote the curvilinear area OX PQA O by S, the triangular area

OQA O by S1 , and the curvilinear area OKPO by S2
SS1-_S2.
oS = 0Lt ,1 (S -Si).

Now, applying the formula for area in polar co-ordinates, i.e., J r 1 dO

and using equations (1) and (3), we get
f 9a2sin2OcosO
I r a' do dO
S (ain30s cos3$)2
LJ0(slnO + cos 9) 2
...!i2 (I -1),say.
4.8 sec1OdO
Now, (sinO + cos9)' - j (1 i- tanO)'
on multiplying the numerator and denominator by sec 2 0)

= J At putting t = 1 tan 91

I I
- I+ tan O'

r1 ______
= 1.i + tan oJ 0 = 1 + tan

sin 1 8 cos 'OdO f tan2Osec2O

C
AgainJ (sIn8 +cos 3 o) 9 J (1 +tan3O)
on multiplying the numerator and denominator by sec '0)

l putting 1+ tan 3OtI

11 1 1
- =- (I + tan 10
3
12 = ( ) [i + tanOI 0 I +-
tan -18 -3.

1 1
$ i+tanOl+tan6J
1g2I2
 240 INTEGRAL CALCULUS

+ tan 2I O- tan $-2]

+ tan 'O
a,[2
= a1 (tanG + I)(tanO-2)
[2 + (1 + tan O)( I tan 0 + tan l0)J
=g1 tan 0 - 2
[2+1 - tan + tan 2 9 1
Now, 0= Lt S=a2.
2
3Z
8—.-
4
the required area = +a 1 + 2a =2a2 = the area of the loop.
Lx. 4. Show that the area between the folium of Descartes and its asymptote
is eqI4aI to She area of its hop, each being
equal to .1al
B
The equation of the tolium is
xl + = 3axv. x'D
Turn the axes through i n ; that is C
substitute (x - y)/'12 and (x + y)/42
for x and y respectively. Then the N Qy
given equation transforms into Fig. I4
I
_x 2 3c-x I
where c = -a.
c+x

C. xl
Here c + x = 0, i e., x = - c is the equation of the asymptote MN
OA =3c,OD = - c.
the required area a between the Folium and the asymptote
•0
2 Li I .0
=2 Lt ydx 1=I(3i
5c +
-t
.0
2 Lt I x(3c-x)
t-+CJ I 'l((x +C)(3c - x ))

Let I x(3c -x)

= ,f ..3c_x))4x=f(1_2cose)(1 +cose)de

on putting x = c - 2c Co. 9, so that Co. 0 = X

2c
 AREAS OF PLANE CURVES 241

= -20
f (cosO + cos2O)dO
= -2c 2 (sinO +sin29)
=_2c( sin( Cos -)+ sin (2 cos - 1--)).

= 2 ( 20 ) LL [sin(cos -' 2c /

+ sin os-
1
2a1
Eo•
putting _2

Again L, the area of the loop )SAC,

= 2 area of the portion OBA
3c .3c
x0c - x)
= 2J ydx
0
(x +c)(3c - x)} dx.
0
Putting, as before, x = c - 2c co. 0,

I. = 2c ) [ ifl (
-_)+(-'
C X c-x\I.,3c
(- 2c h1
4c 1 3 43 a2[on putting c
/a1.
Lx. 5. Find the area between the cissoid r = asin0 and its asymptote.
cos 0
The curve way be traced either from its polar equation or by con-
verting it to Cartesian form,
and the figure will be as
shown. The asymptote is Q
easily found to be the line
x = a or in polar co-ordinates ______________
r cos e = a Now, let 5P—Q— be Q A X
any radius vector at an
angle e to the x-axis, inter-
secting the curve and its
asymptote at P and Q respec-
tivclv. Fig.15
Integral Calculus (main) -18
242 5ITIA3RAL CALCULUS

$
Area OAQPO 2 - r2 2 ) dO I where r = OQ '2 = OP)
2 Jo

_!
- Jt 0
(_.f.2 _
cos 2 0
m2)d9
cos-e
70
(1 +sIn18)d9

a 2 f3 sin28
-2 2 4
Now, the required area bvtwen the curve and the asymptote is c1ar1y
(there being symmetry about the x-axis, and since the direction of the
asymptote is given bye =

OLII [2ç(e_-°)J =a2(4.Jt)1rfl2.

Lx. 6. Find the area common to the Cardiodr r = a (1 + cos 9 ) end the
circler =!a, and also the area of the remainder of the Card bide.
At the common point P of
the two curves, we have
i r
+ cos 9.

) I -- cos 9 = j or,O =1t.

The required area is easily
c
N seen to be

t-ig.lô 2 area OCP 1-area PQO)

= 2 (- (4a) a 2 (1 + cos 9>2d0}

J dO + -

= a 2 .4w i-a 2 ( 4 (w - n)+ 2 (sin x - sin Ix)

+ (n Si 2w - sin! w

(!2.±)a2
 AREAS OF PLANE CURVES 243

Again, the area of the remainder of the Cardioide, i.e., APCR

= 2.area APC = ( y2 - p22 )dO

ri*
(a (1 i- cos O)2 _!i dO
0

(2 cos O scos29 -.)ae

= a2
f 0
I '3 1'3 31 I9 i3 I
=a2t_jE.

Note. The whole area of the Cardloide is evidently the sum of these
two, i.e., = . ra [ See Ex. I thove.j
10.7. The sign of an area.

In the expression $ y dx for an area we tacitly assume that

the ordinate y is positive throughout the range (a, b) and that x in-
creases from atob,i.e.,b>a. y
In this case the area calcu-
lated by the above formula
will be positive. If, however, + C8
y be negative or if Li <a while 0A -
y is positie, i.e., in moving Q
along the curve from x = a to
x = Li we are moving paral- Fig. 17
iel to the negative direction of the x-axis, the calculated area will
be negative.
If, therefore, we proceed to calculate the total area where, in
the range ( a , b) , y is positive for some portion and negative for
the rest, as in the above figure , by using the formula

dx, the calculated result will give us the difference of the

Jay
 244 INTEGRAL CALCULUS

magnitude of the two areas ACP and CQB, which may be posi-
tive or negative or even zero if the magnitudes of the two areas are
equal.
Hence, if our object be to get the sum total of the magnitudes
of the two areas, we should calculate each part separately by the

formula of the type y dx, i y dx, the results being found to

be associated with their proper signs. We shall now discard the
signs and consider the sum of the magnitudes.
In each individual case, therefore, we should first of all have a
Q clear idea of the figure and
_- the area to be calculated, and
then we should proceed. For
S instance, notice that in the
figure -18, area PACR is +
R area CRSD is - and area
SDBQ is +, and that for the
+ range DC of the x-axis, y is
O A three-valued and in calculat-
ing the area PACR we are to
Fig. 19 use one value of y for the
portion in the formula
5 y dx, for calculating the area CRSD

we are to use a second value of y In the formula

5C
y dx, the
upper limit d being less than c for this part, and lastly for the area
SDBQ we are to use the third value of y for this part in the formula
'
f J
ydx. If we take the algebraic sum of the three areas, with their

proper signs, wjet the area bounded by the curve, the x-axis and
the ordinates AP and BQ.
 AREAS OF PLANE CURVES 245
Similarly, in the formula .
5 2
a
r dO in polar co-ordinates
if A < a, i.e., if 0 diminishes in moving along the curve from 0 = a
to 0 = ft , the calculated area
will be negative. Then area Q
OPR is + , area ORS is -,
area OSQ is + • the area
bounded by PRSQ and the
radii vectors OP , OQ being
their algebraic sum. Also for
the range SOR, for eadi value
of 0 , r has three values, and
we must use the right value in
0,4d Fig 19
each case for that part when
moving along PR or along RS or along SQ in the expression r'dO
10.8. Areas of closed curves.

y l Yl

OA
I

KJ1_____
OjA jB X

(i)
cE 2
(ii)
ar
(iii)
x

Fig. 20
In a closed curve given by Cartesian equation, clearly for
each value of x there will be two values of y , say, y, and Y 1
[See figure (i)J. The extreme values of y,say, a and b, are obtained

byputing y1 = yi . Now J (y1 -y2 ) dx will give the positive

value of the required area, provided b > a and y1 > y2 . This
amounts, as it were, to the determination of the area between two
curves having the same equation as the given one, but y being
246 INTEGRA L CA LCULUS

single-valued in each, the proper value being chosen for each part.
The method has been illustrated in Ex. 2, Art. 10.4.
In polar curves, if the origin be within the curve See [figure (ii)I.

- J0 r 2 dO gives the desired area.

If the origin be outside, corresponding to each value of S there

are two values of r, say, r1 and r2 See [figure (iii)]. The extreme
values of 0 , namely a and 0 , are obtained by putting r1 =
Mow, if r1 > r2 and 0 > a, the positive value of the area will be

given by the expression I

- $f ( r1 - r2

In fact, the area OA PB is given by

5a r1 2 dO and is positive,
while r2 2 dO gives the area OBQA with negative sign, the

algebraic sum of the two giving the desired area.

In the case of closed curves there is another method of calculat-
ing the area. Let ( x , y) be the Cartesian co-ordinates of a point on
the curve whose polar co-ordinates are (r. 0 ) , then x = r cos 0,
y = rsin0.
Now, if I be a single variable parameter in terms of which x, y
and, therefore, r, 0 of any point on the curve can be expressed, we
have
dx dr . dO

dr
dt =- sin 0+r cos 0 dO
-
dx
x- -=r
dt dt dt
 AREAS OF PLANE CURVES 247

Hence, the area which is expressed by the integral

well be expressed by the line integral
I r 2 dO can as

J(x ._ y4)dt
dt
along the curve, the limits of t for the closed curve being such that
the point (x , y) returns to its initial position. The rule of signs for
the area is that the above expression is positive when the area lies
to the left of a point describing the curve in the direction in which
I increases.
10.9. Approximate evaluation of a definite integral Simp-
son's rule.
In many cases, a definite integral cannot be obtained either be-
cause the quantity to be integrated cannot be expressed as a math-
ematical function or because the indefinite integral of the function
itself cannot be determined directly. In such cases formula of ap-
proximation are used. One such important formula is Simpson's
rule. By this rule the definite integral of any function (or the area
bounded by a curve, the x-axis and two extreme ordinates ) is ex-
pressed in terms of the individual values of any number of or-
dinates within the interval, by assuming that the function within
each of the small ranges into which the whole interval may be
divided can be represented, to a sufficient degree of approxima-
tion, by a parabolic function.
Simpson's Rule : An approximate value of the definite integral

fa y dx, where y f(x)

-hI(y 1 +Y z.i )+ 2(y 1 + i + ... + y )

+ 4(y + y1 + ...

b 2n
- a
where h = and y , y , y, ,.. . are the values of y when
x = a, a -t- h,a i- 2h,
248 INTEGRAL CALCULUS

In words, the above rule can be written as

4 h ( sum of the extreme ordinates + 2 . sum of the re-
maining odd ordinates + 4. sum of the remaining even or-
dinates ].
Let PQ be the curve y= f(x)andPL,QM be the ordinates
x = a, x = L. Divide the interval LM into 2n equal intervals each

of length h bythep6ints N2 , N3 .... . sothat h = ( b -a)/2n

and let V1 N2 V3 N3 ........ . be the ordinates at N 2 ,N3 ........
Then PL = y1, P2 N 2 = yz. P 3 N 3 y , . . . Through PP2 P3 draw a
parabola having its axis parallel to the y-axis, and let its equation
referred to parallel axes through N2 (a + h , ) be
y=a+bx+cx2.
Then, the area bounded by the parabolic arc PP 2 P3 , the or-
dinates of P. P3 and the x-axis ( such to be called hereafter short-
ly as the area under the parabola)
'I
(a + bx + cx 2 ) dx = 2h(a +4chZ ).. (2)
-h

Since P (-h,y 1 ),P2 (O,yl ),P3(h,y)arcpointsonthe

parabola (1),
Y, =a_bhich 2 , y 2 =a,y3 =a i- bhi-ch2

from which we get a = y2,c = 2h 2 t Y'

 A REA S OF PLA NE CURV ES 249

from (2), area under the parabola = - h (y1 + 43(2 + Y

under the curve
Now, area of the 1st strip (ordinates y,, y,, 3(3 )
x) is approximately area under the parabola
y = f (
_IL I A
+ 'Y z +3f
Similarly, area of the 2nd strip (ordinates ya. 3(4, ys ) under the
curve is approximately = h (ya + 4y 4 + 3(5 ) area of the 3rd
strip (ordinates ys, y ' . 3(7 4 under the curve is approximately
= - It (ys + 4y + y, ) ; and area of the nth strip under the curve
is approximately
= J h(y,. 1 + 4y 1 + Y b ).

summing all these, area under the curve, i.e., y dx is appro

5
xtmately
+ y2.) + 2(3( 3 +3(5 + •.. +3(3-i )
+4(3(2 +3(4 + . + )).
Note. It should be noted that the closer the ordinates, the more ap-
proximate is the value.
Simpon's rule is sometimes called Parabolic rule.
=54.60;verify
Ex. Given e l =Le =2.72.e 2 =7.39,e 3 =20.09.e 4

Simpson's rule by finding an approximate value of £ dx. taking 4 equal

J
intervals, and compare it with its exact value.
Hence,a = 0,b = 4,n = 2,6 1,y =f(x) = c'
by Simpson's rule we get the approximate value

It (( y, ys ) 4 4 ( Y • y0l
= •!,, h C( e" e4 .- 2e 2 + 1(r + e. 3 )l

ii Ii 4 'i + 2 ' 7 39 + 4 ( 2. + 2009 )I

53.87

Exact value =e 4 - 1 54 60 - I = 53.60

[ e'
error = 53.87 - 53.60 = 0.27 ( approximately)
 "U INTEGRAL CALCULUS

EXAMPLES X
1. Find the area of a hyperbola xy = 0
bounded by the x-axis,
and the ordinates x = a,x =b.
2. Find the area of the segment of the parabola y =(x - IM x)
cut off by the x-axis.
3. Find the area bounded by the x-axis and one arc of the sine
curve y = sin x.
4. In the logarithmic curve y = ae',
show that the area be-
tween the x-axis and any two ordinates is proportional to the dif-
ference between the ordinates.
5. Find, by integration, the area of the triangle bounded b
y the
line y = 3x, the x-axis and the ordinate x = 2 . Verify your result
by finding the area as half the product of the base and the altitude.
6. Show that the area bounded- by the parabola 'Jr +
'ly = Va
and the co-ordinate axes is - a,
7. Show that the area bounded by the semi-cubical parabola
y 2 = ax I and a double ordinate is . of the area of the rectangle
formed by this ordinate and the abscissa.
S. Show that the area of
(i) the astrojd x 213 + y 2/3 = a 2/3 is 7La2
.
(ii) the hypo-cycloid
) + 01 ) ... = I is .itab;
(iii) the evolute (ax) 2/3 + (by) 2/3 = (a 2 - b 2)2/3 is - It (a2 - b 2) 2

9. Find the area enclosed by the curves : (a > 0

(I) x (I + f2 ) = I t2 ;y (I + t 1 ) = 2t
(ii) x = 3 cost ;y 2 sift
(iii)x = a cost(I - cost);y = asint(I - cost).
(iv) x a (2 Cost + cos 2t);y = a( sin t + sin 2t).
 AREAS OFPLANE CURVES 21
Ex.X

10. Find the area of the segment cut off from y 2 4x by the
line x.
ii. Find the area bounded by the curve y 2 = x' and the line
y = 2x.
I +y 1
12. Find the area of the portion of the circle x
which lies inside the parabola y 2 = I - x
y 1 = 4ax
13. (i) Show that the area bounded by the parabolas
and x 1 = 4ay is --a2
(ii) Find the area bounded by the curves
 + 4x -4  = 0.
y 2 - 4x - 4= 0 and y t
=4y divide the square
14. Prove that the curves Y 2=4x and x' into three equal areas.
bounded by x = 0 , x = 4 , y = 0, y = 4 
meet at the origin 0 and
15. The curves y = 4x 2 and y 2 = 2x 
forming a loop. Show that the straight line OP
at the point P.
divitles the loop into two parts of equal area.
2 + 2y 2 = a1
16. (i) Find the area included between the ellipses x
and 2x 2 +y t =at
(ii) Show that the area common to the two ellipses
x2
_+ V2
.-.
T2 b2 = 1 and_+=1(l2>t7)

2a
is 2ab tan -' b
(a > 0)
17. Find the area of the following curves
(i) a 2 y 2 = a 2 x 2 -
Ex. 2, Art. 10.4. 1
(ii) (y - x ) = a' -  . ( See
X2

(iii) (x 2 + y2 )2 = a(x -y' ).
(iv) (x 2 + y2 )2 = a'x 2 + b2y2
(Transform (iii) and (iv) to I'olar. I
(v) x = a cos O + b sin O.y = a' cos O +b' sin O.
(vi) z a sin 2t, y a Sin t.
 252 INTEGRAL CALCULUS Ex. X

18. Find the area of the loop of each of the following curves
(a > 0)
(i) y x (x - 1)2

(ii) ay' = x 2 (a - x)
(iii) y 2 = x (x + a).
I- •1 ..t2
(iv) x = 1+t 2,y =t I+12,
______

(v) X = a(1 - t 2 ),y = at(] _i2),(_I 15

19. Find the area of the loop or one of two loops ( where such
exist ) of the blowing Curves (a> 0
(i) x(x 2 + y 2 ) =a(x2 - y 2 ).
(ii) y 2 (a 2 + x ' ) = x 2 (a2 - x 2 ).
(iii) y'(a - x)= x 2 (a + x).
(iv) y 2 = x 2 (4 - x 2 ).

(v) x2 = y 2 (2 - y )

20. Find the whole area included between each of the follow-
ing curves and its asymptote: (a > 0)
(i) x 2 y 2 = a 2 (y 2 - x 2 )
(jj) y 2 (a - x )= x3

(iii) y 2 (a - x) =x 2 (a + x).
(iv) x 2 y 2 + a'b' = 42y2
(v) xy 2
= 4a,(2a - x)
21. Find the area of the following curves: (a > 0)
(i) r = a sin 0

(ii) r 2 = a 2 sin 20;r a = a 2 cos20.

(iii) r 2 (a 2 sin 7 O + b2 cos 2 O) = a2b2
 Ex. X AREAS OF PLANE CURVES 253

(iv) r = a sin 30
(v) r = a( sin 20 + cos 20).
(vi) r 2 = a 2 cos 2 O + b sin 10
(vii) r = 3 + 2 cos O.

22. Show that

(i) the area included between the hyperbolic spiral ro = a and
any two radii vectors is proportional to the difference between the
lengths of those radii vectors;
(ii) the area included between the logarithmic spiral r = e
and any two radii vectors is proportional to the difference between
the squares of those radii vectors.
23. Find the area of a loop of the following curves : (a > 0)
(i) x 1 + y 4 = 2a 2 xy. [Transform to Polar.

(ii) r 2 = a 1 cos 20.

(ill) y2 = a 2 cos 40.

24. Find the area of the ellipse 9x 2 + 4y 2 - 18x-- 16y -1 = 0.

25. If for the curve x ( x 1 + y 2 ) a (x2 - y 2 ) (a > 0),
A be the area between the curve and its asymptote and L be the
area of its loop, show that A + L = 4a2
26. Show that for the curve
y 2 (a + x) -x 2 (3a - x) (a >0),
the area of its loop and the area between the curve and its asy-
mptote are both equal to (3 '13) a2
27. Show that the area Included between one of the branches
of the curve x 2 y 2 a 2 (x 2 + y 2 ) ( d > 0) and the asympi te is
equal to the total area o( the curve (1 1 + y2 )2 = a 2 (2 - y 2 )
(a> 0).
254 INTEGRAL CALCULUS Ex. X

28. If p = f ( r) be the equation of a curve, show that its

area =2j- i -1( r 2p2i. dr
taken between the proper limits.
29. If p f ( w) be the equation of a curve, show that its
area = I C LE
J p ( p + du/ ) du
taken between the proper limits.
30. U) Show that the sectorial area of the equi-angular spiral
p = r sin a included between the zwo radii vectors r 1 and r2
is .( r2 2 - ) tan a

(ii) Show that the area of the lemniscate a l p 0 is a

For half loop rvaries from 0 to a.J

31. Find an approximate value of

0.2
(1 - 2x 2 >113 dx, taking 2 equal intervals.
J o

Given f ( 0.1 ) = 0.99334,f(0.2) = 0.9725

where f(x) = (I -2x2)1/3.
32. Find the approximate value of
2
dx
, taking 10 equal intervals, and calculate the error.
ft X

Given f( 1.1 ) = 0.90909 f(1.6) = 0.62500

1(1 . 2) = 0.83333 f(1.7) = 0.58824
f(1.3) = 0.76923 1(1.8) = 0.55556
f(1.4) = 0.71429 f(1.9) 0.52632
f(1.5) = 0.66667
I
where f (x ) = -;
 Ex. X AREAS OF PLANE CURVES 255

33. Evaluate
2
'( 2 + sin x) dx, using 4 equal intervals,
5
given when x = 00', 2230', 45'O', 6730', 90 0',
J( 2 s- sin x) = 1.414, 1.544, 1.645, 1.710, 1.732.

34. Obtain an approximate value of

-__dx
2 taking 4 equal intervals, and hence obtain an
5
approxima°te value of n correct to four places of decimals.
35. A river is 80 metre wide. The depth d in metre at a dis-
tance x metre from one bank isiven by the following table:
x = 0 10 20 30 40 50 60 70 80
d=0 4 7 9 12 1514 8 3
Find approximately the area of the cross-section.
36. Use Simpson's rule, taking five ordinates, to find approx-
imately to two places of decimal the value of

- 1/x)dx.

A NSW ERS

1. c log .2. 41. 3. 2. 9. (i) R. (ii) 6it.

ita . ( iv) 6ita 2. 10. 4 . 31. 3.2. 32. 47t 1-4.

13. (ii) . 16. (I) 2I2a 1 sin' -. 17. (i) 4a 2 (ii) 7r2

(iii) a 2 (IV) 4 it (a + 6' ) . (v) it (at" - a'b) (vi) a 2

18. (I) -. (ii) j!.a 2 (iii) !-a 5/2 (iv) 2 - 4 it (v) -a 2
2' INTEGRAL CALCULUS

19. (I) 2a 2 (J - I) . ( ii) a 2 2 - 1)(Iii)

. 2a 2 -.1
(1 . (iv) !_
(v) -42 . 20. (I) 4a 2 () !no 2
(iii)2a2 (1 + It . ( v) 2itab
2 21. (i) +ita 2 2;
(ii) a a 2 ( jjj ) nab . (iv) '. ,ta I
(v) na 2 NO 1. 7c (a 2 + b2 ) . (vii) 11 it

23. (i) I na 2 . (ii) a2 (iii) a2 . 24. 6n. 31. 0.1982.

32. 0.69315; error = 0.00001 . 33. 2546. 34. 3.1416

35. 7101711 . 36. 0.84.

 CHAPTER XI
LENC1'I-I OF PLANE CURVES

t Rectification
11.1. Lengths determined from Cartesian Equations.
We know from Diffcrc • iai Calculus that if s be the length of
the arc of a curve measured from a fixed point A on it to any point
1
1 , whose Cartesian "-ordinatcs are ( a, b ) and ( x, y ) respec-
tively, then

ds
= sec, t tartly '. dxl'
Tr
i denoting the angle made by the tangent at P to the x-axis.
Thus, we can wr

s =$Ji ,() dX + C,
dx
2
where - is expressed in terms of from the equation to the curve
and C is the integration constant. If the indefinite integral

J.Ji +(ax dj

be denoted by F ( x ) , then since s = 0 when P coincides with A ,

i.e., when x = a, we get
0 F(a) +C,whcnce C = —F(a).

Thus, a = F(x)— F(a) r f  [i + (1

\ dx/
dx.
*
The process of finding the length of an arc of a curve, i.e., 'of finding a
straight line whose length Is the same as that of a specified arc' is called
Rectification. For the definition of the length of an arc of a curve, see
Authors' Differential Calculus, Appendix.
Inlegral Calculus (main) -19
 zs* INTEGRAL CALCULUS

Hence, between two points having x 1 and x2 as abscissa the

length of the curve is given by

- s
.12
1jTT
=Ja
ix
-' a

=f1
.JdX. (1)

If it be convenient to get and accordingly , in terms

of y, instead of x, from the equation to the curve, we can use
the result

dsI /dx\2
—=.'I1+I--
dy v CL),
whence the length A P is given by

dx 2
=J 1+ (;)
where dx is expressed in terms of y
Also the length of the curve between the two points whose or-
dinates are y1 and yi respectively will be
=f2
III- 1+ ( . )2 dy. (2)

If both x and y are expressed in terms of a common variable

param'ter t and so s is also a function of I, we can write
ic - \ 2

di it NI1 + dxJ
(
dx

(.i? (since
\ ) + ' dt/ di dx dl)
 LENGTHS OF PLANE CURVES 259

Thus, as before, the length of the curve between two points on

it for which t = f and t = 1 1 respectively will be given by

dx )id .
Si - Ii
JN (j) + (3)

All the above cases can be included in a single result in the dif-
ferential form

= Jdx1 _
. dy 1 (4)
where the right-hand side Is expressed in the differential form in
terms of a single variable from the given equation to the curve
This , when integrated between. proper limits, gives the desired
length of the curve
Note In the above formula (1), (2) and (3) • it Is assumed that
dy dx di dy
a- . are all continuous in the range of Integration.
11. 2. Illustrative Examples
Ex. 1. Find the length of the arc of She parabola y 2 r 4ax measured from the
ertex to one extremity of the latus rectum
-
lere 2y
dx
, or, dx = y
= '(4ax)2a
"Jx

The abscissie of the vertex and one extremity of the latus rectum are 0
and a respectively. Hence, the required length

S =Jj 1 +()
di 5\J••-
a
I x+a
+ I"

= ;x (z + Z ., t mIo('x ,(x

= a { I2 t log (1 + 42)).
260 1N I IGRAJ. CALCULUS

Ex. 2. L)ete-,-iine the length of an arc of the cylci.l z a(0 sin 0 ),

y a( I cos 0),mesuredfrom the vcc(ie,thec':in)

I lore,
ds
-- = '
!7'
-.. +
Id'.2
-
dO '4 dO/ \d O

a 2 =
cos 0 ) cos 0

Also at the origin 0 - U Hence the requited length, from 0 -- '' to any
point 0

2a COS-8d0 =4a sin 0

J3

Cur. 1. Since at th -"mity of the cycloid ( i o , at the j =

we have 0 tt there. Thus, the length of a complete cycloid bning : ible th

length irom the vettex to the extremity is?. la sin -In =Fj

Cor. 2. s' = 164 I sin 2 - 0 = 8a.a (1 cos 0) Say

E. 3 F :nd tIn whole length of the loop of the curve

3ay 1 = - a)2.

\e flutiCe here that, for negat i ve values o f x , y is ir.ii nd '

there is no 1,art of the curve on the negative side of the x-a xi \. at
- poin t s where x =
we havey 0. I'i .'v:-"- ..c two
points, for every vaIu.. .i
are equal and oppositi. ot v,
-
-' a loop being thereb y '- - fl ... d. F. r
x —. , each value of x gee -ian a
has two equal a -J
values, and with r -t asing,
'... continually increases n rnag
nttuje. Th, rve is t-* ,s traced
as in the adie' iiy tigsii cx-
i,'mltiei it '. ,op are given
1. , ra. I'd xa.

)tfl the eq tion to th, curve

dv
6ay (a — a) + 2x(x - a) (x — a)(3x- a);
 LENCflIIS 01- PLANE CURVES 261

ds I
(Y / (
dr =
_
1. x - a)2(3x- a)2

= (3r -a) 231 + a

/ * 12ax = 2.! (3
- ax )
the h It length of the loop is
U
*o + a I
x + a 21
a
2 ( . , JX) 2\(3a) { 3
J 10
I F., 1 2a 2
'13a.
The hcie length ol Ike loop, therefore, from the symmetr of the curve,

11.3. Lengths determined from polar equations.

From the formulae

dO
tan =r - -- , cos 0 = dr , sine Lo
in Differential Calculus, where s represents the length of the orc
of a curve from any fixed point A of it to a variable point P whose
polar co-ordinates are ( r 0 ) and 0 dencts the angle between the
radius vector to the point and the tangent at the point, we can write

I ds --
- cosec = I i + cot
=_ji +— (-)
I /Jy\2

ds /dr2
whence '.
do r2 + ()
Again,

ds
sec 0 ' i + tan 2 Q = (2)
dr
From (1) and (2), the length of an arc of the curvc5 can be ex-
pressed in either of the forms

.62 ' dr
(o) dO,
262 INTEGRAL CALCULUS

or, a
=f fZ.
C'
Ji dO 2
+ r 2dr,
dr ,

where r, 0 1 and r2 ,0, are the polar co-ordinates of the extremities

of the required arc. In the first form, r as also j",—
' are expressed in
terms of, 0 front the given polar equation to the curve. In the second
from, - Is expressed in terms of r.

Both (1) and (2) can be combined in a single differential form

do = 1dr 2 + r2d92

Note It is assumed in the above formula that , are continuous in

the range of integration.
Ex. Find the perimeter of the Cardioide r a (1 - cos 0 ) , and show that
the arc of the upper half of the curve is bisected by 0 = I n. C. P. 1949 I
tsr
Here, since r = a (1 - cos 0), - = a sin 0.
do
Hence, the length of any arc of the curve, measured from the origir.
where 0 0, to any point, is given by

5=5 I 4,

p9
a 2 s in'8d0
=J 'La' (1 - cos O)' +
0

0
= 4a cos 0
1] r 4a(1 cos

Thus, the leth of the upper half of the curve, whkh yc, ..nd' .
e=o toe =n, 114a(1_ Cos jR) =4a. I See Figure Fri Art IOt.I
The whole perimeter is clearly double of this, and thus Se
Again, the length of the curve from 0 0 to 6 x is
4a( I - cos n) = 2g , and so the line 0 bisect the arc of the
Upper nail of the curve.
 LENGIIIS OF PLANE CURVES 263

11.4. Lengths determined from pedal equations.

dr
From the formulae cos 0 and p = r sin $ in Differential
Calculus, we can write
ds I 1 1 r
dv cos$'I(1_sin2$)J
2)'
whence the length of an arc of the curve extending from r =
to r = r2 will be given by

rdr
s J .I(i.1 -p2)
where p is to be replaced in terms of r from the given pedal equa-
tion to the curve.
Ex. Find the length of the arc of the parabola P2 ar from r = a
to r = 2a.
The required length is given by
24
rdr F
- vr,2 _p2 1(,2-a,)
fa __ a

= { 1r-
I - a, +i log ('1r +4r

= a'2 • alog(42 + 1) =a 142 +log(1 + 42)J.

11.5. Length of an arc of an Evolute.

We know from Differential
Calculus that the difference be- P
tween the radii of curvature at
two points of a given curve is
equal to the length of the cor-
responding arc of its evolute.
Q
Thus, if Pi and Pz be the 0;
X
radii of curvature at P and Q of Fiji
 INTEGRAL CALCULUS

a given curve PQ, p and q being the corresponding points n the

evolute, the length of the arc pq of the evolute = Pi - P2
In fact p, q are the centres of curvature and so and Q are
the radii of curvature at P and Q of the curve PQ , and if the
evolute be regarded as a rigid curve, and a string be unwound from
it, being kept tight, then the points of the unwinding string describe
a system of parallel curves, one of which is the given curve PQ, of
which pq is the evolute. PQ is called the involute of pq.
Es. Ca(ulate the entire length of the evolule of the ellipse

T2 b' lCP.l98l

U
Fig.3
a , b, a', b' being the centres of curvature of the ellipse at A , B, A', 8'
respectively, the evoluie, as shown in the figure, consists of four similar por-
tions, the portion apb corresponding to the part APB of the given ellipse
Now, from Differential Calculus, it is known that at any point on the
ellipse, the radius of curvature
alb'

where p is the perpendicular from the centre on the tangent at the point.
Thus, the length of the arc apb of the evolute
a l b' a2b2
= Pa - PA
= -. ' = Ta - b1

Hence, the entire length of the evolute of the ellipse

b1
4
 LENGTHS OF PLANE CURVES 265

11.. Intrinsic Equation to a Curve.

If $ denotes the length of an arc of a plane cui y e measured
from some fixed point A on it up to an arbitrary point P. and if
he the inclination of the tangent to the curve at P to any fixed
line on the plane (e.g., the x-axis ), the relation between s and
is called the Intrinsic Equation of the curve.
It should be noted that the intrinsic eqution of a curve deter-
mines only the form of the curve, and not its position on the plane.
(A ) Intrinsic Equation derived from Cartesian Equation.
Let the Cartesian equation to
thecurvebe y = f(x).Then y
denoting the angle between the
tangent at any point P and the
x-axis,

tan 'y = f'(x). ...(1)

dx
Fig4

Also, s = arc AP
=5 a
+ ( Y2 ix
(f'(x))dx = F(x),say, ... (2)
=s:
'a' denoting the abscissa of A, and ' x that of P
Now, the x-elirninant between (I) and (2), ( which will be a
relation between s and Nfl,will be the required intrinsic equation
of the curve.
If the equation to the curve be given in the parametric
form x = f(t),y = Q(t),wc can write
AM - - /d = 4i'(t) ... (1)
t an - dx - dl I dl
T7—( 't)
 26 1TEGRAL CALCULUS

Also dx ' 1 l
s
= f-J (-) + ( ) dt

=5 V IPMP + tG'(t)
= F(t),say, .. (2)
where t, is the value of the parameter i at A
The -eliminant between (1) and (2) will be the required intrin-
sic equation to the curve.
(B) Intrinsic Equation derived from Polar Equation.
Let r =f(G) bethe
polar equation to a curve.
Let 0 denote the angle
between the tangent and the
radius vector at any point
P(r,0), W the angle made
by the tangent with the ini-
x tial line, and s the length of
the arc flp where A (a,a)
FigS
is a fixed point on the curve.

Then, tan L)
= r dO
-
Y =0+t,,
f '( 0)
dr ... (1)
... (2)
8
and s N f (dry dO
dO
a

= 5v (f( 0))' + If' (0)1 1 dO = F(0),say. ... (3)

0

Now, eliminating Q and 0 between (1), (2) and (3), we get a

relation between s and 'v, which is the required intrinsic equation
of the curve.
 LENGTHS OF PLANE CURVES 267
(C) Intrinsic Equation derived from Pedal Equation.
Let p = f( r) be the pedal equation to the curve.
Then, as in Art. 11.4,
V
I
F
rdr( rdr
say. ...(1)
s_J(r2p2)_J(r2_(f(r)l7F(
Also, from Differential Calculus, p denoting the radius of cur-
vature,
ds dr r
Eliminating r between (1) and (2), we get a relation of the
form
ds 1 ( ds
=(s), or, LW --
ds s) =j TT5
7W_
wt'ch, when the right side is inegratcd, wil give the reqired in-
trinsic equation.
11.7. Illustrative Examples.
tt. I Obtain the intrinsic equation of the Calerzary y = c cosh in the
form s =c tan w

Here, tan W ==A'

dx
sanh xc-. (1)
Mao, measuring s from the vertex, where r = 0,

So
JT dx I

=J:tszn1' --
x
cosh .-- dx = [ c srnh - c sinh -
= C
0

Hence, from (1) , s = c tan w

 268 INTEGRAL CALCULUS
Ex. 2. Obtain the intrinsic equation of the cycloid
x = a(O+ sin6), y = a(1J -coO)
taking the vertex as the fixed point and the tangent at that point as the
fixed line.
As shown in Ex. 2, Art 11.2, the length of the arc of the above cycloid
measured from the vertex is given b)
$ =4a sin '
a sin O
Also, tan i =dt, t a n 02-
d 1dx
- =aO- / dO-
dx a (1 cos
cot, 0 )

= e .Hence, front (i),s =4 sin ,

which is the required intrinsic equation
Ex. 3. Find the inS rins:J equation of She Cardtmde
r = a (1 - cos 0
the arc being mea..ured from the cusp ( i.e. u here 0 = 0 ).
Here, V =8+ 4' . (1)
dO 1- co s $
and tan , i.e., r— dr = sin = tan 62-
• =8. . . (2)
Also by the Ex., Art. 11.3. we have
+ (fr)2
S = 4. ( Icos) ...
Since, from (1) and (2), w = 0 +0 = . 0, i.e. 0 =
from (3), s = 4a (1 - cos I W the required intrinsc equation.
Ix. 4. Find the Cartesian equation of the curve for which the intrinsic equa
lion is S=

HereTdx
-
y
dxds
-
= 7s diq
cos s,a
dx=a cos ,d. ..x=a sin W+c ... (i)
Again, dW ds TV
fl

• dy = a sin S41 d. .. y = - a cos w 4 d. . . . (2)

 •(;rHS OF PLANE CURVES

From (1) and (2) ehrninating • i, we get

- - d) the required Cartesian equa z. ,.

EXAMPLES XI
1. Find the lengths of the following
(i) the peri1ntter of the circle x + y = a2

(n) the arc of thi catenary y = - e/a + e -x/a)

3 (
y from the
vertex to the point ( x y

(II) t he .criincter of the atroid x 213 y 2/3 = a 2/3

(Iv) the perimeter of the hypo-cycloid ( )?• ) ' r 1;

Q.
v) the periJ. :tor of the evolute
(ax)"' + (by)2/3 = (a2 - b
2 )

the arc of the semi-cubical parabola ay 2 x 3 from the

cu'todny point (x,y).
.l be -he length ofan arc ol 3ay 2 x ( x -n) measured
=

frunthur lto the point (x.y), show that 3s 2 =4x 2 +3y1.

. ;hw that the length of the arc of the parahoa y 2 = 4ax
w, h k ntercepttd between the points of intersection of the para-
bo mo the straight lino 3y = 8x is a (log 2 +f)
•L that the complete perimctci of the -irve
_f? 2
r =+ 2 Y = is 27r

. If for a curve
sin i- y cos o =f'(0)
and x cosi) y sin O r'
show that s f(0) i-f" (0) + c, where c is a constant
 INTEGRAL CALCULUS Ex X I
270

6. Find the length of the arcs of the following curves

(I) x = e° sin 91 from 9 = () to 0 = - 7j

= cc

(ii) X fl (cos 9 + 0 sin 8 ) } from 0 = 0 to 0 = 0,

y = a(sin 8 O Cos O)

(iii) x = - sin 28 ( l t cos 20

Y r c cos 2O(l - cos 2O)
from the origin to any point.
7. EThow that the perimeter of the ellipse x = a cos 9 , y
= b sin 0, is given by
,' 1\' e 2 il.3
2alttl -)
\2 e 4 /1.3.5\ 2 e'
- i) -
8. Compare the perimeters of the two conies
2 2 2 1
- = I and +L =
9 7 36 28
9. Find the lengths of the loop of each of the following cur% CS:

(i) 9y 2 = ( x + 7)(x -,- 4)2;

(ii) x =t 1 ,y = t -1 t
10. Find the lengths of the following
i) a qiadrant of the circle r = 2a sin
,ii) t , o av, ,f the parabola r (1 + cos 0) = 2 from 0 0
to -ti-
ti . ir (the .i-angut,ir spiral r = ac e ot between
t.e ra.iil v, tors a'
11. It z be the ienth ol : curvi' = a tanh -10 between the
origin and 0 - 2i, azd the area between the same points, show
that A = a - Or
 Ex X I LENGTHS OF PLA NE CURV ES 271

12. Show that the area between the curve

Y =i (e 'i

the x-axis and the ordinates at two points on the curve is equal to
a times the length of the arc terminated by those points.
13. Show that in the ast.roid x" 3 - y 117 = 02/3

(I) 5oe X'1

(ii) p' + 42 = 6as,

s being measured from the point for which x = 0.
14. Show that
(i) in the cycloid x = a ( 0 + sin 0) y = a (1 - cos 0 ),
p 2 + s =16a1,
the arc being measured from the vertex ( where 0 = 0
(ii) in the catenary y = c cosh (x / c),
yl = cp = C 2 + S2
the arc being measured from the vertex;
(iii) in the cardioide r = a (1 + cos 0 + 9p 2 = 16a1
the arc being measured from the vertex (i.e., 0 = 0)

15. Show that the length of the are of the hyperbola xy = a 2

bt ween tht points x b and x = c is equal to the arc of the curve
p (a' 4 y4 ) = a'r 2 be;ween the limits r =b and r =
c.
16. Show that the length of the arc of the cvolute
27ay = 4 ( x 2a )3 of the parabola y 2 = 4ax, from the cusp
to one of the points where the evolute meets the
parabola, is 2a (3 43 - 1)
17. Find the intrinsic equation of each of the following curve.,
the fixed point from which the arc is measured being indicated in
each case
(i) the parabola y2 = 4ax . (vertex),
00 the astroid X1/3 + y2/3 =a 213 .( one of the cusps),
 INTEGRAL CALCULUS t xi
272
(CU,
(iii) the semi-cubical parabola ay 2 =
(iv) the curve y = a log sec ( x / a ) (origin),

() the equi-angular spiral r = ac °' (point 0),

(vi) the involute of the circle, viz.,

[TJu2 a
o cot(point a,O)
a r

18. Find the intrinsic equation of each of the following curves:

r sin a,

(ii) p 2 = __ a1

19. Find the intrinsic equation of the curve for hich thc tongth
of the arc measured from the origin varies as the square roc of the
ordinate. Also obtain the Cartesian co-ordinates of any point n the
curve in terms of any parameter.
20. It s = c tan W is the intrinsic equation of a curve, show
that the Cartesian equation is y = c cosh ( x / c' , given thai when
= O,x = 0 and =c.

ANSWERS

(e x /a-
e (iii) 6a;
i. 0) 2ita; OD-141
-141
a 2 + ab +b2 (v) 4 (a2 !.);
(iv) g+b a

8a 9x \ 2
-
+
(042t'' 2I, (ii)aO2 (iii)Csin
6.
2. 9. (1) 4 43 4 43

( 'a , • 2 s 1); (iii) ( r2 - ) 5CC Q

17. s - co w co I V 4. a tog C osec ji + cot iv);

-, 32'in 2II
 LENGTHS OF PLANE CURVES 273

(Iii) 27s = 8a (sec 3 W - 1); (Iv) s = a log tan (j w +

(v) s =aeca(e(V)c0. 1); (vi) a =!aV2
18. 6) a =Cc V CIt I ; 00S = l aw,.
19. a =4asin; x = a(+ sinO), y =a(I - cosO).

Inlegral Calculus (main) -20

 CHAPTER XII
VOLUMES AND SURFACE-AREAS OF SOLIDS OF REVOLUTION

12.1. Solids of revolution, the axis of revolution being the

x-axis.
Let a curve LM, whose Cartesian equation is given by
y = f (x) say, be rotated about the 'v -axis so as to form a solid of

Fig.!
revolution, and let us consider the portion LL'M'M of this solid
bounded by x = x and x = x 2 . We can imagine this solid to be
divided into an infinite number of infinitely thin circular slices by
planes perpendicular to the axis of revolution 0Y3 . If PN and P'N'
be two adja.ent ordinates of the curve, where the co-ordinates of
Pand P are ( x , y) and (x + A x ,y + A y)respectively, the
volume of the corresponding slice, which has its thickness Ax , is
ultimately equal to ity2 A X .
Hence, the total volume of the solid considered (bounded by
x= x1 and x = x2 ) is given by

- xl
V = L t Z my 'x
-. 0
Ax
=7rj y1 dx.

• Strictly, the volume of the slice between iy1 2 Ax and n Y 2 2 Ax where

and y2 are the greatest and the least values of y within the range PP and
thus equals ny I !.x , where y lies between y and Y 2 and is thus the
ordinate for some point within the range PP' (not necessarily of P). Thus,
Li Iny 2 A x = I y 2 dx. [See A rt. 6.2. Note 2.1
 VOLUMES AND SURFACE-AREAS 275
length measured upto P from any fixed point on the curve LM,
the surface-area of the ring-shaped element generated by rotating
PP' is ultimately 27ty. A s.
Hence, the required surface-area is given by

S = U L ( 2nytss)= 2.

Is , s
5 y ds

being the values of s for the points L , M

=2EfY' ]i+
. (I)i.
Cor. 1. When the axis of revolution is the y-axis, and we consider the por-
tion of the solid bounded by y = y1 and = y 2 respectively,
YZ
V =7E x2dy,
yl
and S 2715 xds 2115 x\j1 + (!)'
Coy. 2. Even if the curve revolved be given by its polar equation ( the
and the portion of the volume con-
axis of revolution being the initial line ),
sidered be bounded by two parallel planes perpendicular tc the initial line,
we may change to corresponding Cartesian co-ordinates, with he initial l'ne
as the x-axis.bywriting x = r cos O, y = r sin 0.
Thus.

V = 71 y2x r2 sin 2 O.d(r cos o),

5 22f _________
S yds = 2n J rsin8.'/dr + rtdO2,
where r is expressed in terms of 8 from the given equation of the
curve, or, if convenient, we may use r as the independent variable and
express 8 in terms of r from the equation, the limits being the cor-
responding values of r
 INTEGRAL CALCULUS
276

A lternative proof of
( i ) V olume of a solid of revolution.
Let a curve CD, whose equation is y f ( x ) , be rotated
about the x-axis so as
/ to form a solid of
revolution. To find the
volume of the solid
generated by the revolu-
tion about the x-axis, of
tte area A BDC bounded
by the curve y = f( x), the
ordinates at A and B an
the x-axis, let a and b be
V
X the abscissa of Cand L).
Fig -
Divide A B into n
equal parts, each equal to h , a nd draw ordinates at the points of
division. Let the ordinates at x a + rh and x = a + ( r + I ) h
be PL and QM, and let us suppose y goes on increasing as x in-
creases from a to b
Draw PN perpendicular on 15m — QR perpendicular on
LP produced . Then the volume of the solid generated by the
revolution of the area LMQP i c es in magnitude between the
volumes gpnerated by the rectang!os LMNP ad LMQR
i.e., between ht[f(a+rh)Jlh and n If( a+(r+I)h}J2h
Hence, adding up the volumes generated by all areas like
LMQP, it is clear that the required volume lies in magnitude
between

it L [f(a +rh)l i h and n 1 If{ a t (r '.- l)h)12h.

a . r_a
Now let n -, , so that h —+ 0 ;thenas the limit of each of the
above two sums is
 VOLUMES AND SURFACE-AREAS 277

ie.,$byzdx,
flfa1f(X12dX
it follows that the required volume is also equal to this
definite integral.
ii) Surface area of a solid of revolution.
Let the length of the arc from C upto any point P ( x, y ) be s
and suppose that the surface-area of the solid generated by the
revolution of the arc CD about the x-axis is required. As in the case
of the volume, divide flL into n equal parts, each equal to h , and
erect ordinates at the points of division. Let the ordinates at
= a -, rh.arid x = a + ( r + 1) h be PL and QM, and let the
ar PQ be equal to 1 The surface-area of the solid generated by
the revolution of 1.MQP about the -axis lies in magnitude be-
tween the cuo\ed surf,,ce of two right circular cylinders, each
of thickness I no of radius PL and the other of radius QM,
i.e. , between
2irf(a + rh)l and 2itf(a + (r + 1)h)l.
Hence, adding up all surface-areas generated by elementary
areas like PQ, it is clear that the required surface-area lies in mag-
nitude between

2. f(a +rh)h and 2nff(a + (r + 1)hjh.

Now let n -- oo, so that h -4 0; then I I h tending tot, the

limit of each of 'he above two sums is
bb
ds yds.
f(x) —dx, i.e., it
j dx -
Hence, the required surface-area is also equal to this definite
ral.
 278 INTEGRA L CA LCULUS

12.2. Illustrative Examples.

Ex. I. Find the volume and area
of the curved surface of a paraboloid of
revolution formed by revolving the
parabola y 2 = 4ax about the x-axis,
and bounded by the section x = x.

Here, y = 2.f.

dxV x
Now the required volume Fig.3

xI Pr1
V = , y 2 dx= U. d. =2*ax, 2 =*xy12
0 0
(where yi Is the extreme ordinate, so that y1 2 = 4ax1 2 )
=2 y1 =
(the volume of the corresponding cylinder, with the extreme circular section as
the base and height equal to be abscissa).
Also, the required surface-area

S =2*5 Y\/I +()2 dX =21

Jo
= 41tIa5'Ia
+xdx = . x4a((a+ x )3/2 -a311).
Ex. 2. The part of the parabola y 2 = 4ax bounded by she latus rectum revol-
ves about the tangent at the vertex. Find the y
volume and the area of the curved surface of the L
reel thus generated.
Here the axis of revolution being they- X
axis, and the extreme values of y being evi-
dently i2a, L

the required volume Fig.3

•.2a .21
v=xI z?d_xf -1dy Esince y2=4ax)
J2I '-21 j
 VOLUMES AND SURFACE- p S 27
R (g)S
= 2 r- = -4 JT
Also the required surface-area

S =2nf xds = 21tf 'tJl +(y

1I
4a 4a since dx = 1
=2JJ dy2aJ

= 4xa2f tan 2 0 sec 0 dO I putting y = 2a tan 01

= (sec30-
4ita2f sec 20)dO

=4a 2 [+tanosec2o_taneseCo_.logtafl(
+ +O)J
r42 I42 -logcot . ,J =g2 [3'12-Iog('12 + 1)1.
Ex. 3. Find She vol ume and the surface-area of She solid generated by reolv-
ng the cycloid x = a (0 + sin 0) y = a (1 +
cos 9 ) about its base.
The equations, show that the cycloid has the x-axis as its base, the ex-
treme values of are given by 0 = * v, i.e.. x =±ax.
The required volume
- ax x
V=,t I Y 2 dxia 3 j (1+ Cos O)'4g
-x

=8&l3jCOS'!Od9_..a35R523

The required surface-area

5 =2ifyds = 2x fy .ldx I +dy2
 210 INTEGRAL CALULUS

= 21t a( + Cos 0) 4(a(1 +cose)de) I+(-aslnede)'

=2lra l J (I + cos 9)42(1

- + cos O)dO

=8jta 2 j 0
d
cos' 1 =87ta'.-8 = 6.4

Ex. 4. Find the volume and the surface-area of the solid generated by evolv-
ing the cardioide r = a (1 - cos 0 ) about the initial line.
Here, since the curve is symmetrical about the initial Iir,e, the solid of
revolution might as well be considered to be formed by revo ving the upper
half of the curve about the initial line. The extreme points of the curve arc
given by 0 = 0 and 0 = n.
The required volume
v=,tfYzaxiJr1sin2ea(rcoso)

=ira 3 f(1_ cos9)'sin 2 O.d((1 - cos$)cos0)

753JO (1- cosO) 1 sin 2 B(- sinO + 2sin0co8)d9

I x increases as 0 diminishes from n to 01

= 3f (1- z)(l - z 2 )(1 - 2z)dz [putting z= cose)

= . ,r43
The required surface-area

S =2nJY ds = 21rJr sin e.Idr 2 +

=2nf a(j_ cos 0) sin Q.(a sin 9dO) 2 +a 2 (1_ cos U)2d02
Jo

=2,w2 f" ( I - cos 0) sin 0'i2(i -cosO)dO

0
 VOLUMES AND SURFACE-AREAS 281

2
= 242ITa2 z31 dx I Putting x = I - cos 01

= 2I2,a2 . .(2) 5 / 2 _32,ta2

12.3. Solids of revolution, axis of revolution being any line

in the plane.
lfthegivencu LM be revolved aboutany line A B its plane,
Y p and the portion considered
of the solid of revolution
formed be bounded by the

I
planes perpendicular oA B
through the points A and
respectively, then RN
being the perpendicular on
X AB fom anoint P on
Fig.4 the curve, P'N' the con-
tiguous perpendicular, the volume of the portion considered
given by
AB
V =Lt En.PN2.NN' = PN2 (A N).
0
Also, the surface-area of the portion consiciez.d is given by
S = LtE2n.PN.( elementary arc PP' )= 2n PN.ds.

From toe given equation of the curve and of the line, AR, PN,
as also AN and ds are expressed B
in terms of a single variabc.
and the correspo'-ding values
of the variable for the points A
and B are taken as the limits
of integratior
Ex. A quadrant of a circle, of
a, revolves round 115 chord.
volume and the surface-area o A
id spindle thus generated. Fig.5
282 INTEGRAL CALCULUS

ly AP= J.
P being any point on the quadrant
2a sin Oand mLPAN =. LPOBwhere
APB,
m LAOP = 0 dear-
! !
PN = 2a sin O sin ( it - - 9) a (Cos
= ( . ii 0).

AN 2a sin !O Cos (!it - O)

(0 -) - Cos.!ltJ;
a 1sinit + sin(e -
Elementary arc PP = a .4'

Also, for the solid formed limits of 8 are 0 and it

Hence, V it fPN 1 .d(AN)
respectively.

pfR
=irasJ
0
{co5(o Cos !,t)1 Cos (9l)de

=ita3j [Cos i ( O_! it)_12 Cos 2(0I)+l(Ol

Cos (39 - f it) +cos(O -iit

)

- T2 (Cos(20 -fir) + U1d

= ita [sin(39
12 4 - sin (20 -fit)
5 1 i
T2 0
= (0-3it) +

.2
Also, S =2itJ
Jo
PNadO
f
2ita 2f 2 ( Cos (o - f it) _ Cos !it)de

r2ira1 [sin (8 -f ir) - 0j

'2ita
1 (-)
 VOLUMES AND SURFACE-AREAS 283

12.4. Theorem of Pappus or Guldin.

If a plane area bounded by a closed curve revolves through any angi#
about a straight line in its own plane, which does not intersect the curve,
then
(I) The volume of the solid generated is equa to the product of the
revolving area and the length of the arc described by the centroid of the
area.
(If) The surface-area of the solid generated is equal to the product of
the perimeter of the rez'olvirg area into the length of the arc described by
the cent roid of that perimeter.
Proof.
(I) Let 8A be any element of the area whose distance frorr the
axis of rotation is z Then, 0 being the angle through which the

Fig.6
area is rotated, the length of the arc described by M is zO
and hence the elementary volume described by the elk rnent
M is zO . &A.
The whole volume described by the given ar therefore
= r.z' 5A = OLz 5A = OZA t rorn Elementary Statics)
(where A is the total area of the curve and z is the distance of its
centrold from the axis of revolution I
= AZO = area of the closed curve x length of the arc
described by its centroid.
 284 INTEGRAL CALCULUS

(Ii) Let &s be the length of any element P1" of the perimeter
of the given curve, and z' its distance from the axis of revolution.
The elementary surface traced out by the element 8s is ultimately
Z ' 0 . 6s.

The total surface-area of the solid generated is therefore

Ez'0.6s = 9z . s Oz's (from Eler.cn'.ry Statics)

where s is the whole pernetcr of the curve, and z' the distance
of the centroid of this perimeter from the axis )

= sz'O = perimeter x length of the arc described by its

cent ro i d.
Note. The at,ve results hold even if the axis of rotat ii touch -s the
closed curve.

Es. 1. Find the volume and surface-area of e solid tyre, a being the
radius of its section, and b that of the core.
The tyre is clearly generated by revolving a circle of radius a about an
a,us whose distance from the centre of the circle is b
ihe centre of the circle is the centroid of both tt'r area of the circle as
also of the perimeter of the circle, and the length of the path described by it
is evidently 21tb
Hence, the required volume =na 2 x 2b z 2 2 a 2 b
and the required surface-area = 21ca . 21tb = 47t 2ab
Es. 2. Show that the volume of the solid formed by the rolason about the line
0=0 of the area bounded by the curve r =f(0) and the lines 0 = 0, .0 = 92 is

•0
r' sin OdO.
0,

Hence, find the volume of She solid generated by revolving the car-
dioide r = a (1 - cos 0 ) about the initial line.
Dividing the area in question into an infinite number of elementary
areas (as in the figure, § 10.5) by radial lines through the origin, let us con.
.,dr one such elementary area bounded by the radii vectors inclined at
angles 0 and 9 + dO to the initial line, their lengths being rand r 4 dr,
 VOLUMES AND SURFACE-AREAS 285

say. This elementary area is ultimately in the form of a triangle, whose area
is 4 r ( r + dr) sin dO , i.e , 4 r 2 dO upto the first order. Its C.G. is, neglect-
ing infinitesimals, at a distance 4 r from the origin and its perpendicular
distance from the initial line is ultimately 4rsin 8. The elementary volume
obtained by revolving the elementary area about the initial line as therefore,
J,y Pappus' theorem, ultimately equal to
21c. 4r sin O 4r dO =4tr 3 sin OdA.
Hence, integrating between the extreme limits 0 = 0 and 0 =9,, the
total volume of the solid of revolution in question is

l-sin 0dO.
0
In case of the cardioide x = a (1 - cos 8), the extreme limits for 0 are
easily seen to be 0 and it , and so the volume of the solid of revolution
generated by It is

4n fa I (1 - cos 0) 1 sin 0 do, which on putting I - cos 8 = z

easily reduces to
2
2_ 6
41t4 3 z3dz' . ita 3
3 4

EXAMPLES XII
1. Find the volumes of the solids generated by revolving, about
the x-axis, the areas bounded by the following curves and lines

(i) y = sinx;x = O;x =it.

(ii) y = 5x— x 2 ;x = O;x = 5.

(iii) Y2 = 9x.y =3x.

(iv)'lx +'Iy ='Ja;x= O;y = 0.

2. Show that the volume of a right circular cone of height It

and base of radius a is4ira2h.
 286 INTEGRAL CALCULUS EX — A ll

3. The circle x 2 + y 2 = a 1 revolves round the x-axis ; show

that I he surface-area and the volume of the whole sphere generated
are respectively 4ita 1 and na3
4. Prove that the surface area and the volume of the ellipsoid
formed by the r2volution of the ellipse x 2 fa 2 + y 2 lb 2 = 1
Ci) round its major axis are respectively

21tab( .J1 -e l +e -1 sin le)and7tab2,

and (ii) round its minor axis are respectively
I 1i+ei
2,t t a, + -- log' j-- j and
4
ia2b.

S. Show that the curved surface and vourne of the catenoid

formed by the revolution, about the x-axis, of the area bounded by

the catenary y = _2( a +e ' ), the y-axis, the x-axis, and an

ordinate are respectively

t (sy +az) and -7ta(sy +ax),
s being the length of the arc between (0, a ) and ( x , y)
6. The arc of the astroid i = a cos 3 0 ,y = a sin 1 0 , from
0 = 0 to 0 = -in, revolves about the x-axis ; show that the volume
and the surface-area of the solid generated are respectively 7ta'
and .na2
7. A cycloid revolves round the tangent at the vertex ; show
that the volume and the surface-area of the solid generated
are n 3 a l and ! L ita I respectively, a being the radius of the
generating circle.
8. The portion between the two consecutive cusps of the cy-
cloid x = a ( 0 + sin 0), y = a (1 + cos 0 ) is revolved about the
x-axis ; show that the area of the surface so formed is to the area of
the cycloid as 64 : 9 . I Nagpur, 1934
Ex. X11 VOLUMES AND SURFACE .AREA5 27

9. Show that the surface-area of the spherical zone contained

between two parallel planes = 2na x the distance between the two
planes, where a is the radius of the sphere.
10. Show that the volume of the solid generated by the revolu-
tion of the upper-half of the loop of the curve y' = x' ( 2 - x )
aboutOX is 4x.
11. Show that the volume of the solid produced by the revolu-
tion of the loop of the curve y + x) = x - x) about the
x-axis is 2ira' (log 2 - .) . I P.19351
12. Show that the surface-area and the volume of the solid
generated by the revolution about the x-axis of the loop of the curve
x = t' , y = I - 1 I are respectively 3n and is 1
13. The smaller ijf the two arcs i,to.vhich the parabola
Y 1 = Sax 'ivjdes the circle x 2 + y 2 = 9a' is rotated about the
x-axis. Shot-, that the volume of the solid generated is 11 Ica 3
14. If the curve r = a + b cos 0 ( a > I') revolves about the
initial line, show that the volume generated is . ise (a' + I"
15. The following curves revolve round their asymptotes; find
the volume generated in each case
(i) y' (2a - x) =
(ii) y (a' + x' ) = a' ; I P. P. 1933
(a - x )y ' = a'x.

16. An arc of a parabola is bounded at both ends by the latus

rectum of length 4a . Find the volume generated when the arc is
rotatz'd about the latus rectum. I Nagpur, 1935 1
17. Show that the volume of the solid formed by revolving the
ellipse x = a cos 0 , y = b sin 0 about the lifle x = 2a is 4,t 'a 'b
18. Show that, if the area lying within the cardioide
r =2a(1+ cos 0) and outside the parabola r(1+ cos 0)=2a
revolves about the initial line, the volume generated is 18ia'
 288 ENTEGRAL CALCULUS Lx X I.

19. Show that the volume of the solid generated by revolution

about OY of the area bounded by OY , the curve y 1 = x' and the
line y = 8 isic.
20. The arc of a par,, boIa from the vertex to one extremity of
the latus rectum is revolved about the corresponding chord. Prove
that the volume of the spindle so formed is (245 / 75 ) ia3
ANSWERS

•(j) i n 1 (ii) fl (iii) • It. (iv) ir

15. (i) 2n 2 a 3 +tl 16._Ica3.

 CHAPTER XIII

CENTROIDS AND MOMENTS OF INERTIA

13.1. Centroid.
It has been proved in elementary statics that if a system of par-
tides having masses m , m 3 ..... . have their distances paral-
lel to any co-ordinate axis given by x 1 , X2 , X3 . . . , then the
corresponding co-ordinate of their centre of mass will be given by
- m l X 1 + m2X2 +... Emx
m1+m2+...

Similarly, y= Em , etc. -
Now, if, instead of a system of stray particles, we get a con-
tinuous body, we may consider it to be formed of an Infinite num-
ber of infinitely small elements of masses, and in this case it may
be shown, as in the other cases, viz., determination of lengths, areas,
etc., the summation, 1, will be replaced by the integral sign.
Thus, if 3m be an element of mass of the body at a point whose
co-ordinates are (x , y) (or, in three dimensions, x, y, z) the posi-
tion of the centre of mass of the body will be given by
-jxdm -J v din
Jam'
the limits of integration being such as to include the whole body.
In practice, the elementary mass 8m is proportional to the ele-
ment of length 5s, or element of area, or element of volume of the
corresponding element, according as we proceed to find the centr-
old of an arc, or area, or volume, and the limits of integration then
will be the limits of the corresponding element.
13.2. Centroid of a thin rod.
( j ) When the rod is uniform.

Let be a rod of length a and let us take as x-axis.

Inlegral Calculus (main) -21
290 OflPGR.AL CALCULUS

- Let P. Q be two neighbour-

0 pQ A x ing points on the rod at distances
x and x + Sx from 0, so that
Fig.l PQ
= 5x. Let p be the density
and a be the uniform cross-section - " , he rod. Then the element of
mass 5m at P = a . 5x p. where t.- . rid p are constants.
Let x be the distance of Its C. C. from 0. Then taking moment
about 0, we have
xEu&xp = £a6xp.x,
i.e., x 16x = Lx&x (on dividing both sides by the constants a, p).

x2J
_____ [4 ... (1)
dx [x]
5
The limits of integration are taken as such, since for the whole
rod xvaries from C) to a.
Thus, She C. G. of a uniform thin rod is at its mid-point
(ii) When the rod is of variable density.
Suppose the density p at the point P be a known function of
Its distance from one end, say, 0 . Then p = f ( x ) .
Here, proceeding as above, the element of mass 5 m at P
---6xp= a5xf (x).
a&xf (x) -a6xf(x).x,
i.e., x Ef( x) 8x = £ xf( x) 8x, dividing by the constant a.

xf(x)dx
- 5 -
.5 ... (2)
 CENIROIDS AND MOMENTS OF I?4tT1A 291
Substituting the known value off( x), In any case, and Integrat-
ing, the final value of x Is obtained.
For example, lithe density at any point of the rod varies as the
distance from the extremity 0, then f(x) - kx, where k
constant, and therefore
Is
x 2 dx /fadx a. ... (3)
Note. If a be the cross-section 01* rod it & point P on It and p be the
density there, then a p (i.e., mass per unit length ) is called the line-den-
sity of the rod at P. By the single word 'density' isuauaIy mearI volume-
density, i.e. , mass per unit volume.
If in the case (II) It Is given that the line-density X at any point P varies
as its di:tance from 0, then 8m (the element of rnasa) at P would be
).. 8x . Now we can proceed as In (3).
13.3. Centroid of at arc.
Let (x, y ) be the co-ordinates of any point P on the arc AB,
and p be th' density at P. Let s be
the length of the arc CP measured p
from a fixed point C on the arc.
Then 6s = elementary arc PQ at P,
and hence I
ç' Ss = element of mass
at P ( = m)
hg.
Let (x , y) be the co-ordinates of
the C.G. of the arc A B . Then, as in (4) of A rt . 10.1, we have
- ixdm Jpxds -
Jpas Y f-
Iydm fpyds
fpds
the limits of integration extending from A to B.
When p is constant, the formula (1) becomes
- Jxds - 1 da ... (2)
141 .
292 INTEGRAL CALCULUS

The 1orrnuh (1) and (2) are fundamental formula' for the th
minatiori of the C.G. of an arc: and this c all h' eisilv tran5formd
when the equation of the curve is given in Car&esiart coordinates
(general or parametri), or in polar co-ordinates.
Note 1. In the application of the abovc * ntegral9 the following results
should be no'ed. When the equation of the curve is

(I) y = [(x), ds = dx
's 2r Y
(ii) x = f ( y). d -J+( 4 . ) dy.

(iii) x =t). y = 14,(f), as = %J('

(iv) r.Oj Ods

,-

and x =rcos9V = rs,rt.

Note 2. The C. C. in such cast is generally not on the ac .4B

13.4. Centrwd of a plane area.

— use 1. Car',ir:.
Suppose the area k bounded 5y ftc curve y f( x) the axis
cix and the ordinates x = x

Let (.ry).(r + 5x.y +

be the co-ordinates of P ad a
neighbouring point Q on the
curve. Divide the whole area Into
elementary strips like PMNQ , by
drawing lines parallel to the y-axis.
The area of the strip = y dx ul-
F.3 timately, since S x is vary small.
 cL:ROIDS AND MOMENTS OF INERTIA

Let the area be homogeneous and let p be the surface-density

of the strip PMNQ . Then 5m , the element of mass of the strip
PMNQ - yöx p and the C.G. of the strip PMNQ is ultimately at
the point ( x , - y ) ( with sufficient accuracy for our purpose). Let
(x4 y ) be the C. C. 01 the area AI.LB .Then, taking moments about
öI and t3 respectively, we have
xpy.&pyöX.X; y.py.x=Lpy&x -y.
Cancelling out the censtant p from both sides, we get in the
limit

2x dx 5yldx
J
!-
ydx ydx
J J
where y has to be expressed in t'rms of x from the equation of
the curve. -
Note. The surface -density p at any point of an area is a, whore a is
the volurne density and A is the thickness at the point.
Lase 11. Pear
Let the ar&a j1Q5 be hounded by the curve J(0) and the
radii vectors OA, OB ( 0 = and 9 = so that mZ XOA = a,
,nZXOB
4—
Let C) he the origin, O- the initial line and O'i' the y-axis.

Let the whoie ard be divided into elementary triangular strips

like OI-Q by raW vectors drawn y
from C) Let the cu-ordinates of

area the + strip

k-,f

OPQ = r 2 50 ultimately, since 0 - X

80 is very small Then the C.G.of
 294 INTEGRAL CALCULUS

the strip OPQ is a point C in OPQ, i

vhose co-ordinates are ul-
timately (3 r cos e, 3 r sin 0) (with sufficient degree of accuracy
for our purpose). Let p be the surface-density of the strip. Then
elementary mass 3m of the strip OPQ Is -1
at 0 1 . Let ( x, y) be the co-ordinates of the C.C.rof60the. p. situated
area A OB.
Therefore, taking moments about the y-axis and the x-axis, we
have
Z+r1P602rap3r cog e5e;
y.Z.}r2p&0 - Z-rp.4rsin 0.50.
Cancelling out from both sides .. p,sInce p is constant, we have
fina!ly in the limit

fa r 3 cos ed
fa -.
where r -f(
5 rd0
faa
0 ) from the equation to the bounding curve.
13.5. Centrojd of the volume and surface of revolution of a
uniform solid.
Suppose s solid is formed by the revolution of the curve
y - f ( x)about the x-axis t3 and suppose It is bounded by two
ordinates A L, BM corresponding to x - x 1 and x
(I) The volume generated by the element of area PNN'P' ,where
M (x , y) are the co-ordinates of
L P, Is the area of the circle
described by ) x (the
• thickness between the two
X
circles described by
and ')and y26x ul-
timately I since PN y, and
Sits very small I . If p be the
Pits density of the slice bounded
 CEN1ROIDS AND MOMEN1S OF INMThk 295

by the two circles, then 6m the element, of mass of the strip

= p.ny 2 8x. The C. C. of the element from symmetry, lies on
and is ultimately at a distance x from 0. Hence, If ( x, y) be the
co-ordinates of the C. C. of the volume generated by the area
A LMB, then, taking moment about the y-axis, we have
x.Epny z 8x =Eply t 3x.x.
As the solid is of uniform density, cancelling out pn from both
sides, we get

ylxdx
= Eyx8x J=
a2

J
and from symmetry, y = 0.
(ii) The area of the surface generated by the revolution of the
arc PP' ( = 8s ) about 153e is (the circumference of the circle
described by TN_) x (length of the arc PP) and . 8s ultimate-
ly, since PN = y and S s is small. If p be the surface—density,
then 8,n the element of mass of the belt p 2xy . bs.
The C. C. of the belt from symmetry lies on 153 and Is ul-
timately at a distance x from 0 . Hence, if ( x, y) be the co-or-
dinates of the C. C. of the surface generated by LM, then, taking
moment about the y-axis, we have
x.1 p.2xy Bs £piisy&s.x.
As the surface is of uniform density, cancelling out 2xp from
both sides, we get
- Ey s.x _ Jyxd.
- Ey&s - Jyda
In the integration, the limits for a correspond to x xt and x2.
Cor. When the equation of the curve Is given in polar co-ordinates, say,
f ( 9), the above formulae can easily be transformed into the following
forms by the relations between Cartesian and polar co-ordinates, viz.,
x = r cos 0, y = r sin O.
 296 ThITECJRAL CALCULUS

,Sf 12

Solid: Jx
.1
OcosOJ(rcoo).de

d ,y=o
1 frI sin' OcosO—(rcos9).de
taken between proper limits
C ds
r 2 sin 8 cos o—.dO
Surface: x = -----
f '" = 0
I ds
Jrsin9a.dO
taken between proper limits.
13.6. Illustrative Examples.
E. 1. Find the cent raid of an wire in the form of a circular arc,
Let A B be a wire In the form of a circular zirc ,dius 'a ' , which
subtends an angle 2a at its
Y1 B
centre 0.
-
Take 0 as origin, and OX
which bisects the arc A B , as
X x-axis
Then, by symmetry, the
centroid G lies somewhere on

A
Fig.6 Now, 0 denoting the vec-
torial angle of the point P ozi the
arc, the element PP' there has a length a do, and the abscissa qJ/ is a cos
8 Also, to cover the whole arc, 0 extends between the limits - a to a
Hence, the abscissa CC of the centroid C is given by

- Lxdm
I a
acosO.a1L)

J a
(p denoting the linear d-iist. . h- .
 CENTROIDS AND MOMENTS OF iNERTIA 297

Cos Od
2 sin a = sina
a-
2a a
do
a
Cor. The distance of the centroid of a semi-circular arc from the centre
is 2a /it -
Ex. 2. Find the centre of gravity of i uniform lamina bcunded by parabola and
a double ordinate of it.
Let the lamina be bounded by a parabola y' = 4ax and a double or-
dinate RMR' given by x = x1.
i pp' R
By symmetry, the centroid lies on
the x-axis and hence y = 0
a
Divide the lamina into elementary
strips by lines paralel to the y-axis. QQ'R'
Consider the strip PQQP' , where the
co-ordinates of P are ( x , y ) The
length PQ is 2y and the breadth NN'
is &x Hence the area of the strip is ultimately 2y 5x The limits of x , to
cover the area considered, are clearly 0 to r
1-knee, for the required centre of gravity,

xdm z 2ydxo
5
=J
J.m 2ydx.a
5
where ci is the surface-density of the lamina
pX1
z.2 dx. x3'2 dx x
Jo JO 3
= =
eli =

2'ixdx a x 11 dx
Jo Ja
Thus, this centre of gravity d.vides the length OM in the raiio of 3 5
 298 INTEGRAL CALCULUS

of the ellipse x 2 Ia 2 sy' lb = I 2

• Ex. 3. Find the centre of grarity of a uniform lamina in The
form of a quadrant
(P.P. 7935)
Let A OB be the quadrant considered. Divide It into elementary strips
B' p by lines parallel to they-axis. The area
of the elementary strip corresponding
to the point P, whose co-ordinates
are (r,y), is ultimately y8x,and

ED the centrojd of this element Is at the

middle point of the strip (which Is
supposed infinitely thin ) and thus
Fig.8 has its co-ordinates (x, .y / 2) . The
limits of x for the quadrant con-
sIdered are evidently 0 and a.
Hence, the C. C. of the area considered will be given by, (x ', y'
denoting the co-ordinates of the centrold of the element dm which Is taken
here as the strip,

f x'am
j
a
x.ydx.a

f dm
- t a being the surface-
1. ydx.a
density of the lamina
j 0
.5 _______
j x!ia — xdx.cj
o a

Ia2_Idr.o
I sinceg 2-+ b l
J
J0 a

J 0 x Ia - xdx 0
sinOcos1
= a
= f 5 a2 —x2dx
5
2
ode
o

I Iputting x = asin 0)
3
= a - = 4a-
In 3n
22
 CEN11OIDS AND MOMENTS OF INERTIA 299

fy'am f5.Y ydx.a f . .T (a 2 -x')dx

Jdm
f )4x.a J -xdx

3
COS 3 0d 2
1 fo 1 3 4b
= 1' -

f 'I
2 3n
lx
cos" OJO - -
o
Cor. The cenlroid of half the ellipse bounded by the minor axis is on the major
axis at a distance 4a/3n from the centre.
Also, the centroid of a semi-circular area of radius 'a' Is on the radius
bisecting It, at a distance 4a / 3s; from the centre.
Ex. 4. Find the centre of gravity of a solid hemisphere.
the hemisphere may be supposed io be generated by evolving
a circular quadrant A PB about one bounding radius OA , which we may
choose as x-axis. By symmetry, the B
I'
centre of grav, IT of the hemisphere
will be on t5x" . Now divide the a!
hemisphere Into ix1tItely thin cir-
cular slices by planes perpendicular C i4 A X
to the axis of revolution t5 An ele-
ment of such slice, corresponding to
the point P, has its volume ultimate-
ly equal to it y I 8x (x. y being the
Fig.9
Cartesian co-ordinates of P), and the
x co-ordinate of its centre is x.
Hence, If p be the density of the solid hemisphere and a its radius, the
position (,'the C. C. Is given by
p5

x J J
x.xy2dx.p x(a'
0 0
- ( sincex 2 +y1

50
iry1dx.p
$(
0
a'
300 INTEGRAL CALCULUS

1 a 2a4
3
a = —.
a3 8

13.7. Moment of Inertia.

If a system of particles have masses m 1 , m2 , m . . and if r,
• - be their distances from a given line, then L mr 2 is defined a
the moment of inertia of the system of particles about the given line
If M be the total mass of the system m 3 , m 1 , etc., it is usual
to express the moment of inertia of the system about any line in the
form Mk 2 , where k represents a length and is called the radius of
gyration of the system about the given line.
If, instead of a system of particles, it is a body in the form of a
thin wire or a lamina or a solid of which we want to find the mo-
ment of ine,tia about a given line, we may consider the body to be
made up of an infinite number of ifinitely small elements of mas-
ses, and then the summation Zmr reduces to the integral I r 2 din,
where the limits are such as to cover the whole body.
13.8. Two important theorems of moments of inertia.
(a ) If a thin lamina (thickness negligible) has its momcnk of in
ertia about two perpendicular axes in its plane respectively equal to 1
and 12 , then the moment of inertia about a normal to the plane through
their point of intersection is 1 + 12
Take the given perpendicular av as x-axis and y-ax. I and
1 being the respective moments of inertia. Consider u clement of
'nass riM adxdy at the point P ( cs = surface-dewity). Its mo-
ment of inertia about = d1 2 y 1 odxdy Simnilarl , i1 2 ( that
about O1 ) = x I cdxdy . Also, dl = moment of inertia of the ole-
c nt about a normal to the plane at the oiigiri ( point of intersec-
tion of the given axes) = ( x y' ) odxdy for toe distance of
P ( x , y ) from the origin C or the normal to the plane
= J(x 2 + y 2 ) Thus, dl = dl, + d11 is true for every point P
Thei fore, integrating between proper limits over the whole area,
 ( . FP r IROiDS AND MOMENTS OF INERTIA 301

J=fl(2 t-y 1 )adxdg = lIx' c4xdy + fly 2 .adxdy =12 +1

N. B. This theorem is true even if a is a junction of ( x, y)
(I') The theorem of parallel axes in the case of a lamina is The mo-
nent of inertia of a thin lamina (thickness negligible ) about any given
line in its plane is equal to that about a parallel line through its C. G.,
together with the moment of inertia of the whole mass concentrated at the
C. G. about the given lute
4--s . -
C is the centre of gravity. Take Cx parallel to thel re Bit)
as x-axis. Ley (y-axis ) cut BA at B . Let GB = h = distance of
the given line from G . Consider an element of mass adxdy at
P ( x, y ), a = surface-density. PN = perpendi:ul4r on BA = h - y.
= (h — y ) 2 .Then,
1( moment of inertia about BA) = Jia(h — y)'dxdy
= hi 11 cydxd y - 2h II yadxdy + ii oy dxdy
= Mh 2 —0 + 1,, . (i),where M =.adxdy.
= mass of the lamina,
= JJay I dxdy = moment of intertta about4?',
also, y (distance of C. C. from' Ux = 0, here)
= Si aydxdy / M.
.Iiaydxdy = 0.
Thus, from (I) we get the theorem.
13.9. Illustrative Examples.
Lx. 1. Find the moment of inertia of a thin nift'r'vi straight rod of mass M
and length 2a about its perpendicular
bisect
An inf!nitesl ial element of length B IPA
a t P. whose distance from the mid-
de point of the rod i9 x, ha! Its mass
V 6x I 2a) Hence the moment of inertia
; the od about the perpendicular bisector OY is given by

I= -=M
f 2a .2a 3
302 INFEGRAL CALCULUS

Er. 2. Find the moment of inertia of a thin uniform lamina in the form of a
rectangle about an axis of symmetry through its centre.
Let 2a and 2b be the lengths of the adjacent sides i5 and A B of the
rectangular lamina A BCD, and
the axes of symmetry
through its centre 0, which are D
parallel to them.
M being the mass of the U X
lamina, the surface-density is
clearly M 1(4ab) Now, divide C B
the lamina into thin strips pa.-al-
lei and considei any strip
PQ at a distance y from 09
whose breadth is &y. The mass of the strip Is then evident-
ly (M/( 4ab )) 2a Sy Every portion of it being ultimately at the same dis-
tance y from 0- , the moment of inertia of the whole lamina about the
x- axis Is given by

2 M ' 2ady
i-b "

Smiiaxly, the moment of inertia of the lamia about 15V is given by

I, =M---

Er. 3. Find the moment of inertia of a thin uniform elliptic lemma about it
axes.
Let x 2 /a 2 + y 7 / 0 = I be the equation to the ellipse. Its area Is known
to be itab, if M be its mass, the surface-density Is M /( irab) Dividing the
lamina into thin strips by lines parallel to the x-axis, an elementary strip at
a distance y from the x-axis has its length 2x = (2a / b )'I( b 2 - y 1) from
the equation of the elliptic boundary. Thus. 6y being the breadth of the
strip, its mass is

.Jb2 -
- y' 6y

Hence, the moment of inertia of the lamina about the x-axis is

given by
 CEKIROIDS AND MOMENTS OF INERTIA 303

=I 2--
iab
2 ! irj dy
b

- sin 2 0 cos 1 0 dO I putting y = b sin 0

7
J-

- 2Mb 2 ' - M 1'

- it 8 4
Similarl y , the moment of inertia of the lamina about the y-axis is gien
by
ly =
M
Cor. The moment of inertia of a thin uniform circular disc of mass M
and radius a about any diameter is M (a 2 / 4
E. 4. Find the moment of inertia of a thin uniform circular plate about an
axis through its centre perpendicular to its plane.
Let M be the mass and a the radius of the circular lamina, so that its
surface-density is M IC ,t22)
Divide the lamina into infinitely thin concentric rings by circles con-
COfltr j c with the boundary. Any
elementary ring between circles of
radii r and r + & has its area ul-
timately equal to 2ttr Br and so its
mass is IM /(lta2 )J 2trr.As every
part of the ring LS ultimately at the
same distance r from the axis in
question which is perpendicular to its Figl2
plane through the centre, the moment
of Inertia of th ring about the axis is ultimately (M /( ira 2 )] 2itr &r. r 2 -
Hence, the requited moment of inertia of the disc about the axis
is given by

I = I - 21trdr.r2

2M r1
r5dr= a rM a
2 4
304 INTEGRAL CALCULUS

Lx. 5. Find the moment of inertia of a sphere about a diameter.

If Al be the mass and a the radius of the sphere, the volume of the
sphere is known to be 1 ita and hence its density is M /( na 3)
Take the diameter about which the moment of inertia Is required to be
the x-axis. Divide the sphere into infinitely thin circular slices by planes per-
pendicular to this axis. An elementary slice between the planes x and
r + öx has its volume ultimately equal to it (a 2 - x 2 ) 8x,since its radius
is 4( a 2 - x 2) 1 Sce Fig Lx. 4 , Art 13.6. 1 Hence the moment of inertia of
this slice about the r ..xis, which is perpendicular to its plane through its
centre, is ultimately
M a2-xt
• (g 2 2)2 X • (see Er. 4 above. I
Hence, the required moment of inertia of the whole sphere about the
diameter is given by
M a1 -x2
= J5(a' -z' )dx
2
S-a
*1
- (a4 '-2a 7 r 2 + x 4 )dx
8 a3 .
3
3M
T(a4.2a--2a2.+-
-'- - 2a5\
--) = 2 Ma2

EXAMPLES XIII
1. Show that the C.G. of a thin hemispherical shell is at the mid-
dle point of the radius perpendicular to its bounding plane.
2. Show that the C.G. of (I) a solid right circular cone is on the
axis at a distance from the base equal to -! of the height of
the cone ; (ii) a thin hollow cone without base is on the axis at a
ditane from the base equal to L of the height of the cone.
3. Find the centroid of the whole arc of the cardioide
r = a(1 + cosO).
4. Find he centroid of th.3 area bounded by the cycloid
x =a(O + sin G), y = a (1 .- cos O) and its base.
 Ex. X 1II CENTROIDS AND MOMENTS OF INER'IlA

S. Find the centroid of the sector of a circle.

6. Find the centroid of the arc of the parabola y 2 = 4ax in-
cluded between the vertex and one extremity of the latus rectum.
7. Find the position of the centroids of the following areas:
(I) A loop of the curve y 1 (a + x) = x 2 (a - x ) .

(ii) Area bounded by the curve y 2 (2a - x) =

and its asymptote.
(iii) Area bounded by y 2 = 4ax and y 2x.
(iv) One loop of r = a cos 2.
8. Find the C. C. of the arc which Is in the first quadrant of
thecycloid x = a(0 +sinO),y a(1 —cosO).
9. Find the centroid of the area of the astroid x 21' + y 2(3 =
2/3 lying in the first quadrant.
10. Find the centroid of the area between the sine curve
y = sin and = O, where 0 ^S x
11. Find the C. C. of the area of the parabola
IxV '2 IV \1/2
- i +t p 1 between the curve and the axes.
D vbi

12. Find the centroid of the area of half the cardioide

r =a(1 + cos 8) bounded by 0 = 0.
13. Find the centroid of the area of the right loop of the lem-
niscate r' = a 2 cos 20.
14. Find the C. C. of the solid formed by the revolution of the
quadrant of the ellipse ( x I /a 2 + y 2 /b ) I about its (I) major
axis, (ii) minor axis.
15. Find the centroid of the (I) surface and (ii) solid generated
by revolving half of the card iolde r a (1 + cos 0) bounded
by 0 0 about the Initial line.
Inlearal Calculus (main) -22
 INTEGRAL CALCULUS Ex. X III

16. Find the C. C. of the surface formed by the revolution of

the parabola y 2 = 2x cut off by the line x = 4 about the axis of
the parabola.
17. An equilateral triangle of si 'a revolves round its base
which 15 fixed. Find the volume of th '.olid generated.
18. Find the moment of inertia of a solid right circular cylinder
of radius a about its axis.
19. Obtain the moment of inertia of a solid right circular cone
of height h and semi-vertical angle a about its axis.
20. Prove that the moment of inertia about an axis through the
centre perpendicular to the plane of a thin circular ring whose outer
and inner radii are a and b is M (a 1 + Li ), where M denotes
the mass of the ring.
21. Find the moment of inertia of a rectangular parallelopiped,
the lengths of whose edges are respectively 2a, 2b, 2c about an
axis through its centre parallel to the edge 2a.
22. -Show that the moment of inertia of a thin hollow
spherical shell of radius a and mass M about a diameter
is M(2a2 /3)
23. Show that the moment of inertia of a parabolic area of latus
rectum 4a, cut off by an ordinate at a distance h from the vertex,
is .Mh2 about the tangent at the vertex, and -!Mah about the axis,
M being the mass of the area.
24. Show that if a thin lamina has its moments of inertia about
two perpendicular axes in its plane respectively equal to 1 and 12,
then the moment of inertia about a noimal to the plane through
their point of intersection is I + 1
25. Prove the Theorem of parallel axes in the case of a lamina,
namely, that the moment of inertia of a thin lamina about any given
line in its plane is equal to that about a parallel line through its C.
C. together with the moment of inertia of the whole mass con-
centrated at the C.G. about the given line.
 Ex. X III CENOIDS AND MOMENTS OF INERTIA 307

26. Find the moment of Inertia of the circumference of the irde

of radius 'a' about a diameter.
27. Find the radius of gyration of a circle of radius 'a' about a
diameter.
28. Find the moment of inertia of the surface of a sphere of
radius 'a ' about a diameter.
29. Find the moment of inertia of a truncated cone about its
axis, the radius of its ends being 'a' and b'
30. Find the moment of inertia of an isosceles triangle, each of
whose equal sides is 'a' about the perpendicular from the vertex
upon the opposite side.
31. Find the moment of inertia of the area bounded by
= a'cos 20 about its axis.
32. Find the moment of inertia of a circular area of radius 'a'
about the line whose perpendicular distance from its centre is d.
33. Find the moment of inertia of a rectangular parallelopiped
whose sides are 2a, 21; , 2c about its edge Za.
34. Show that the moment of inertia of a thin uniform rod of
length 2a and mass M about a line through one end perpendicular
to the rod is M.a2
35. SLw that the moment of inertia of a thin uniform lamina
in the form of a rectangle whose sides are 2a and 2b about an axis
perpendicular to the plane of the lamina at the point of intersection
of the diagonals of the lamina is - (a' + b' ) M
36. Show that the moment of inertia of a thin uniform elliptic
lamina whose semi-axes are a and b about the line through the
centre of the ellipse and perpendicular to its plane is -1 (a' + b') M.
(See Ex. 39. 1
37. Show that the moment of intertia of the area of a lemnis-
cate of Bernoulli r' = c' cos 20 about the line in its plane through
the origin and perpendicular to its axis is Mc2 ( 3 + 8 )/48
308 INTEGRAL CALCULUS Ex XIII

38. ABC is a uniform triangular lamina and I is the length of

the perpendicular drawn from A on BC If M be the total mass of
the lamina, then show that the moment of inertia of the lamina
about BC ls.M1t
39. (i) Show that the moment of inertia of a uniform el-
liptic lamina of mass M, the equation of the ellipse being
x 2 /a 2 + y 1 lb 2 1 , about a diameter making an angle 0 with
the major axis is M (a 2 sin 2 O + b 2 cos20)/4
(ii) If r be the length of the semi-diameter of the ellipse
in the above case, then show that the moment of inertia
Is M a 2 b 2 /(4r' ).
(iii) if, in th above case , p be the length of the perpendicular
from the centre of the ellipse on the tangent parallel to the semi-
diameter, then show that the moment of inertia about the tangent
is . MP.
ANSWERS

3. x =a, y = 0. 4. x = 0, y

in -"
S. On the radius, bisecting the sector, at a distance Ia from the
centre, 2a being the angle of the sector at the centre, and a the radius.

- a 342 - log (12 + 1) - 49 242- 1

6. X =4
• '1 + log C '12 + 1) ' 3 '12 + log ( '12 + 1)
3s-8 ,y0.- - -
(1i)x=3S-m ,y=0.
7.(1)x=1 41

'2842a
=a,y=m.,
1
256a
8.-x=(x-1)a,ya.
4 - 2
9. - -

10. ;,y - i s .

--Sm- 16m . xa42-

12. X I y
T
L. x —i-- , y 0.
 CENThOIDS AND MOMENTS OF INTLk 309

14.0)=a, yO;(il)=O,=b.

15. (I) For surface x— 50—

,a , y 0; (U) For volume z -4a—Y - 0-

- 149 -
l6.x= .g . ,y=O.

18.M f. 19.j j Mk2t an2u. + C2

21.M

26. . 27. 28. . 29..

( as - S
10 - b3
Ma
31.!1-(*).

32. M(d +_a2). 33.!(b2 +c2).

'1.B. M is the mass in each case.
 CHAPTER xiV

ON SOME WELL-KNOWN CURVES

14.1. We give below diagrams, equations, and a few charac-

teristics of some well-known curves which have been used in the
preceding pages in obtaining their properties. The student is sup-
posed to be familiar with conic sections and graphs of circular func-
tions, so they are not given here.
14.2. Cycloid.
The cycloid is the curve traced out by a point on the circum-
ference of a circle which rolls ( without sliding ) on a straight line.
A

P
;I , ^
3 1 /

Fig. I
x = a(O - sin 0) y=a(1 - cosO)
Let P be the point on the circle MP, called the generating
circle, which traces Out the cycloid. Let the line OMX on which the
circle rolls be taken as x-axis and the point 0 on öY', with which
P was in contact when the circle began rolling, be taken as origin.
I.ct a be the radius of the generating circle and C its centre,
P the point (x.y)on it, and let mzPCM =9.Then Ois the angle
through which the circle turns as the point P traces out the locus.
OM=arcPM — aO.
Let PL be drawn perpendicular to
x=OL=OM— L.M=aO— PN=aO— a sin O
= a ( 6 - sin 9)
 ON SOME WELL -KNOWN CURVES 311

y = PL = NM = CM- CN = a - acosO
a(l - cosfl).
Thus, the parametric equations of the cycloid with the starting
point as origin and the line on which the circle rolls, called base,
as x-axis , are

x = a(e - sin O),y Cos 0)

a(1- ... (1)

The point A at the greatest distance from the base 09, is

called vertex. Thus, for the vertex, y, i.e., a(1- cos 0) is
maxi-
mum. Hence, cos 0 =- I , i.e., 0 it.

AD = a ( I - cos it 2a. .. vertex l,( an, 2a).

ForOandO',yO - :. cos 0I. :.0=O and 2it.

As the circle rolls on, arches like OA O' are generated over and
over again, and any single arch is called a cycloid.

x=a(Os sin O) y=a(1— Cos O)

Since the vertex is the point (an, 2a ), the equation of the cycloid
with the vertex as the origin and the tangent at the vertex as the
x-axis can be obtained from the previous equations by transfer-
ring the origin to (air, 2a ) and turning the axes through it, i.e., by
writing
and 2a + it' sinit + y' cos it for it
an + x' cos It -y' sin it

and y respectively.
Hence, a ( 0 - sin 0) = an - x',
 312 INTBGRAL CALCULUS

or, x'=a(n-O)+asinO=a(('+sjnO')
where 0' i— 0,
and a (1 - cos 0) • 2a - y',
or, y'2a-a+a cos 0a+a cos O
a - a Cos (i - 0) . a(1 - Cos O').
Hence, (dropping dashes) the equation of the cycloid with the ver-
tex as origin and the tangent at the vertex as x-axis is
x =a(0 + sine), y= a(1- cQsO). ... (2)
In this equation, 0.0 for vertex, 0 = scfor 0 and e = - n for 0'.
The characteristic properties are:
(1) For the cycloid x = a ( 0 - sin 0 ) , y a (1 - cos 0),
radius of curvature twice the length of the normal.
(ii) The evolute of the cycloid is an equal cycloid.
(iii) For the cycloid x a (0 + sin 0), y = a (1 - cos 0),
w . 0 and s 2 = gay , s being measured from the vertex.
(lv) The length of the above cycloid included between the two
cusps is 8a.
(v) Intrinsic equation Is $ = 4 a sin iy.
Not.. The above equation (2) can also be obtained from the
Fig. (1) geometrically as follows:
U (x ' , y') be the co-ordinates of P referred to the vertex as origin
and the tangent at the vertex as x-axis,
L V OD- OL -x - $)+ m sin e,
y A D - PL • 2s - y • 2a - a(l - cos 0) a(l + cos 0).
Hence, writing e' (or0) for - 0, etc.
14.3. Catenary.
The catenary Is the curve In which a uniform heavy flexible
string will hang under the action of gravity when suspended from
two points. It Is also called the chainette.
 ON SOME WELL - KNOWN CURVES 313

Its equation, as shown In books on Statics, Is

x C l -sat
y c cosh - - lc + *

C is called the vertex; DC - c .r is called the directrix.

The characteristic properties are:
(I) The perpendicular from the foot of the ordinate upon the
tangent at any point Is of con-
stant length.
(ii) Radius of curvature
at any point length of the Ap
normal at the point (the centre \ 4'
of curvature and the x-axis c
being on the opposite sides of o N x
the curve). Fig.3

(iii) y 2 = c 1 + s 2, s being measured from the vertex C.

(iv) s = c tan v,y=c sec w. (v) x=c log ( sec w+ tan W).
14.4. Tractrix.
Its equation is
a - 1a- y1
X = '/ a- y3 +
log a + I(a - y2)
or, x = a(cost + logtant),y a sint.
Here, OA a . The characteristic properties are:

(i) The portion of the tan- Y

gent intercepted between the Al
curve and the x-axis is con-
stant. ___________
(ii) The radius of curva- o
j T
ture varies inversely as the
Fig.4
314 INTEGRAL CALCULUS

normal (the centre of curvature and the x-axis being on the opposite
sides of the curve).
(iii) The evolute of the tractrix is the catenary
y = a cosh(x/a).
14.5. Four-cusped Hypo cycloid.
+ (.)2/3
Its equation is (x) lJ

IY
or, x=a B
y = b sin',
A' 0 AX
Here, OA = OX =a;
B'
OB = OB' =b.
Fig.5

1 +ab+b2
The perimeter of the hypo-cycloid A BA 'B' is 4 a
a +b
The astroid is a special case of this, when a = b
14.6. Astroid.
Its equation is x 2 /3 + y 113 = a213
or, x = acos 3 O, y = asin3O.
Here, OA = 08 = OA ' = 08 = a.
The whole figure lies completely within a circle of radius a and
centre 0 . The points A , A ' ,B, B' are called cusps. It is a special
'1' type of a four-cusped hypo-
B cycloid. [Sees 14.5J
The characteristic property of
A' 0 AX this curve is that the tangent at any
B' point to the curve intercepted be-
tween the axes is of Constant
Fig.6 length.
 ON SOME WELL -KNOWN CURVES 315

The perimeter of the astroid x 113 + y = a 213 is 6a

14.7. Evolutes of Parabola and Ellipse.

(i) The equation of the evolute of the parabola y 2 = 4ax is
27ay 2 = 4(x - 2a)3

The curve is called a semi-cubical parabola.

Transferring the origin to ( 2ii , 0 ) , its equation assumes the
form y 2 =kx 3 where k=4/(27a),
which is the standard equation of -
the semi-cubical parabola with its
vertex at the origin. ________
C
Hence, the vertex C of the
evolute is ( 2i, 0 )
(ii) The equation of the
evolute of the ellipse
Fig.?
x t / a 2 + y 1 / b 2 = I is

( ax ) + (by) 1/3 = ( a2 - b 2 ) 1/3

which can be written in the form
(x)2.'3 =

where a =(a 2 - b 2 )/a, = (a 2 - b 2 )/b.

(a 3 - b2 )1
The area of the evolute is i ab
V

Fig.s
 316 INTEGRAL CALCULUS

a2 bl
The length of the evolute is 4 (
r - )
Hence, it is a four-cusped hypo-cycloid.
14.8. Folium of Descartes.
Its equation is x 3 + y 3 = 3axy.
It is symmetrical about the line y = x.
The axes of co-ordinates are tangents at the origin, and there is
a loop in the first quadrant.
It has an asymptote x + y + a = 0 and its radii of curvatures at
YJ the origin are each • a.

The area included between the

curve and its asymptote
= the area of the loop of the
curve
=a
Fig.9
14.9. Logarithmic and Exponential Curves.

:( x

Fig. 10
(1) y = log (ii) y =

(i) xis always positive; y = 0 when x = 1, and as x

becomes smaller and smaller, y. being negative, becomes numeri-
cally larger and larger. For x > 0, the curve is continuous.
 ON SOME WELL -KNOWN CURVES 317

(ii) x may be positive or negative, but y is always positive

and y becomes smaller and smaller, as x, being negative, becomes
numerically larger and larger. The curve is continuous for all
values of x.
14.10. Probability Curve.
The equation of the ?rob
ability curve is y e -,
The x-axis is an
asymptote.
The area between the curve
and the asymptote is
Fig. 11

= 2f e 1 dx = 2. -1 ='In.

14.11. Clssoid of Diodes.

Its Cartesian equation is
Y ' (2a - x) = -
OA = 2a ; x = 2a is an
asymptote. X O} A
Its polar equation is
2a sin 2O
Fig. 12
Cos 0

14.12. Strophold.
The equation of the curve
Is

y2=x2 a+x
a— x B OHX
011 = OB = a.
Fig. 13
318 INTEGRAL CALC LUS

OCBPO is a loop.
x = a is an asymptote.
a - X
The curve y t = x2 is similar, just the reverse of
a +x
strophoid, the loop being on the right side of the origin and the
asymptote on the left side.
14.13. Witch of Agnesi.
The equation of the curve is

xy2 = 4a (2a - x)
Here, OA = Ia.
This curve was first dis-
O>A cussed by the Italian lady
mathematician Maria Gactaua
Fig. 14 Agnesi, Professor of Mathe-
matics at Bologna.

14.14. Logarithmic (or Equiangulax) spiral.

Its equation is r ae e 101 ( or, r = ae
where cot u or m is constant.
(i) The tangent at any point makes a constant angle with the
p radius vector ( = u)

/ ) T (ii) Its pedal, inverse,

polar reciprocal and evolute are
X all equiangular spirals.
(iii) The radius of curva-
Fig.15 ture subtends a right angle at
the pole.

Note. Because of the property (U, the spiral is called equiangular.

 ON SOME WELL -KNOWN CURVES 39

14.15. Spiral of Archimedes.

Its equation is r = aO . Y

/01
Its characteristic prcperty
is that its polar subnormal is
constant. ------

Fig. 16

14.16. Cardioide.
Its equation is (i) r = a (1 + cos 6), or (ii) r = a (1 - cos 6 ).

In (i), 0 =0 for A , and 6 =n for 0.

In (ii), 0 = it for A ,and 0 = U for 0.

x ri

Fig, 17
(i) r =a(1 + cosO). (ii) r =a(1 - cos 8).
In both cases, the curve is symmetrical about the initial line
which divides the whole curve into two equal halves and for the
upper half 0 varies from 0 to it, and OA = 2a.
The curve (ii) is really the same as (i) turned through 180.
The curve passes through the origin, its tangent there being the
initial line, and the tangent at A is perpcndicuhr to the initial line.
The evolute of the cardioide is a cardioide.
320 Th1TEGRAL CALCULUS

The perimeter of the card ioide is 8a

Note. Because of its shape like a human heart, it is called a cardiojde.
The cardloide r a (I + cos 9) is the pedal of the circle r = 2a cos 0 with
respect to a point on the circumference of the circle and inverse of the
parabola r = a / (1 + cos 9)
14.17. Limacon.
The equation of the curve is
r = a + b cos 0.
When a > b, we have the outer cuve, and when a b, we
have the inner curve with the
loop.
When a = b the curve
reduces to a cardioide. I See
fig. in § 14.16..J
Limacon is the pedal of a
circle with respect to a point
outside the circumference of
Fig. IS the circle.

14.18. Lemniscate.
Its equation is r 2 = a I cos 28,
or, (x 2 + y2 )2 a2 ( Xl - Y )
It consists of two equal
loops, each symmetrical about
the initial line which divides
each loop into two equal halves.
X
OA =OA '— a.

Fig. 19
= 2 cos29.
 ON SOME WELL - KNOWN CURVES 32

The tang2nts at the origin are y ± x.

For the upper half of the right-hand Loop 0 varies from
0 to
A characteristic property of it Is that the product of the distan
ces of any point on it from (± a I 42, 0 ) is constant.
The area of the lemniscate is

resented by r a2 sin 20 is
also sometimes called lemnis-
'0
The lémniscate is the pedal of the rectangular hyperbola
r2 cos 20 = a 2 . The curve rep- rY rA

Cate or rose temniscate,to distirt- x

guish It from the first
lemniscate, which is sometimes
called Lemniscate of Bernoulli
after the name of the mathe- Fig.20
matician J . Bernoulli who first
studied its properties. r2 a 2 sin 20
The curve consists of two equal loops, situated in the first and
third quadrants, and symmetrical about the line y x . It is the
first curve turned through 450.
The tangents at the origin are the axes of x and y.
The area of the curve is a2
14.19. Rose-Petals (r = a sin nO, r = a cos no).
The curve represented by r = a sin 38, or r = a cos 30 is
called a three-leaved rose, each consisting of three equal loops. The
order in which the loops are described is indicated in the figures
by numbers. In each case, OA = OB OC = r. , and mZ AOB
= mLBOC = m,COA 120°.
The curve represented by r = a sin 20, or r = a cos 29 Is called
a four-leaved rose, each consisting of four equal loops. In each case,
OA =OB OC OD a and mLA OB = mLBOC mZCOD
= mZDOA =90°.
Inlearal Calculus (main) -3
 NTrrRAL CALCULUS

IY Y
CO)

7 J<iiila)
W
Fig,2

r . sin 34 r a cos 3$.

x )

(2)

E,fl
r r a cos 20.

The class of curves reprLwnted by r = a sin no, or r = a cos no,

I ,,- here n is a positive integer, is called rose-petal,
there being n
or 2n equal loops according as n is odd or even, all being arranged
ymrnetrically about the origin and lying entirely within a circle
whose centre is the pole and radius a.
12O. Sine Sipral (r'. a z sin nO, or, r' a a cos nO).
The class of curves represerited by (i) Y = a sin no,
or (:1) 1' a I cos no is celled sine spiral and embraces scveral
important and well-known curves as particular cases.
 ON SOME WELL - KNOWN CURVES

Thus, for the values a 1 , I , - 2, 4 2,— .i and - the se

sp is respectively a straight line, a circle, a rectangular hyper-
hut, a lemniscate , a parabola and a caduidc.
For (1) nO; for (ii) = in 4 no

The pedal equation in both the cases is

P = r • I /"
 DIFFERENTIAL EQUATIONS
CHAPTER XV
INTRODUCTION AND DEFINITIONS

15.1. Definition and classification.

A differential equation Is an equation Involving differentials (or
differential coefficients) with or without the variables from which
these differentials (or differential coefficients) are derived.
The following are examples of differential equations:
Cx (1)
dx
(.1)2=axx +bx + c ... (2)

yo
dx
Y (4)
V dx'/ dx
2
ty
dx'
+ 5 (L) +
dx
=0 ... (5)

xaz + az- = 0 ... (6)

a'u d'u
0. ... (7)

Differential equations are divided Into two classes, viz., Ordi-

nary and Partial.
An ordinary differential equation is one in which all the differen-
tials (or derivatives) involved have reference to a single indepen-
dent variable.
A partial differential equation is one which contains partial dif-
ferentials (or derivatives) and a; Involves two or more inde-
pendent variables.
 IN11ODUCI1ON AND DEFINITIONS 325

Thus, in the above set, equations (1), (2), (3), (4) and (5) are or-
dinary diferential equations and equations (6) and (7) are partial
differential equations.
In order to facilitate discussions, differential equations are clas-
sified according to order and degree.
The order of a differential equation is the order of the highest
derivative (or differential) in the equation. Thus, equations (1) and
(2) are of the first order, (3) and (5) are of the second order, and (4)
is of the third order.
The degree of an algebraic differential equation is the degre;
of the derivative (or differential) of the highest order in the equa-
tion, after the equation is freed from radicals and Iractiofls in its
derivatives. Thus, equations (2) and (4) are of the second degree.
Note. Strictly speaking, the term 'degree' Is used with reference to
those differential equations only which can be written as polynomials in the
derivatives.
We shall consider in this treatise only ordinary differential
equations of different orders and degrees.
15.2. Formation of ordinary Differential Equations.

Let f(x,y,c1 ) = 0 . . . (I)

bean equation containing x, y and one arbitrary constant c1
Differentiating (1), we get

x ydx (2)

Equation (2) will, in general, contain c 1 . If c1 be eliminated

dy
tween (1) and (2), we shall get a relation involving x, y and
which will evidently be a differential equation of the first order.
Similarly, if we have an equation
f(x,y,c1 ,c) = 0 ... (3)
containing two arbitrary constants c, and c3 ,then bydifferentlat-
326 INTEGRAL CALCULUS

Ing this twice we shall get two equations. Now, between these two
equations and the given equation, in all three equations, if the two
arbitrary constants c1 and c2 be eliminated, we shall evidently
get a differential equation of the second order.
In general, if we have an equation
f (x ,y ,ci , 2 ...... c,, ) 0 ... (4)
containing n arbitrary constants c1 , , . . . , c,,, then by differen-
tiating this n times we shall get n equations. Now, between these
n equaions and the given equation, in all ( n t- I )equations, if
the n arbitrary constants c 1 , c1 , . , c be eliminated, we shall
evidently get a differential equation of the nth order , for there
being n differentiations the resulting equation must contain a
derivative of the nth order.
Note. From the process of forming a differential equation from a given
primitive, It Is clear that since the equation obtained by varying the arbitrary
constants In the primitive represents a certain system or family of curves,
the differential equation (in which the constants do not appear) expresses
some properties common to all those curves. We may thus say that a dif-
ferential equation represents a family of curves all satisfying some common proper-
ties. This can be considered as the geometrical interpretation of the
differential equation.
15.3. Solution of a Differential Equation.
Any relation connecting the variables of an equation and not
involving their derivatives, which satisfies the given differential
equation, i.e., from which the given differential equation can be
derived, is called a solution of the differential equation. Thus,
y = e + C, where C is any arbitrary constant,
and y = A x + B, where A and B are arbitrary constants,
are respectively the solutions of the differential equations (1) and
(3) of Art. 15.1.
From the above, it is clear that a differential equation may have
an unlimited number of solutions, for each of the different relations
A relation containing n arbitrary constants may, in certain cases, give rise
to a differential equation of order less than it.
 IN11ODUC11ON AND DEFINITIONS 327

obtained by giving particular values to the arbitrary constant or

constants in the solution of the equation satisfies the equation and
hence, is a solution of the equation ; thus y x - ' 111, y 2x - 3,
y = - x, etc. are all solutions of the differential equation (3) of
Art. 15.1.
The arbitrary co n stants A , B, C appearing in the solution are
called arbitrary constants of integration.
The solution of a differential equation in which the number of
independent arbitrary constants is equal to the order of the equation
is called the general or complete solution (or complete primitive ) of
the equation.
The solution obtained by giving particular values to the ar-
bitrary constants of the general solution is called a particular solu-
tion of the equation.
Thus, y = A x + B is the general solution, and x
y = 2x - 3, y = - - x are all particular solutions of the equation
(3) of Art. 15.1.
There is another kind of solution called the singular solution,
which will be discussed in a subsequent chapter. I See A rt. 17.5 1
By a proper manipulation of the arbitrary constants in the
general solution of a differential equation, the general solution is
very often written in different forms ; it should be noted, however,
that each of these forms determines the same relation between the
variables. This will be subsequently illustrated in the worked out
examples.
When an equation is to be solved, it is generally implied that
the complete solution is required.
It sometimes happens that the process of solving a differential
equation leads to integrals which cannot be evaluated in terms of
known elementary functions. In such a case, the equation is con-
sidered as having been solved when it has been reduced to an ex-
pression involving integrals and it is then said that the solution of
the equation has been reduced to quadrature.
328 DTEGRAL CALCULUS

Note. 1. The arbitrary constants In the solution of a differential equa-.

tion are said to be independent, when it Is Impossible to deduce from the
solution an equivalent relation containing fewer arbitrary constants. Thus,
the two arbitrary constants A. B In the equation y = At' I B are not Inde-
,endent since the equation can be written as y = At . e' = Cc
Note. 2. In this elementary treatise, we shall not concern ourselves with
the question whether a differential equation has a solution or what are the
conditions under which it will have a solution of a particular character; in
fact, we shall assume without proof the following fundamental theorem of
differential equation, viz.,
An ordinary differential equation of order ri has a solution involving n in-
dependent arbitrary constants, and this solution is unique.
15.4. Illustrative Examples.
Er. 1. Find the differential equation of all straight lines passing through the
origin.
Let y=mx ... (1)
be the equation of any straight line passing through the origin.

Diflerentiatif%g (1),
dx m. ... (2)
Eliminating m between (1) and (2), we get

y r dx , the required differential equation.

Er. 2. Find the differential equation from the relation

x = a cos t + b sin I,
a and b being arbitrary constants.
Differentiating the given relation twice with respecj to t, we get
- a sin t +bcost -and
x2- acost -bslni -(acost + bslnt)=-x.

+ x = 0, i.e., + x = 0 Is the required differential equation.

Ex. 3.Eliminate a and b from y a tan x+ b.

Differentiating the given relation with respect to r,

" .. (I +x2)y1
=.
 INIRODUCI1ON AND DEFINITIONS 329

DifferentiatIng. (1 +Z2 ) ya + 21yj = 0.

This Is the required eliminaflt.
EXAMPLES XV
1. Show that the differential equation of a system of concentric
circles having the centres at the origin is x dx + y dy = 0 . Inter-
pret the result geometrically.
2. Prove that the differential equation of all circles touching
the x-axis at the origin is (x 2 - y 2 )dy - 2xy dx = 0.
3. (i) Show that the differential equation of all parabolas
(a) having their axes parallel to the y-axis is Y 3 = 0;
(b) with foci at the origin and axes along the x-axis is
yy + 2xy1 - y = 0;
(ii) Show that the differential equation of the family of circles
x 2 +y 1 +2gx + 2fy + c= 0 is(l +
y12)y3_3y1 Yi 2 = 0;
(iii) Show that the differntial equation of the family of car-
dioidesr a(1 + cosO)is(l + cosO)dr+ rsin9dO = 0.
4. Show that the differential equation of the system of rectan-
gular hyperbolas xy = c 2 is x dy + y dx = 0 ,and interpret the
result geometrically ; deduce that the tangent intercepted between
the axes is bisected at the point of contact.
5. Vertify that y + x + I 0 is a solution of the differential
equation (y— x)dy - (y 2 — x 2 ) dx = 0.

6. Show that V = (A / r) + B is a solution of the differen-

tial equation
d2V
- +2 - -dV= 0.
dr 2j ' d r

7. Find the differential equation from the relation

(J.E.E.'89)
(i) y = A sin x + B cos x +x sin x;
I). E. E. '84)
(ii) y = A c + Be -' ;
 330
1EGRAL CALCULUS Ex.XV
(iii) y A cos x + B sin x + C cosh x + D
where sinh r,
A .. B, C, D are arbitrary constants.
S. Eliminate a and b from each of the relations
(i) y = alogx + b; (ii)xy = ae + be
(iii) ax 2 i by2 = 1;
EC.P.1945J
(iv) r = a + b cos e.
9. (I) Show that the differential equation, whose genera
solution is y C1 x + c 2 x , is y = xy1 _I x2y2

(ii) Show that

Y = cosx,y = Sinx,y =c 1 cosx,y =c 2 Sinxareall

solutions of the differential equation y2 + y = 0.
In (i) and (ii), c1 , c 2 are arbitrary constants.
I
10. Show that the differential equations, whose general solu-
tions are

(i) y = A sin x + B cos x,

C. P. '88
B cosh x,
(ii) y = .4 sinh x +
where A and B are arbitrary constants, are respectively

= 0 and —'— y = 0.
dxl

ANSWERS
I. The radius vector and the tangent at any point are mutually perpen-
dicular.
4. The radius vector and the tangent at any point are equally inclined
to the X-axis.
7. (i) y + y 2 = 2 cos x.
(ii) Y2 - y = 0 (iii) y - y = 0.
8. (I) 13'2 + y' = 0. (U) xy 2 + 2 Yi = y.
(iii) x ( yy, t y ' ) = yy1 (iv) r2 = r1 cot 9
 CHAPTER XVI
EQUATIONS OF THE FIRST ORDER AND THE FIRST DEGREE

16.1. A differential equation of the first order and the firt de-
gree can be put in the form
M dx + N dy = 0,
where both M and N are functions of x and y. or constants not
involving the derivatives. The general solution of an equation of
this type contains only one arbitrary constant. In this chapter we
shall consider only certain special types of equations of the first
order and the first degree.
16.2. Separation of Variables.
If the equation M dx + N dy = 0 can be put in the form
f 1 (x)dx + f (y)dy = 0,
then it can be immediately solved by integrating each term sepa-
rately. Thus, the solution of the above equation is
if, (x)dx + if, (y)dy C.

The process of reducing the equation Mdx + N dy = 0 to the

form fi ( x ) dx + f 2 ( y ) dy = 0 is called the Separation of the V ari-
ables.
Note. Sometimes transformation to the polar co-ordinates facilitates
separ-ation of variables. In this connection it is convenient to remember the
following differentials
if =r cos O, y= r sin O.
(I) x dx + y dy = r dr, (ii) dx 2 + dy 2 = dr 2 + r 2 dO 2

(iii) x dy - y dx = r I do

For illustrations, see Ex. 8(u) and (iii) of Examples X V I(A ). I

16.3. Illustrative Examples.
Ex.I. Solve (l + y )dx i. (1 + x 2 )dy = 0.
Dividing by C I * x )( 1 + y 2 ) we get
332 INTEGRAL CALCULUS

dxdy + + =0.
I+ x2 y
Integrating, tan -' x + tan - = C. ... (1)

Note. Writing the arbitrary constant C In the form tan - I a, the above
solution can be written as tan'x + tan 1 y = tan-'a,

or, lair' 1!..t1. = tan-'a, or, +y=a(1-xy) ....(2)

Both forms of solutions (1) and (2) are perfectly general and any
one of these can be considered as the complete solution of the given equa-
tk,n. [See Art. 253.
Ex.2. Solve x(y' +1)dx + y(x 2 +1)dy = 0.
Dividing both sides by (x' + I )(y' + 1), we have
X ___
x'+l dx+ y' +I dy=0.
integrating, we have
jlog(x 2 +1) + log(y1 +1) = C.
Writing 1 log A In the place or C, the above solution can be written in
the form
(x' + I ')( y 2 + I)= A.
Note. In order to express the solution In a neat form, we have taken
log A (A being a constant) in the place of the arbitrary constant C.

Ex. 3. Solve (x + y)' A

dx =a'
Putx+y=v, i.e.. y=v-x. .
dx 7x1
the equation reduces to
dv a2 a'.,'
dx iv
) =a',or,--I+_-1=
dx 2
"(--2 D

(
dx
= a' + a
dv
Integrating, fax =fav - a2
5 a' +
 FIRST ORDER - FIRST DEGREE 333

+
or, x + c = v - a .-a tan' -a = x + y - a tan ' a

y = a tan - + C Is the required solution.

a
Ex. 4. Find the foci of the curve which satisfies the differential equa-
tion (I + y t )dx - xydy = Oand passes through the point (1,0).
Separating the variables of the equation, we have
dx - ydy .,
x l+y2

Integrating, logr — flog( I + y 2 ) = log C,

or, log log C. .. x = C 4iy 1

+1)=

This is the equation of any curve satisfying the given differential equa-
tion. If the curve passes through (1,0), we have I = C.
the equation of the required curve Is x 2 - y 2 =
It is a rectangular hyperbola. and Its foci are evidently (± 42 .0)
Ex. 5. Show that all curves for which the length of the normal is equal to the
radius vector are either circles or rectangular hyperbolas.
Since the length of the normal = y 4(1 + yt 2 ) and the radius vector
=4(x 2 +y2).
... y 2 (I+y 1 2 )=x 2 +y l , or, y 2 yj 2 x 2 , or, yy=±X

±. :. xdx ±* y dy = 0.
Integrating, x 1 ± y 2 = a , a 2 being the arbitrary constant of in-
tegration.
Thus, the curves are either circles or rectangular hyperbolas.
dy Lx. 6. Show that by substituting ax + by + c = z in the equation
= f ( ax + by + c) the variables can be separated.
die dz
Since ax+by+cz. ..a+b dx dx

dx b\dx
334 INTEGRAL CAL(ThLIJS

Hence, the given equation transforms into

\
=f(z),
dx
i(dz
dz
a + bf(z) dx.
Thus, the variables are separated.

EXAMPLES XV! (A)

Solve the following differential equations ( Ex. I - 10

1. (i)
.I: y = x 2 + x + I
dx y 2 + y + I (ii)x2t
x-+v = 1.

I =
(y -
dx + yx(x-1)
2. (i) ydx +(I + x2 )tan I xdy = 0.

(ii) eYdt +eY - 'dy = 0.

3. (i) xJi-
- y ' dx + y Ti_ . x 2 dy = 0.

(ii) x 2 (y - I )dx + y 7 (x - I)dy 0.

4. - + + 1
dx xt+x+I 0.

5.(j) ..Y+\/y 2 -0. (ii) x(I +LJ

dx v l_x2 dx y(1 +x2)

d
—^
'I( _IXy 2 -I)
dx =0,
xy

6. (i) sec x tan y dx + sec y tan x dy = 0.

(ii) x cos 2 y d x - y cos 2 xdy = 0. I H.S.'85J
log ( sec x + tan x)
dx - +_ tan y)
cosx - cosy
7. (x 7 - yx' )dy + (y 2 + zy' )dx = 0. IC.P.'88)
 x. X V !(A ) FIRST ORI)LR - IRST DEGREE 335

S. (I) y dx - x dy xy dx.
(u) x'(xdx 4 ydy) + 2y(xdy - ydx) 0.

(1 - X
(••) xy, x2 +y
-y

c. (i)41 = e'V . (ii) ='Iy-x.

10. i sin' () x
+ Y.(ii)log
(f)
a = x - by

11. Find the particular sciution of

cos ydx + (1+ 2e )sin ydy = 0.
;shenx = 0.y =-n.
12. Find the equation of the curve for which
(I) the cartesian gubtangent is constant.
(ii) the cartesian subnormal is constant.
(iii) the polar subtangent is Constant.
(iv) the polar subnormal is constant.
13. Show that the curve for which the normal at every point
passes through a fixed point is a circle.
14. Show that the curve for which the radius of curvature at
every point is constant is a circle.
15. Show that the curve for which the tangent at every point
makes a constant angle with the radius vector is an equiangular
spiral.
16. Show that the curve in which the angle between the tangent
and the radius vector at every point is one-half of the vectorial
angle is a cardloide.
17. Show that the curve in which the angle between the tangent
and the radius vector at every point is one-third of the inclination
of the tangent to the initial line is a card bide.
336 INTEGRAL CALCULUS Lx. XVI(A)

18. Show that the curve in which the portion of the tangent in-
cluded between the co-ordinate axes is bisected by the point of con-
tact is a rectangular hyperbola.
ANSWERS
1. (j)(x -y 3 ) +j. (r l --y2 )+x - y C. (Ii) y I +Ce".

xy = c (x - 1 )( y - 1) . 2. (i) y tan -' x = C.

(ii) e 2x + e 2 V = C 3. (0 ./T? + 11 =

(ii) (x +1)2 + (y + 1) 2 + 2 log (x - 1 X -1) = C.

4.2xy+X+y+C(X+Y+I) = I . 5.(l) sin- I x +sin'yaC.
(ii) I + y2 =C( I +x 2 ) . (iii) 4(x2 -1 )-sec-'x+ 'J(y' —1) = C.
6. (I) tan x tan y = C.
(ii) x tan r - log sec x = y tan y - log sec y + C.
(iii) [ log ( sec x + tan x)P - l log ( sec y+ tan y)P C.

1. log =C. 8.(i)ye , =Cx. (Ii) ( x 2 +y 2 )(x+2) 2 =Cx1

(iii) X + y 2 =sln 2 a, where at =tan' (y/x) + C.
9. (I) eY +e'^Ce . (it) ^log('l-1)jX+C.
10.(i)tan(x+ y)sec(x + y) = C +x. (ii)aeY+ be"= C.

11. (e' + 2) sec y = 3 42 . 12. (i) y = Cc'

(ii)y 2 =2aX+C. (iIi)r(C_O)=a.(iV)aO+C.
16.4. Homogeneous Equations.
If M and N of the equation Mdx + N dy 0 are both of the
same degree in x and y and are homogeneous, the equation is said
to be homogeneous. Such an equation can be put in the form

dx
Every homogeneous equation of the above type can be
easily solved by pjltting y = vx where v is a function of
, whereby it reduces to the
dv
and consequently ax- = v ^ x ()
 FIRST ORDER - FIRSF DEGREE 337

and consequejtly jdyy = 0 + x ) ,4whereby It reduces to the

form v + x ( -) = f(o), i.e., - in which the variables
are separable as shown below.
Ex. Solve (x' +y' )dx - 2rdy 0.
The equation can be written as
.x =1
dx 2xy
Putting y = ox,sothat A = v + x ,wehave
dv x + vx 2 -.
-
I + 01
dx 2vx2 2r
dv 1+01 1v2
-v= 2v
dx 2v
dx 2v dv =0.
x Iv2
integrating, log + log (l -v 2 ) log C.
r(I-v3)=C.
Re-substituting y / x for v and simplifying, we get the solution
- y 1 = Cx.
16.5. A Special Form.
The equation of the form
yalx+biy+cl(abi (1)
dxa 1 x+b2 y+c1 a2 b1
can be easily solved by putting x x' + h and y + k,where
It and k are constants, so that dx dx' and dy = dy' and choos-
ing It , k in such a way that
a1 h+b1 k +c1 =0
and a 2 h+b2 k +c2 =0 2
For, now the equation reduces to the form
'= b, y'
a, x' -s-
+ b,y'
dx ' a2 x'
which is homogeneous in x ' and y' and hence solvable 1w the
method of the previous article.
Inlegral Calculus (main) -24
$38 INTEGRAL CALCULUS

Note 1. The above method obviously fails, ifs; / 5 = l' / b ; for In

this case h and It cannot be determined from equa v ion (2).
Let the equation be

ajz+b1 y+c1 ( '1'

At + b 3 y + c2 a
dz
a b I
Let - 1- - .. atm, b, =
2m
'here in Is a non-zero constant.
Assuming this to be the case, let he common v!ue of these ratios be
Ienoted by 1/re • so that l = a t M and b2 = b, in.
The equation (3) becomes
a1x+b.y +c1
dx m(a 1 r t bay) + c2
Now, putting a x s. b1 y = o, the variables can b- 7as s ly seporat . J ;J
hence the e q uation can be solved. [See Ex. 2, i'eiou'.]
Note 2. If In the equation (1), &2 , then the ution can be
solved more easily by grouping th, terms suitably. j See Et. 1(i) of Ex-
am ples XVI(C .)

16.6. Illustrative Examples.

Ex
6x - 2y -7
dx 2x+3y--6
Putting x =x+h,y =y' +k, so that ix=d,4y=dy',w ehave

= 6x' - 2y'
dk' + 6h - 2k - 7
dx' 2x'
+ + 2.J, + 3k - 6
Putting 6k- 2k - 7 0 and 2h + 3k -6 = 0, and solving these
two equations, we have h ,k I
' 6x' - 2y '
the equation becomes
dx' = +
Since the equation is now homogeneous, putting y' = ox' an.! henc.
x' ., and simplifying, the eq.istion reduces to
dx' =-
1 6o+4
+ 4V - do, which on integration giv
- log AY =.log(3v2 +4v -6).
(Ax')-' (302 +40 - 6)112.
 1W ORDER - PST DGRBE 339

Now, restoring the values of x' and v', where x ' x- and
v y'/Y' 2( y - 1) / (2x - 3) • we get the solution in the form
3y +4xy- fix 2 -12y+14xC.

Lx. 2. Solve -.

Since here a, /a b/b, , .. putting 3x- y v,weget

y dv, and hence the given equation gives
3 - ,
= - 2r-7 p + 19
Z v 'i'4 p+4
____ dv.
di rp44 dv = 1- + 19
)
+ 19 (
x+c=v-l5 log (v+19).
On restoring the value of v, we get the solution In the form
2x- y - l5 log (3x - y + 19) =C.
Lx. 3. Show that i, an equation of the form
yf 1 (xy)dx i xf2 (xy)dy -0,
the variables can be separated by the substitut ion x y 0.

Since xy v,y and d(xy)r dv,i.e.. ydx + idy = dv

p dx i.e., x dy = dv - I dx.
arid dy ..

(v)dx + f1 (o)(dv 1 dX ) 0.

• b(_
v (f (v )f z (V )) + x -
Thus, the variables are separated.
(See Ex. 14,25,16 of Examples X V I(B) .)
We car as well form an equation In v and y, by taking xy v, x Illy
yd -iv
and
(For Illustration see A lternative proof Lx. 4 of A rt. 16.7. J
 340 INTEGRAL CALCULUS

EXAMPLES XVI(R)
So!ve(Ex.1 - 15):-

x+y = 2y. (ii) - + = i.

dx x x2

2. (1) y - 2y )
(ii) y
dx x ( x - 3y) dx x ' + y3
3. (x 1 +y 2 )dy xydx.

4 (1) = (ii) y ( y + x)
dx X+ y dx x(y - x)

5. (1) (3xsi'h (y/r )+5ycosh (y/x)dx-5xcosh (y/x)dy = 0

(Ii) (1 + 3e - / Y ) dx + 3e4 ( 1 - xfy)dy = 0.

6. (x2 -2xy)dy + (x 2 -3xy + 2y 2 )dx = 0.

7. v2dx + (x 1 +xy)dy = 0.

8. W -41 = Y
dx x + tan xI. HS.'81, '83, '39 1 (ii) dx 2x.-3y
9. (6x _5y +4)dy +(y _2x _1)dx 0
10. (x .- + 4)dy + (7y - 5x)dx = 0.
11.
12.
23. y(2xy +1)dx + x(1 + 2xy + x 1 y 7 )dy = 0.
14. x 2 y 3 d + 3xydy + 2ydx = 0.
15. (1 +xycosxy)dx + x 2 cosxydy = 0.
16. Show that ( 4x + 3y + 1 ix + (3x + 2y + 1) dy = 0
represents hyperbolas having as asymptotes
x + y = 0,2x .- y + 1 = 0.
 FIRST ORDER - FIRST DEGREE 341

ANSWERS
1. ( j ) y = x + CcI t Y''. (ti) 2x - y = Cx1y.
2. (i) y 3 ', = Cx 2(II) y' CC"/Y'

3. y =Ce'/Y 2 . 4. (i)y 2 + 2xy - = C.(u)xy = Cc Y1

5. (I) x 3 = Csinh 3 (y/x). (it) r + 3ye' / Y = C.

6. y = xlog(Cx ). 7. xy2 C ( x +2y).

8. i> x = c sin (ri) 3 log (x 2 +y 2 ) = 4 tan -' I + C.

9. (Sy-?%-3 ) 1 =C(4y-4x-3).10. (3y-5x+ 10) =C(y-x+1 )

11. 2y-x+C=Iog(x-y+2). 12. 6y-3x =log(3x+3y+2)*c.

13. 2x 2 y 1 log y .-4xy — I = Cxw' 14. x(xy-.. 2)' =C(xy-1 >3

15. xe'''Y = C.

16.7. Exact Eqdation.

The differential equation Mdx+ Ndy = 0, where both Al and
N are functions of x and y, is said to be exact when there is a func-
tion uof x, y, such that Mdx+Ndy =du, j2., when Max+ Ndy
becomes a perfect differential.
Now, we know from Differential Calculus that M dx + N dy

should be a perfect differential if -- = - . Hence, the condition

ay ax
that M dx + N dy = 0 should be an exact differential equation is
àA I aN
ày ax
The method of solving an exact equation of the type
Mdx + N dy = 0 is as follows:
"First integrate the terms in M dx as if y were constant, then in-
tegrate the terms in N dy considering x as constant, and, rejecting the
terms already obtained, equate the sum of these integrals to a constant".
This will be the solution of the required equation.
 INTEGRAL CALCULUS

Ex.1 Solve (20 + 4y)ix •(4x +Y - 1)dy.

Here M2x' +4y, N • 4x +y- 1.
4 hence it Is an exact equation.
JMdx J (2x 3 +4y)d •2!. +4yx . j x +4ry,
JNdy..J(4x+y-i)iy•4y+Y2Y
Rejecting the term 4xy in (2) which already occurs In (I) and then ad-
ding (1) and (2) and equating the .un to a constant, we get the general solu-
tion to be
ijX4 + y 2 + try - y • C.

An exact differential equation can often be solved by inspec-

tion, by picking out the terms of M dx + N dy that obviously form
a perfect differential and then integrating this. This Is Illustrated in
the following worked Out examples. While grouping the terms
suitably to form a perfect differential it will be convenient to
remember the following differentials.

vdx+xdy-d(xy).
ydx - xdy
Q) xdy .- ydx
2
Iy

16.8. Integrating Vac tors.

If a differential equation when multiplied by a factor ( a func-
tion of x, y) becomes exact, this factor Is called an Integrating Fact
tor of the equation. An integrating factor Is sometimes shortly
writen as I. F.
Integrating factors can often be obtained by inspection. This is
illustrated in Ex. 2 below. An equation can have more that one in-
tegrating factor, this is also illustrated in Ex. 3 below.
16.9. Rules for determining Integrating Factors.
Let the differential equation be
Mdx +Ndy- 0. ... (1)
The condition that It should be exact • is
aMaN
(2)
YX-
 343
FRZr ORDER - FIRST D&W.EE

.. (I). II is a finctInn of x only, say f( x), tile?"

N
ef'' will be an integrating factor of (1).

If M dx + 14 dy 0 , be an uct equation, when it is multi-

plied by e" we must have

-- (Me') = ax
ly
+ Ne l l()' f(x)
ay ax

i.e.,
aN am
Rule (II). If f(y)(a function OI y alone)
M
ef," is an integrating factor.
Pzoof is similar to that given above.
Rule (III). If M and N are both homogeneous functions
in x, y of degree n (say ) , then

(Mx + ?Jy* 0)
is an integrating factor of the equation M.
W. care easily .how that
dj M Jj N
y\Mx+NyI UMx+Ny
if we remember that M and N are homogeneous functions of degree
6M
a and hence x .- + y -- am.

a aN
and x +y aN.
 344 INTEGRAL CALCULUS

If Mx + Ny -0, then - and the equation reduces to

y dx - x dy 0 which can be easily solved.

Rule (IV ). If the equation (I) is of the form
yf(xy)dx + xg(xy)dy 0,

then Mx Ny ,(Mx - Ny-* 0)

is an integrating factor of (I).
We can easily show that
a M \ a N
Mx-Ny).1..Mx-wy

ar yf(xy) xg(xy) 1
[xy(f(xy)_g(xy)J - xyi.f(xy)- g(xv))J
provided we remember y T F(xy) x F(xy).

If, however, Mx - Ny = 0, then = - and the equation

reduces to xdy + ydx 0 which can be easily solved.

16.10. Illustrative Examples.

ELI. Solve :(2x2 + y 2 + x)dx + zydy 0.

M
Here2y; - - y. the equation is not exact.

am aN
Now, ay = -
N xy x
by Rule (I), IF. =glI) =e 1 5 = x.
Multiplying both sides of the given equation by x, we hdve
(2x + xy 2 +x')dx + x'ydy 0,
 FIRST ORDER - FIRST DEGREE 345

or, 2x3dx + x 2 d.x + xy(ydx + xdy) 0,

or, 2x3dx + xdx + xyd(xy) z 0,
or, 2x' dx + x3 dx + z d x 0, where z = xy.
ry Z2
Integrating, -!^ mci.

the required solution is 3x 4 t 2x 3 + 3x 2 y = C.

Ex.2. Solve: (x 5 +y ' )dx - xy 2 dy = 0.
Here, -. = 3y -y2 . .. the equation is not exact.

1 1 1
Now, =
Mx + Ny x 4 + zy I - xy 3 x'
The equation Is homogeneous.
by Rule (III), I / x 4 Is an integrating factor.
Multiplying both sides of the given equation by I I x 4 , we have

(. -

This Is exact.

Mdx f
+--)
dx = log 3 x3

3 X3

by Art. 16.7, the solution Is

log = c,i.e., y 3 = 3. 3 logx +cx3

3-ix
Ex. 3. Solve (y + r) dx + x dy = 0.
The equation can be written as
(ydx + xdy) + xdx = 0, or, d(xy) + xdx = 0.
Integrating, xy + 1 x = C is the required solution.
 46 INTEQRAL. CALCULUS

E. 4. SOIDC v (1 s x dx - x dy 0.
The (qua!ica zan bi c'1tten as
ydx-. zdy + yzdx = 0.

Div; d' ng b, j_d rd.i 0 'jr, d xdz 0.

Hcnc iregrang, tie aolution is 1 + x c

Note. l-iere I / y' 'n integ rating facter.

Ex. 5. Soe y dx - x dy = 0.
Multiplying the given equation by this can be written as

Y dx4ijr =0, i.e. i(-!)=o. i.e., = C, i.e.,x=Cy... f)

A gain, multiplying the equation by I / a 1 , this can be written as

xdy-ydx = o, i.e..d( 1 ) .0, i.e., 1 c, i.e.,y=cx....(2)

Again, multiplying the equation by I I xy, this can be written as

y d — x dy -
xy i.e.,
=!
a A o .. log a - logy = log C,
y
i.e., log - = logC.

- = C, i.e.. a = Cy. ... (3)

Further the equation can be written as

(4)

This belongs to the linear (cnn (See Art. 16.11).

Note. Thus, we see from (1), (2) and (3) that the number of integrating
factors of an equation is w,o,e then one and from (4) we find that a differen-
tial equation can be solved by different methods.
 FIRST ORDER - FIRST DEGREE 347

Ex. 6. Solve
(x 3 y' •x 1 y 2 + xy+1)ydx+(x'y ) - x 2 y 2 - zy + l)xdy O.
Now, (x-* Y' + x 2 y 7 + xy + 1)x'y 2 (Zy + l)+(xy + 1)
(xy + l)(xy 2 + I)
and zS y s _xlyl_xy+1x2y1(XY1)_(xY_l)
= (xy-! )(xy' -1)=(xy-1)2(xy+l)
the given equation become,
(xy + 1)(x 2 y 2 + I )ydx + (ry- I ) 2 (xy + I )xdy 0,

or. ( z 2 y 2 + I)ydx+(xy- I) I xdy O, (I)

cancelling the common factor (xy +
or, ( x 2 y' Zxy + I)xdy 0,
+ I)ydx + ( x 2 y 2 -

or, xy (ydx + xdy) + (ydx + xdy) -2xydy 0,

or, x 2 y 1 i(xy) +d(xy) _2x'ydy = 0,
or, d(xy) s' >_ dy = 0 [on dividing byXYL

or, dv + -dy
1 =0 Iputtiigzy =

or, v - - - l log y = C,

or, xy- 1 - 2 log y = C.

zy C

Alternative MethL4

xy = v, so that x dx = '
dv ty
Putting
ye get, on simplification,
(i+ VI dv_31 0.
Y
integrating, . - . - 2 log y C,

i.e., zy--2logyC.
348 INTEGRAL CALCULUS

EXAMPLES XVI(C)
Solve
1. (i) (2x-y +1)dx +(2y- x -1)dy = 0.
- + ax + hy + g = 0.
dx hx+by+f
(iii) (1 -x 2 ) Zxy = x- x3
dx
(iv) , y =
dx --6y-s-2

2.(i)x + y =y 2
dx log x.
(ii) x - =y + cos
dx x
3. (i) xdx + ydy + (x 2 + y 2 )dy = 0.
(ii)x'y 1 + xy + 2 '(1 - x 2 y 2 ) = 0.
4.(i)x dy _y dx +a(x 2+y 2)dx .0
(ii) xdy - ydx - 24(x 2 - y 2 )dx = 0.

S. (I) xdx + ydy+ xdy - yd x 0.

(ii)(x +x 5 +2x 3 y 2 + xy 4 )dx + ydy = 0.
6. (i)(x +y )dy +(y _)dx Q.
(ii) (x + y)(dx - dy) = dx + dy.
(iii) (2 - xy)ydx - (2 +ry)xdy = 0.
(iv) 2xydx- (x2 - y 2 )dy = 0.
(v) (xy 2 + 3e' )dx - x 2 ydy 0.
7. (i) (x 3 + 3xy2)dx +(y 3 + 3x 2 y)dy = 0.
(ii) (x 3 y 2 -y)dx - (x 2 y+ x)dy = 0.
(iii) (x' + y 1 +2z)dx + xydv= 0.
 Lx. X V I(C) FIRST ORDER -FIRST DEGREE 349

8. (i) sin - y cos x

+ y' = 0.
dx
(ii) ( xy cos xy + sin xy ) dx + x 2 cos xy dy = 0
(iii) xcos(y /x)(ydx+xdy)y sin (y/x)(xdy_ydx).
(iv) (cosy + ycosx)dx + ( sinx - xsiny)dy = 0.
9. (I) y(l + xy)dx + x(1 - xy)dy = 0.

(ii) (x + 2y') = y.
(iii) ( x 2y 2 + xy)ydx +(xy- 1)xdy = 0.
10. (x 2 + y2 + 4)xdx +(x 2 - y 2 + 9)ydy = 0.
11. (1 + 3x 2 + 6xy 2 )dx +. (1 + 3y 2 + 6x 2 y)dy = 0.
12. Solve (I + x 2 ) y1 + 2xy = 4x' ,and obtain the cubic
curve satisfying the equation and passing through the origin.

ANSWERS
1. G) X ' +y 1 xy+xy=C.(ii)ax 2 +by 1 +Thxy+2gx+2fy=C.
(iia)y(1 - x 2 ) = I - x 4 i-C.üv)5xy-3y 1 s-2y--2x 2 -3x=C.
2. MI + Cxy = y (I + log X). ii .Y + sin = C.
3. 0)x 2 +y' =CeY . (ii) sin-, x(ry) +x log x = C.
4. (i tan - ' Y .- ax = C. 60y r sin log (Cr2

5. (i) (x2 +y2 ) + tan-' I = C. (ii) (C+ x 1 )(x2 +y 2 ) =

6.(i) x 1 -y' -2xyC. (iA)x -y +C= log (x +y).

(iii) log (x/y)- xy= C. (iv) x 2 's- y =Cy.
(v) y' /r 1 = 2e'/ C.
3 +
7. (i)r 4 1-y 4 +6x 1 y 2 = C. (ii)r 2 -- y 2 + 2 x = C.
(iii)3r' +$x +6r1y
S. (1) sin x = y (x 1. C). (ii) x sin zy = C.
353 ThTEC.RAL CALCULUS

(Iii) xy cos (y / x ) C. (iv) x c's y + y Mn z = C.

9. i) log (2 ,'y)-1 fry C.(ii) X y + Cy.(lIi)xy- log y C.
10. x 4 - y 4 i 8r 2 + iy + 7x 2 y 2 = C.

11. + y + x' + + x 2 y 2 = C.
IX, + C; 3y(l + x 2 ) 4x3.
fl. y(l + x 2)

16.11. Linear Equations.

An equation of the form

I + Fy = Q
dx
in which P and Q are functions of x alone or constants is called a
linear equation of the first ordcr.
The general solution of the above equation can be found as fol-
lows. Multiply both sides of the equation bye IP

+ Pye1" =
ax

JPd ) = QeJP l.
( YO
dx
integrating, ye 1 = I Qe 1 P d dx + C,
or, y = e-1' d, I I Qe dx + CI is the required solution.
Cor. 1. If in the above equation Q is zero, the general solution Is
y =C'1"
Car. 2. ii p be a constant and equal to - m, then the solution Is
Y= e' tJ" Qdx +C).
Note. Here the factor e - , which renders the lefthand member of
the equation a perfect differential, is called an Integrating Factor. It Is some-
ti-les shortly written as 1. F.

Es. Solve COSX dx + yam x= I

Dividing by C°S I dx . y tan x sec x. ... ()

 FIRST ORDER - FIRST DEGREE 353

Here Pdx =J tan xdz log sec x.

the integrating factor Is e °5 '' sec x
Mult.plying both side, of (3) by sec z we have

seer iy + y sec r tan s = sec x, or, (yseex) =sec 2x.

1. negrairg. y seer = tan r • C 4q the rouarcd

16.12. Bernoulli's Equation.
An cquit'n of the form

+ Py = Oy,
dx
where F and Q are functions of x alone, is kflown as Bernoul(s
e.juatidn. It is easiy reduced to the Iinc.r form of Art. 16.11
shown below.
Div : 3 b'th ids by y' • , w get

-.-Py' 1 = Q.
dx
dv dv
Putting y-, v, and hence ( - )y -'
the Pquation reduces o

dv
- n)Pt, = (1 - n)Q.

This being linear in v cn be solved by the method of the

vious article.
Note. I-lre ,i s a ratIona nurvcr

16.13. !liust.raive Examples.

Ex. 1.
-. di,
d+ --v
x
-
d.,1-x 2
Divi,;jng both sides of the eqeation by y w
- d x
Y — ______ = ( 1)
dx I - x2
Put ylfl = .!.-/2 =
dx T.
352 INTEGRAL CALCULUS

(1) reduces to + 2(1- r 2

V .. (2)
which is of the linear form.
Here JPdx
I. F. = CI0(l
5 dr
2(1-x 2 ) = - log(1 - x) = log (I
1/4
= ( 1 - x )
-x2 )-114

multiplying (2) by( 1 2 ) - 1/4 and integrating, we have

(1 - x P'4 =52(1 - r2 )l/i dx = -( I - x 2 +

substituting the value of v, the required solution Is

= - .1 ( I - x I ) + C(1 - x 2 )I/4
Ex. 2. Solve + xsin2y = x' cosy.
Dividing by cos 2 y, we have

sec 2y'+
dx
2x tan y (1)

Putting tan y = z, so that sec 2 y . = dz

dx
, we have

dz
-+2xz=x. ... (2)
This is of the Irnoar form. Here I. F. = e t 2 z Ax =

in (2) by e x and integrating,

ze'- 1 = fxe I 2 dx + C = (x' -1) + C (integrating by parts)
e tan y = - e (x2 - I ) + C Is the required solution.

EXAMPLES XVI(D)
Soive ( Ex. I - 14)
1- Zr
1. (I) dx += -.
dx x2

2. •- +y cot x = 2 cos x.
dx
Ex. XVJ(D) FIRST ORDER - FIRST DEGREE 353

3. cos 2x_+y=
dx
tan x.
4. () (1 xZ) xy = 1, (ii) - + xy x.
dx dx
(ii) +
dx x
5. ( x 3 - x)y 1 - ( 3x- l)y =x I - + x.

6. (xcosx)y 1 + y(xsinx + cosx) = 1.

7. ( i + y 2 )dx - ( tan-'y - x)dy 0.

8.(i)y2+(x-!)=0(li)(x+y+1)X=l.
y dx dx
(iii)(x 2 y 3 + 2xy)dy = dx.
Write as a linear equation in x

9 Y+iy=yt. 10.y 1
dx
-2y tan x-sy' tan x=O.
1+ 1L1
dx x x 1dx x x

12. sin 2y =x 3 cos 'y . I Put tan y z

dx
logy = 1(logy)2
14. --
dx I +x
= (1 + x)e' sec y.
15. Solve = x 2 , given y 1 when x = I
.! + Y
16.Show that the equation of the curve whose slope at any
point is equal to y + 2x and which passes through the origin
is y = 2(e' - x 1).
17.Find the curve for which the sum of the reciprocals of the
radius vector and the polar subtangent is constant.
18. Show that the curves for which the radius of curvature
varies as the square of the perpendicular upon the normal belong
Inlegral Calculus (main) -25
"4 INIF.GRAL CALCULUS Ex. X V I(D)

to the class whose pedal equation is

P3 = + +
k being a giv,en constant and A an z,. trary constant.

ANSWERS

I. (i)y = x- 1 + Ce'. (6)y=x1(1 + Ce11).

2.'y sin + C (sin x)! 3.y = ( tanx -1) + Ce'

. = C + sin -': iii) yr I + Ce l,2

(itO zy +- . cosx' =C.
5. y (x 3 - x log Cy. 6. yr sec x = tan x+ C.
1y
7. x ,. 1 = Ce - + tan y (I) x-1= y I + Ce11Y
(ii) r + y + 2 = Ce)' (ill) 2r I =l_y I .Ce'

9. y ' = Cx — x log x. 10. 5scc 2 x = y(tansx .. C)

11. i) x=y + Cx2y. (ii) 2r e? (1 * Cx ).
12.6i tan y=r' +c. 13. x '. log y(Cx 2 +

14. sin = (1 + x)(e' +C) 15. 4xy = x4 +3

17. r r + Ct°
 CHAIflR XVII

EQUATION OF THF MST (W DEf BUT NOT ( THE FIRST

DEGREE

17.1. The typical equStion of the first order and the nth degree
an he written as
F +Pi p" 1 -P1 p' +P,. r 01 , ... (1)
where p str.is for and P P2 .....,F,, are fuctons o
X ,., nC y.

The complete solution of such an equation would involve only

one arbitrary constant.
Two cases may arise
Case 1. When the left side of the equation can be resolved into
rational factors linear in the derivative.
Case II. When the left side cannot he thus factorised.
17.2. Left side zesolvable into factors.
In this case, the equation (1) takes the form
(p_f,(x,y),p— fz(x,y))..(p— f(x,y))=O. ... (2)
It is evident from above that a solution of any one of the equa-
tions

P -ui (x.y) O,p -1 (x,y) = U,etc.

is also a solution of (1)
Let the solutions of the equations (3) be
(x,y.C1 )=O,, (x,y,C2 )=O,...,+,. (x.y.C.. )=O,
where C, , C2 . . , C. are arbitrary constants of integration.
These solutions are 6ident1y just as general, if C, C2 . .= C,.,
since the individual solutions are all independent of one another
and all the C's can have any one of an Infinite number of values. All
 356 INThGRAL CALCULUS

the solutions can thus, without los of generality, be corn

bined Into
, (x,y,C)*I(x,y,C) ..... .,1(x,yC) = 0.
17.3. Left side not resolvable into factors.
Here, we shall consider only the following cases:
(A) Equations solvable for y.
Suppose the equation Is put in the form
Y = f (x,p).
Differentiating this with respect to x, we shall get an equation in
two variables x and p; suppose the solution of the latter equatior
is(r,p C)=0.
The p-e!iminaut between this relation and the original equa
tion gives a relation between x, y and C, which is the requirec
solution.
(B) Equations solvable for x.
X f(y,p).
dr
Differentiating this with respect to y, and noting that = lip,
we shall get an equation in two variables y and p . If p be
eliminated between the solution of the latter equation ( which cor-
tains an arbitrary constant) and the ociginal equation, we shall get
the required solution.
(C) Equations In which either x or y is absent.
In such cases, the variables are easily separable, or they nay be
solvable for y or for x.
Note. In case the elimination of p annoI be effected, or It leads to com-
plicated expression., It Is customary o express x and y separately in
terms of p. and these values of x .1 y. as If parametric equations, con-
stitute The solution of the equation.
C1.1raut'sEquation y = px +f(p)[A rt. I7.5Jas also its extendedf
y =x$(p) + f(p)[ Ex. 2, A rt. 77.5j be',ngto this type.
 FIRST ORDER - }UGHER DBGREB

17.4. Illustrative Examples.

Er. 1. Solve p 2 + 2px + py + 2xy - 0. C. P. 'SS
The equation can be written as
(p+ 2xX p + y) 0.

..either p+2z=O,ix, -+2x.0,i.e.,iy,2xix'0,

dx
C;
whence, integrating, y + r 3

otherwise, p + y = 0. or, A + y • 0, i.e., + dx . 0,

SIX
whence, Integrating, log y + x .= C.
the required soIutiirls (y .' x t -CX x + lo y- C) - 0.
Ex.2. Solve 4xp7 - -x 0.

From the equation,

y 3 x (4-
differentiating with respect to x, we have

I i I

which on simplification reduce, to p dx x dp the Integral of which is

P =Cx. (1)
Now, eliminating p between (I) and the given equation, the required
solution is 4Cix2 - 8Cy - I = 0.
Ex.3. Solve x = y -p2
Differentiating with respect to y, we have

! =1 - 2p
P dy

dy =2 P P dp 2{ p + i

integrating. y = p 7 + 2p + 2 log (p - I) + C ... (1)

and hence, x = 2p + 2 log ( p - I) + C. ... (2)
(1) and (2) taken together, or the p-eliminant of them constitutes the
general Solution of the given equation.
358 INTEGRAL CALCULUS

Ex. 4. Sote 4yp - ?pr y 0

We can write the equation a

r=2yp+- ... (1)

Differentiating this with :vspect to we have

f Ei
2 dy2p dy

,
2p
- 1
1 + . 1_f. j = o.
dy + yip = 0, ic.. py C. ... (2)

Sb..tttuting the value of p obtained from (2) in (1), we get the solution
y I = 2Cr - 4C1
Note. It will be noted in this connection that. In solving examples of
this type, the factors containing derivatives which are omitted often give
rise to other solutions of the differential equations which are not included
in the general soiuion. Such SOtUUUflS are termed 'ingu1ar solutions. I See
Art. 17.5 1
17.5. Clairaut's Equation.
An equation of the form

dv
y p + f ( p where p = dX
is called C1airaut equation.
Differentiating both side s, of the equation with respect to x, we
have

pp+xd+,p)d,or.dtx+ttpfl=O.

either - 0
dx
or, x + f'• (p) = 0. .. (2)

From (1), p C. .-. (3)

 FIRST ORDER - HIGHER DEGREE 39

Now, if p be eliminated between (3) and the original equation,

we get y = Cx + 1(C) as the general or complete solution of the
equation.
Again, if p be eliminated between (2) and the original equa-
tion, we shall obtain a relation between x and y which also satis-
fies the differential equation and as such, can be called a solution
of the given equation. Since this solution does not contain any ar-
bitrary constant nor can it be derived from the complete solution
by giving any particular value to the arbitrary constant, it is called
the singular solution of the differential equation.
Thus, we see that the equation of Clairaut's form has two kinds
of solutions,
(i) the complete solution (linear in x and y) containing one
arbitrary constant;
and (ii) the singular solution containing no arbitrary constant.
Now, to eliminate p betwen
y = pr + f(p) and 0 = x + f'(p)
is the same as to eliminate C between
y = Cx +f(C) and 0 = x +f'(C).
i.e., the same as the process of finding the envelope of the line
Y = Cx + f(C) for dilferent values of C.
Thus, the singular solution represents the envelope of the family of
straight lines represented by the complete solution.
Note, It I. beyond the scope of the present treatise to enter Into the
details of the theory of singular solution..

Ex. 1. Solve y = px + p - p'

Differentiating both sides with respect to r,
F
dr+-2
dx dx'

dx (+1-2p)O.
36, DIITEGRAL CALCULUS

either 0, i.e., p. C, ... (I)

SIX

or,x+l-2p.O,i.e.,p.4(x+l).
Eliminating p between (I) and the given equation, we get
y Cx + C - as the complete solution
and eliminating p between (2) and the given equation, we get
Y = 13 (x + I) x ++(x + 1)-.+(r + 1) 2 -!(x +
i.e.. 4y = (x + 1) 2 as the singular solution.
Note. It can easily be verified that the family of straigh.lines repre-
sented by the complete solution touches the parabola represented by the sin-
gular solution.
Ex.2.Solee y=(1 +p)x+ap2.
Differentiating with respect to x, we have

p = (I . p)+ (x +2.ap) dr
dx
• 'ex -2ap.
This is a linear equation in x and p. (See A rt. 16.11. 1
multiplying both sides bye I d0l , i.e., t , we get
eP +t' .x= - 2ap.eP
TP
or, /(xeP)=_2ap.eP.
integrating.. reP =- 2alpe rd p + C =- 2ae' (p-I) + C,
or. x 2a(I -p) '. CeP
y = 2a - ap 2 + (1 + p) Ce P from the given equation.
The p-eliminant of these two constitutes the solution.

EXAMPLES XVI
Solve The following and find the singular solutions of Ex. 5 to
only
1;(i)P2+p_6=O. (ii)pB+2xp_3x1=O.
Lx. X V II FUW ORDI - mciIER DEGREE 361

2. (1) p - p(e +e ) +1 = 0.
(ii) p l y - p(xy + 1) + x = 0.
(iii) p4p 2 + xy) =p 2 (r + y).
3.0)p -(a +b)p + ab = 0. (ii) p(p+x)= y(x +y).
4. (I) xyp 2 - ( x 1 - y 2 )p - xy= 0.
(ii)p 3 - P( X , + xy i-y 2 ) -s-x t y+ xy 2 = 0.
p3_(x2+xy+y2)p+(X3Y+X2Y2+XY3)P_t3Y30.

5. (I) y = p1+ alp. (ii) y = px

y = px +p.
6.0)y=px+ap(1-p). (ii)py=p'(X _b)+Q.

7. (x . - a)p 1 + ( r - y )p- y = 0.
8. (y i- l)p - xp 3 2 = 0.
9. (i) p'x - p l y - 1 = 0. (ii) y ryp l + 2px.
10. sin y cos px - cos y sin px - p = 0
11. (1) x = 4p + 4p' (ii) p 1 - 2xp + I = 0

12. (i) e - 3 p=0 (ii) y = p cos p - sin p

13. (I) y =px + p. (ii) y = (p + r 2 ) x +

14. (i) i yp = ap 2 . ( ii) y = 2px +

lS.p' -p(y + 3) + x = 0. 16. y =A p 3 +Bp2.

ANSWERS
1. (I) (y +3x- c)(y-2z- c) = 0.
(ii) (2ys-3x-c)(2y-x3-c)0.
2. (i) (y - - c)(y+ e - = 0.
(ii) (2y - x - c)(2x- y 2 - C) = 0.
(iii) (y- c)(2y -r 3 - c)(y - cc . 0.
362 INTEGRA L CA LCULUS

3. (i) (y — x — c)(y — br-c) 0.

(ii) (y- ce')(y +x — ce -1) = 0.
4. (t) (ry -c)(x2 - y' — c) 0.
(U) (2y_x7_c)(y_ce')(yfx_J_cea)=O
(fit) (x 3 -3y + c)(e' 2fl + cy)(xy + Cy +1) =0.

5. (0 y=cx+.!;y2 =4ax.

(ii) y = cr + I(aIcl + b1 );. + j...

y =cx + c ;pyI + x (n — 1)'' =0.

6. (I) y = cx + ac(1 — c);(x +a) 2 = 4ay.
(It) Cy =c (x — b) + g ;y = 4a(x — b).
7. (x-a)c' + (z -y)c - y =0;(x + y) 2 = 4ay.
8. (y+ I)c -c 2 r+2 = O ;(y + 1) + 8x =0.
9. (I) c'x - c'y - I = 0. (ii) y' =2cx + c1

10. y = cr + sln'c.
11.() x= 4p + 4p 1 (ii) x =.(p + p')
y 2p 2 i- 3p 4 + c. y . p2 -Iogp + c

12. (') x 2 tan' p - p . c (ii) x = c + cos p

y = Iog(p 3 +p). y = p cog p - sin p.
I3.0)y=p 2 r+p (iI)y =(p+p2)r+p-'
_Iogpp+c _1+ce"P
- (p-I)2 — p2
14.(i) x+ yp = ap2
x(I + p 2 )" 2 = plc + ilog(p +(I + p 2 ) 1")J.
(ii) (3xy+ 2x' +c)' _4(x2 + y) 3 = 0;
15. y (1 - p2 ) 1/1 + ( 1 - p 1 ) = c, with the given relation
16. y A p' +Bp2
X = . Ap' +28p + c.
 CHAPTER XVIU

LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS

18.1 Equations of the Second Order.

Ye shall first consider linear differential equations with con-
Stan coefficients of the second order, Since they occur very fre-
quently in many branches of applied mathematics The typical form
of such equations is

LZ
dx2
+ P Py = X, ... (1)

or, symbolically D I + I', D P1 1 y =

where P , P are constants and X is a function of x only or a con-
stant. Two forms of this equation usually present themselves, nam-
ely, when the right-hand member is zero, and when the right-hand
member is a function of x We shall first consider the first form and
then the second.

18.2 Equations with right-hand member zero.

Let the equation be

+ p1 + Py = 0. .. . (2)

As a trial solution of (2) , let us take y e . Then, if we put

e in the left side of (2), it must satisf y the equation, i.e., we
must have
(,,1 + Pm + P2 )e = 0,

or, since e" * 0, ni 2 + P1 m + P2 = 0 ... (3)

The equation (3) is called the Auxiliary equation of (2)

Let m, , be the two roots of the equation (3)

Then. y = e ' and y = '' are obviously solutions of (2).

Also, it can be easily veiied by direct substitution that

This trial solution is suggested by te solution of the first order linear

equation v Py = 0, which is of same form.
 364 INTEGRAL CALCULUS

y = Ce , y C 2 e-2 and y = Cei' + C 2 e satisfy the

equation (2), and, as such are solutions of (2).
':0w consider the nature of the general solution of the
equatio 2) according as the roots of the auxiliary equation (3) are
(i) real and distinct, (ii) real and equal and (iii) imaginary.
(1) Auxiliary equation having real and distinct roots.
If m1 and m1 are real and distinct, then y = C,eIZ + C2e2
is the general solution, since it satisfies the equation, and contains
two independent arbitrary constants equal in number to the order
of the equation.
(ii) Auxiliary equation having two equal root..
If the auxiliary equation has two equal roots, the method of the
preceedlng paragraph does not lead to the general solution. For, if
in, = mz = a say, then the solution of the preceding paragraph as-
sumes the form
y = (C, + C1 )e I I = Cc ,when C, + C, = C,
which is not the general solution, since it involves only one inde-
pendent constant and the equation is of the second order.
A method will now be devised for finding the general solution
in the case under discussion. Since the auxiliary solution (3) has
two equal roots each being equal to cx, it follows that the differen-
tial equation (2) assumes the form
Li
dx 2 2a
dx
+ a'y = 0.
Let y = v, where v is a function of x, be a trial solution
of this equation. Substituting this value of y in the left side of the
above equation, we have
d-1 v . d'v
dx2 = 0, i.e.,dx-1 = 0, since ^ 0.
Now, integrating this twice, we get o = C, + C, x.
Hence, the solution of (2) in this case is
y = (C, + C,x)e
 LINEAR EQUATIONS 365

This is the general solution of (2), since it satisfies (2), and con-
tains two independent arbitrary constants.
(iii) Auxiliary equation having a pair of complex roots.
If m = cx + ip and m 2 = a - i, then the general solution
of (2) is
y = C1e * + C2

The above solution, by aájusting the arbitrary constants, can

be put in a more convenient form not involving imaginary expres-
sions; thus we have
y=c iC1 e' + C 2 e- 1 ' I
=e [C1 (cos Px + i sin Ox) + C2 (cos Px - i sin Px )J
=e [( C 1 + C1 ) cos Ox + i ( C1 - C2 ) sin 13x I
= e' (A cos x + B sin xJ,
where A = C1 + C 2 and i(C1 - C1 ) are the arbitrary con-
stants which may be given any real values we like.
Again, by adjusting the arbitrary constants A and B suitab-
ly, Le., by putting C cos c for A and - C sin E for B, the general
solution can also be written in the form
y = Cecos(x +
where C and r are the two arbitrary constants.

Ex. 1. Solve dxl dx

Let y = e ' be a solution of the above equation;
then e(m 2 t3m+2) =0. .. m 2 +3pn+2 =0,since e* 0.
m=-1, or, -2.
the general solution Is y =C1 e + C2
Ex.2. Solve ±!Y - + a2y = o.
dx 2dx
Let y = e ' be a solution of the above equation;
366 INTEGRAL CALCULUS

)=O, or, ,,t-2a,,-a1 =O,ser''O

-. 0.
Since the auxiliary equation has repeated roots herr,
the general solution is y (C + Cx ) es-'
I.3. Solve (D 2 4 2D + 5) 0.

The equation +2-41 + Sy = 0.

Let y ell be a solution of the equation

then e " ( m 1 42m+5) 0 in +2m+5 0, since e s 0,
:. m=-1±2i;
the general solution - is y = Ce •
which, as shown in Art. I82(iii), can be put in the form
Y = r' (Acos2x + i3sin2x).

EXAMPLES XVJII(A)
Solve

1.
dx2 dx

2. 12 0.
dx 2 dx

dx dx

4 --'+(a + b + aby = 0.

5. (i)2---3+y=0.
dx 2dx
(i) -'+2+y=0.

6. y - 4 Y' +4y 0.

7. (i) (D + D)y = 0. (ii) (D'+6D+25)y = 0.

8. (02 - 2mD + m 3 + n2) y =0
 XV1II(A) LINEAR EQUATIONS 367

9. (i) (D — 411) + 13)y = 0. (ii) (D — n 2 )y = 0.

d25
10. () -
dti + 4+13s0.(ii)(D+3)2y=0.

11. Solve in the particular cases :-

+ Y_ 2y = 0; when x = O,y = andy = 0.

.-4' dx
d2x + y
(ii) 0; whefl x= O,y = 4; when x . it.y 0
dxI
d'x dx
(iii) - — 3+ 2x =0; when t = 0,= 0 and -= 0.
d
x dx
(iv) d nx = 0,when = 0,1 = O and x =
di I
12. Find the curve for which the curvature is zero at every
;)Oint.

13. Show that -if I

d20dO
. gO = 0 ,and if a and = 0,

when t = 0. then 0 = a cos I 'Ig/i))

I4 Show that the solution of

dx dx
+ k + ux =0

is x = efl(Acosnf + Bsinn1),ifk<4i and nt = jx _k2.

ANSWERS

1. y = c,e + c1e ' 2. y = c,e 3-

3. y c1e' + c2 e" 4. y = ce ' +

5. (0 y c,e' + ce ll 2 (ii) y (A + Bx)e'

6. y=e 2 '(A.Bx 7.(0y=A+Re'.

(ii) y = e I ( A cos 4% 4 R sin 4x

8. y C ' ( A cos ,ix s H sin nx
 INTEGRAL CALCULUS

9• () y ell ( A cos 31 + B sin 3x ) (ii) y = + Be -

10.(i) s=e'(A cos 3t+R sin 3t). (II)yu''(A+t,x)
11. Ci) y = 2e • e 22 (I) v = 4 cos x . (iii) i = 0

(l Y) x = a co&, n: 12. A straight line.

18.3. Right-hand member a function of x.

We shall now consider the solution of the general form

+ r, -41 + P2 y = x. .. w

If y = ( x) he the general solution of

d
+ P2 y = ( 1... 2)

and y w ( x ) he any particular solution of (1), then

y= (x) + i (x)is the general solution oI(1).

This result can be established by direct substitution.

Thus, substituting y C ( x ) + i ( x) in the left side of (1),

we have

(+ , .?- + + , +
dxl } + }
The first group of terms is zero, since y = 0(x) is a solu-
tion of (2), and the second group of terms is equal to X
since y = ( x) is a solution of (1)

Hcnce,y =4i(x) +i,(x)isa solution of(1), and itisthe

general solution, since the number of independent arbitrary con-
stants in it is two, 0 ( x) being the general solution of (2).

Thus, we see that the process of solving equation (1) is natural-

ly divided into two parts : the firstis to find the general solution of
(2), say 0 ( C ,C 1 , x ) , and the next is to find any particular solu-
tion of (1), say V C x) not containing any arbitrary constant. Then

y = $(C1 ,C,x)+W(x)
will be the general solution of (1).
 LINEAR EQUA1IONS 369

The expression 0 (C, , C2 , x ) is called the Complementary func-

tion and i.e., any particular solution of (1) is called the Par-
ticular integral of the equation (1).
18.4. Symbolical Oprators.
We have already shown in Art. 18.2 how to obtain the Comple.
m?ntary function now we shall consider how to obtain the Par-
ticular Integral. In order to discuss methods of finding a particular
integral,it would be convenient to introduce certain symbolic?'
operators and their properties.
With the usual notations of Differential Calculus jn , etc. will
be denoted by the symbols D, D', etc. Also lID (or, D ),
I / D I (or, D 2 ) ,etc. will be used to denote the inverse operators
i.e., the operators which integrate a function, with respect to x,
once, twice. etc. Let us write the equation
._!1p ' +P2y=X ... (I)
dx2 dx
in its symbolic form
(D 2 +P1 D-+-P2 )y=X. ... (2)

or, more briefly as f ( V )y = X. . . . (3)

The expression X will be used to denote a function of x
not involving arbitrary constants, such that the result of operating
upon it with f(D) is X, and as such and f(D) denote
two inverse operators.
Thus, the function p!. , X is clearly a Particular Integral of the
equation f ( D ) y =X .
As a particular case, when f ( D) = D, -- X will denote a
function of x, obtained by integrating X once with respect to x,
which does not contain any arbitrary constant of integration
similarly X will denote a function of x, obtained by integrat-
ing X twice with respect to x, and not containing any arbitrary
constant of i ntegration. For example,
Inlegral Calculus (main) -26
 INTWJRAI. CALCULUS

1 1
X 4 =x'; 512 x S =x;
-
1 1
1j1=x; 1
.1 =x2
1
.
DI 2
Important Result. on Symbolical Operator..
If F ( D) be an y rational integral fwct*on
La., UF(D) - D + a,D 1 + .... + c I) + a,tben
Ci) F(D)e'
(II) F(D)e'V = eF(I) 4• ; .•,v bcThgafuncuonof
(Ill) F(D2)fshC* i ( arn(ax + I,)
Cos
[ • b)
(AX cos(ax+ b).
By actual differentiation, we can easily verify the above result,.
18.5. Methods of finding Particular Integrals.
We shall discuss here the methods of obtaining particular in-
tegrals, i.e.4 the methods of evaluating X, when X has spe-
cial forms.
(a) X = x , m bring a positive iakger.
Expand I / f ( D ) , i.e.. (I( D)) 'in ascending powers of 0
and operate on x - with the result. It is dear that in the expansion
no terms beyond the one containing D need be retained • since
D' x = 0.
Note. The justification of the above method lies In the fact that the func-
tion of x which we shall get by operating on x - by the series of powers of
D obtained by expanding (f(D)' ),when operated upon by/( D), will
give r"' . For example,

= 0 + D2) -' x 4 = j - + D- . . . )x 4
( 12x1 +24.
+I = -

Now, (LJ 2 +fl(x 4 -lZx' +24) 12x 2 -24+x' _12x? +24x'.

b ) X = e'V ,where V isa function of x,or a constant.
If V is a function of x • we have from Art. 18.4(u),
f (D)eV ,=ef (D+a)V e u V , say ,
For proof see Authors' Differential C.elcsdus.
 lINEAR EQUATIONS 311

o that f(D4 a) V, = V, i.e., V1

= /( 0+ a)
Thus, 1
ell f(D+ 1)V.
Again, noticing that I ( P + a) k where k' is a constant is
evidently a constant = k say, and proceeding exactly above we
can show that

--
F(D)
e' Ic f(D
- ek
41)
--
UP + a) I
c ) X = e" , where a is any constant.

if f 0, f f(a)'1'
) lila))
I From A1L 184 (I)
I ell =
I ,provided f(a 0.

iff(a) = 0, then (D - a)isa factor off(0).

either f(0)=(D—a)i(D),where •(a);t0,... (I)
Or
else, =( 0—a) 2 . .. ( 2)

I , I I I e' .1 e'
(l)/)e
(D—a)(a)(aD—a
1 Xe"
[by(b)l =
t(a)D

(ii)rJ5)e - I ell =e" 1 by(b)i =e"-

(d) X =sin(ax + b)orcos(ax + b).

If f(D) contains only even powers of 0, let us denote it by
( ) Then, 11 0 ( —a' 0,we get, by Art. 18.4.0ii),
srn(ax +b) O(—a 7 ) sin (azi-b) sin (ax +b)
Ø(D 2 )

sin( Ax + b) - Sin(aX 4 b), if(—a 2 0

372 INTEGRAL CALCULUS

Similarly, coo (ax + b cos ( ax + b), ifo (-a sO

M
If •( — a 1 ) 0, or if f( 0) contains both the first and the
second powers of 0, the method of procedure that is to be adopted
in such cases is illustrated in Ex. S and Ex. 6 of § 18.7 below.
(e) X = x' sin( ax + b) or xm cos (ax + b).
In evaluating particular intgrals of this type, it is convenient
o replace sin (ax + b ) and cos ( a: i- b ) by their exponential
,alues and then proceed as in case ( b)
(f) X =X V, where V is any function of x
I I I 1 1
I(D)XV 1(D)j
t f(D) f(D)''
Proof:
We have 0 (xV) = xDV + V,
D 2 (xV) = D(xDV)+ DV xD'V + 2DV,
and similarly, 0 (xV) xD V + nD -' V

= xDV+ (D)v ....(I)

Hence, f(D)xV xf(0)V + f'(D)V. ... (2)

Now, putf(D)V = V 1 ;hence V
(2) becomes

f(D)xf_Vl=xVl+f'(D)f(,)VI
I
i.e., V1 - ) xV * f(D)HD) )VI
' f(D)
I f
Transposing, we get

715)XVi ={ X_71j ) f (D)} 7.15)Vl.

Dropping the suffix, we get

f(0)XV
=( 7(0)' (0))
 LINEAR EQUATIONS 373

Note. It should be noted that, when X Is the sum or difference of

two or more functions of x, say X = X, ± X 3 X then the particular in-
tegral
(X1 ± X2 ± X3 ) = p45_>x tfJ) X ±
= 7Tö

18.6. Alternative method of finding X.

When the auxiliary equation has real and distinct roots, cor-
responding to each such root m, there will he a partial fraction
of the form A/(D - m ), where A is a known constant and hence
I
X i n the form
: :a: A 2
X
each term of which can be evaluated by the method shown below.

Now, D x=D e -x = e -- e X.
D
-1-- - X dx. ... (1)
D -m x - e
This method is illustrated in Er. 8 of An. 18.7.
18.7. Illustrative Examples.
Ex.1. Solve([)' + =x2
Here, the auxiliary equation m + 4 = 0 has roots in =± 2i
the complementary function = A cos 2x + B sin 2x
Particular Integral = = ) I'

!(1
4
!4 D')

=(1 +D 2 + - D 1 -... ) X _!)

the required general solution is
Y = A cos 2r a. B sin 2x + 24 (z' _2).
 374 INTEGRAL CALCULUS

Es. 2. Solve (0 -3)'y = 2e1'

Here, the auxiliary oquatlon (m - 3 ) = I) has roots 3,
C.F.= (A + Bx)e3
I PI- 2e4
2c' -- (4 3)1 --2e4'
(D - 3)2 -
the general solution is y = (A + Br)e' + 2ell
E'i.3. Solve (1)- 2)' y =6e"
Here, the auxiliary equation( m - 2) 2 = C) has roots 2, 2.
C.F.= (As Bx)e'

Pl (e2!j. 1 = 6e' =3x' e1'

(0 -2)' (Se'
the general solution is y = (A Bx)e' + 3x' e'
Es. 4. Solve + y = cos 2x.
The equation can be written as C , s I ) y cos 2x
The auxiliary equation m 2 = 0 has roots ± I.
C. F. = A cos x + B sin x.
I
P.1. _D2+l__22+l3 Cos 2x cos 1cos2x 2x
the general solution is y = A cos r + B sin r - cos 2x.
Es, S. Solve (D + I )y cos x
As in Ex. 4, C. F. = A cos x + B sin
But the method of obtaining the particular integral employed in Ex. 4
fails here. We may, however, substitute the exponential value of cos r and
proceed. Alternatively, we may proceed as follows:

Let Y
= DI I cos x and Z = 0' + sin r.
Y +iZ = 5...L.1 ( Cos s + iSin r) = Dl

— C"
(D + i)' + 1 Ie"
2i + 0'
=t-i_(1+-2Y
2W..\ /
1_1..t!1
2i i 0'
 LINEAR EQUA11OI4S 313

x x
=T (coax + t sin x).
=
equating the real part. Y = x sin X.
the general solution is y = A cos x + B sin x + in x.

il - 2 -!Y
Ex. 6. Solve dx + Sy = 10 sin x.
dx
The equation can be written as ( D' - 20 + 5) y = 10 sin X.
The auxiliary equation m -2m + 5 = 0 has roots I ± 2i.
C . F. =e a (A Cos 2X + B sin 2x);
(0 2 + 5)+ 20 10 sin
= 0 2 _ 2D + 5 lOsInx =
- 02+20+5 4 IOsinx = (0 + 20 + 5)sinx
1' + 5)2 +
!(... sinx + 2cosx + 5sinx) = 2 sin x + coax.
the general solution ts y =e'(ACos 2x+Bsin 2x)+2 s inr+ coax.
Ex.7. Solve (0 2 - 41) + 4)y =
The auxiliary equation m I - 4m + 4 = 0 his roots 2, 2.
C.F.= (A x +

P 1 -- 0 2 _ 40 + 4 x3e2 (D- 2)
-e2a .1.
0 2- 20
the general solution is y = ( A x + P1) e 2 ' +

Ex. a. Evaluate D +30 + 2

I 5" . . . (I)
Given expression = (0 + i)(D+ 2)

r i ___
= [1) + I - 0 + ]2"
I
= 0.1
ix _s 2 ! Ju s t . dx. .. .(2)
.j
376 INTEGRAL CALCULUS

Let

Put e =z.
1 = f el e dx and]: = f

.. eZ d xiz .
et e dx

i t =fe' iz= e l =

12 =fzc s dx
f
= u3 - 02 dz= ze - C' sc' (z- 1)= a '(es -1)

from (2), the given expression

g 2 e'
= e-2-e''

18.8. Two special types of the Second Order equations.

d2w
(A) 2 = f(x).
dx2
Integrating both sides with respect to x. we have
dy =f(x)dx +A •(x) + A, say.
dx
Integrating again,
y4 $(x)dx+Ax+ B =W(x)+Ax+B, say.
Note. A. a generalization of the above method, we can solve the equ-
ation 44 . f (1 ) and in particular - 0, by successive integralon.

(B)
Multiplying both sides by 2 , we get
21=
dx 2 2f(y dx

d
or, - ,.dx
dx
Now, integrating both sides with respect to x, we have

k ax)
2Jf() A d. + C 25f(y)dy + c,.
 LINEAR EQUATIONS 377

Let 21f(y)dy =(y).

dx
dy
,whence, integrating,
dx = ± 'Ij$ (y) + ,
x= y, C, ) + C,( say ).

18.9. Illustrative Examples.

Ex. 1. Solve fy
dx
= cos nx.
Integrating both sides with respect to x, we have
dv I
= —sin,tx + A.
dx ,i
Integrating again, y - cos nx + Ax + B,
which Is the general solution.

Ex. 2. Solve
dx l = y

Multiplying both sides by 2 , we get

dx

o ( =a-y-d.
--
dxdx 1y'dx' 'dxVdx/
Now, integrating both sides with respect to x, we have

(4)2 2.15 - dy + C,

- 2a1 C-C
SI'

ydy
"dx
.X =±lSaJ
or, = J
y

x =± '?- .1 + C2

x - C,

C, ' (r - C )' = C,y' - A.

This Is the general solution.
37$ INTEGRAL CALCULUS

Net.. An alternative method of procedure for solution of the equation of

the above type, i.e.. of the type d Y = f( y) is indicated below.

P ut AE
dx = dx2 = dx dy dx ' dy

,or, pdp=ay- 3 dy.

' dy -
integrating.-} p2 = - - y-2

P , i.e., =c I 1
( dx
Y) Y
Now the rest is the same as before.

Ex.3. Solve x2
a1 a
x
dx tdx
+ 'i 2 y = 0.
Put r = C 1 .so that z= log x;

then- !Lx = e 1 =x.

dz

dz dx dx dx dz2 dx \ dxl dx '.. dx t dxl

i.e., x 2 Li
dx tdx
+ x.1 = LX
dzt
the given equation reduces to

,t ly=0
dz2
Multiplying by 2 and Integrating with respect to z,
dz

n t y 2 = constant = na (say).
Adz

. =±n4a--y
dx
or. t=ndz.
4(92-y'
integrating T. cos' - = nz + E,

whence y = a cos ( ,iz + £ ) , or, y = a cos ( n log x. +t) is the required

solution, a and £ being arbitrary constants of Integration.
 LINEAR EQUATIONS 379

18.10. Equations of the types

-Z a,
(A) d xl ..... . dx

(3) ,y ) = 0.
dx m 136
(A) These equations do not contain y directly. The substitu-

lion Is Ll (derivative of the lowest order) = q.

(B) These equations do not contaln.x directly. The substituion

dy
is
Then --=
dx' P dy ' dx'
Li. = ' + dy
dy'
p () ,etc.

18.11. illustrative Examples.

Ex. 1. Solve- 2x .-!1 W - (44)' + o.
dx'
d ld'Vi
q. .. dx
the given equation become,

2x - +I = 0.
dx q2
dq ,or, log(q' - I)log
= (c1x)
q -1=c1 x.

q. i.e., = '1(1 + c1x).

4 .--
AIX3C,
(I + c1 x)" +

22
y..__(1+CiX)5"tC2X+C3
3c, Sel

(1 + C1 x) 11' + dx + C,

:. 15c, 'y 4(1 + ci x)" + CZ + C3.

30 INTEGRAL CALCULUS

)2 ) i/2
- (-!
Lx. 2. Solve y
+ (( )2 - (vi) ' 0.

Put AY = p. ..
dxdy =p
the equation transforms into

yp_p2 +{p 1/2

-'
p 0dy
p = qy + ( I-q' where =
)1/2
dy
This Is C1airut's form.
= Ay +(1- A' )" = Ay + k(say).where k =(1- A')11'
dy
Ay + k
x + B =!. log (Ay + k) log (Ay + (1 -A')").

EXAMPLES XVflI(B)
Solve the following equations
d'
2x + 3. 00 dxl + X
dx' + 4y

2. (ii)f-+--6y=x.

3. (i) (0+3)2 y = 25e 21 . (ii) (02 + 9)y = 9e'

4. (i)-'-a'y=e'. 00.1,y e1.

(iii) .-4 AY + 3y = 2e"

S. (i) (0' - 4) y = sin 2i (ii) (0' + 4) y = sin 2x

6. (I) sin
2+Y= X. (ii) + = x cos x.

(Iii) +y cos' 7 (lv) + = x sin 'x

Lx. X V III(B) LINEAR EQUATIONS 381

7. (1) (DZ_1)y =x e' . (ii) (D' - 9)y = e3z cos x.

8. (I) (D 1 + 2D + 2)y = xe

(ii) ( D 2 - I)y= e' sin 1 x

(iii) ( D + 1) y = sin x sin 2x.
(iv) ( D 2_ D_2)y sin 2X.
(v) (D-2) 2 y = x 'e' . (C.P.'861

d' d
9. (') dx +2 dx + =e+e

(ii) + k' y .(iii) -2 A +i =

(jy) = cosh x. (v) - -y =x e sin x.

10. X2 dx2 - x dx y =logx. tPutx =c'J (C.P.'85J

11. (x 2 D 1 + xD + I )y = sin (log x 1 ). (Puix =e')

12. (i) Show that the general solution of the equation for
S. H. M., viz.
d Ix
x,is x =A cos(ni +

(ii) Evaluate e" cos bx and hence show that

5 e 11
COS
bxdx = -1--- (a COS bx + b sin bx).

13. Solve in the patticular cases

Ci) 12Y +y = sin 2x, when x = 0 y = 0 and = 0.

(ii) y - 5y1 +6y = 2e ; when x 0, y =land y, I
(iii) (0' -4D +4)y=x' ;when x=O,y=and Dy=1
(iv) (0' -I)y=2;given Dy=3,when y=I;and x=2,
when y - I
3*2 INTEGRAL CALCULUS Ex. XVIII(R)

Solve : -

14. (i) X =I. (ii)

di 2
15. (i) y, Cos 'x = I . (ii) yiy, 4
16. y = tan y scc'y ,given i = 0, when y = 0.

17. (1)_Y= I (jj) + 0.

dx x2 y'
d2x
18. (I) x sin ()i)
dx'

d'1 Iv
19. (i) X 4..! = 2
dx dx dx' dx

20. + = e.
dx 1dx

21. (1 + x' )y + 2xy, = 2.

22
dx'
(.4v
dx)
+ y 2 logy = 0.
)1 +
dx' +
23. .4_i ( fi
dx
=o
dx

24. y2 — (y, )1 = 0.

25. yy2 + (y1 )2 = 2.

26.
dx 4 dx1
ANSWERS
1. (i) y = A cos 2x + B sin lx +(2x + 3).
(ii) y = A co x + B sin x * ( x — 6x)
2. (I) y i4e' + B +x _..r2 +x.
(ii) y = Ae'- + Be 3 ' - (x
 LEAR EQUATIONS 383

3. (I) y (C1 + C 2 z)e 3 ' +

(ii) y -A coo 3x + B sin 3.x

4. (I) y CI," + C,U"

+

(ii) y .4e' + Be'

(iii) y =C,t' + C2 e'' +
S. (I) y A e' +Bu" -sln2x.
(Ii) y - A coi2x + Bsinx -'jxco2x.
6. (1) y = A cos x +B sin x •- -xcosx.
(H) y = C1 cca 2% + C2 sin 2x + Lx co, x + !sin X.
(iii) y A cos x + B sin x + .. - 1. cos 2x.

(iv) y = A cos 2x + B sin 2x +x _ j -x Cos 2x ---x 1 sin 2x.

7. (I) y = C1 e' + C2 e' .t- . e' (3x - 4 ) .

(ii) y = C, e ll + C,e'-' +-e'' (6 sin x - cos x).
8. (I) y (A cos x + B s in z + x).
+ u'

(ii) y = / e' + Be' --e' (sin 1. x + 4cosx).

(iii) y = A cos x + B sin x + . x sin z + -cos 3x.
(lv) y = At -' + Be 2 + -.( cos 2. - 3 sin 2x).
y e'' (A + Bx
(v) ).
9. (I) y = e' (C1 +C2x+x)+,'
(ii) y = ( A + Bx)eb +e' (1 -k)2.
(iii) y = (A + Bx +x2 )e'
(iv) y =A e +Be +xsinhx.

(v) y = Ae'+Be'-4--{(1Ox + 2) cos x + (5x-14) sin x),

10. y = (A + B log r)x + log 42.
11. y = A cos log x + B sin log - sin log x'
 384 INTEGRAL CALCULUS

13. (I) y = sin x -j sin 2x. (Ii) y =e'

(lu) y =..xe 2
+fx +is (iv) y + 2
+ x1-2
14. (i)y=x log x+Ax+B. (ii)y=(x_2),z++
IS. (ii y = log sec + Ax sB. (ii) C 1 2y 2 =A + ( C2 ± C1 I X )2

16. (siny + Cc I )(siny + Cc') = 0.

17. (I) 3x 2 ( 'ly - 2C 2 )( 'y + c 1 ) " + C2
(ii) 'IC 1 y 1 +y - ('l Y + 'Il + C, y) = aC, 42 x +
log C2
TCT
18. (I) y = C 1 + C2 x
(6 - x 2 ) sin r - 4x cos x.
t

(1) x = .e2I + C,t + C3 . 19. (1) y = +B.

(ii)a log (y+B)=r+C. 20.y=C,e-+c3+e.
21. y = log (l+ x 2 ) + A tan 'r + B.22. y =
23. e' (C 1 - eY ) = C2 . 24. c (C,x + C2 ) = 1
25. y 2 =2x +C1 x+C2 , 26.y=Cie'+C2e-+C3x+C4

18.12. Equation of the nth order.

The linear differential equation of the nth order with constant
coefficients is

•4+PI d"-'
d-'
+ P,,.. 1
d
+ P.=X, ... (1)
or, symbolically (D" + P,D" + P2 D 2 + P,) y X, (2)
or, more briefly f ( D ) y = X, .. (3)
where P1 , P2 ......P,, are constants, and X is a function of x
only, or a constant.
The method adopted in the case of the solution of the second
order equation admits of easy extension to the above case. Thus,
the general solution of (1) consists of two parts (I) the CompL ien•
iary Function and (ii) the Particular Integral, the complementar3
function being the general solution of
f(D)y = 0 ... (4)
and the particular integral being the value of . X.
 LINEAR EQUATIONS 385

Assuming, asbefore, y = e as a trial solution of (4), we shall

find that y = e" will be a solution of (4),
if f(m)=O,iz.,ifm'4-Pi m +..+P,, = 0 ....(5)
Equation (5) is then the auxiliary equation of (4).

If the auxiliary equation (5) has n real and distinct roots

viz., ,n 1 , m 2 ,..., m, then the complete solution of (4) is
y Ci e- 1 + Ce"2 + ....
If the auxiliary equation has a multiple real root of order r, and
if this root be a, then f( D) contains ( 0 - a)' as factor, and
the corresponding part of thç complementary function will be the
solution of( D - a)' y = 0.
Assuming, as before, y = e
(0- cx)'y = (0- a)'e'v =
and the solution of D' v = 0 is, by successive integration,
(CO + C1 x+ C,xi+ . .. + C,., x'' ),
whence y = (C0 + C1 x s- C 1 x2 + ... + C,., x'' )e —
is the corresponding part of the complement4ry function.
If the auxiliary equation has complex roots a ± i , the cor-
responding part of the solution is, as before,
y =e' (A cosx + B sin x),
and if a ± ip are double roots of the auxiliary equation, the cor-
responding part of the solution will be
e ((A, + A 2 x) cos x + (B, +B,x)sin OxJ.
The method of obtaining the particular integral of (1),.when X
has those special forms I See Art. 18.5 1, is essentially the same as
shown in the case of the second order equations.
18.13. Illustrative Examples.
E.1. Solve (D 3 + 313' + 3D+ 1)y = e'
Here, the auxiliary equation Is m 3 + 3m' + 3m + I = 0 of which
therootsare - 1,-i.- I. C. = e(C.., + C,x + Cir').
Integral Calculus (main) -27
386 LNTEGRAI. CALCUILS

1 1
= (D' + 3Tr4 3D
+ fl = (V •

-e s— ' ----1-c '--1--c z 1

0'
the general solution is y = cC0 • C, r + C, x • x' ).

Ex. 2. Solve (0 4 + 2D 3 + 30' 4 40 + 1 ) = xc'

The equation can be written as (0' • I.' I )'y xc'
Here, the auxiliary equation is ( in •m • I 1' = 0; it has double
complex roots - - t I j- '13, - ± '13
C.F.is e l(A,+A,x) cos (-13x)+(8, +B,x) sin (.'.'3x)J
1 ____ 1
= e
= ( +0+1)' j(1)1) 2 +(VI)i)'
I IF 1
= e 3)1 X = C'
IIiT D(i 5Pi1
=e'( 1 + L)( I + 21))} = ,' I -20 - . • . Ix

,e' r - 2).
the general solution is
Y e'" UA, + A,r) cos (- 43x)
• R,x) sin (f'13x)J +e'(x 2)

Ex. 3. Solve - 2._ a'

4 t y sin (2x • 3)
The equation can be written as
(D 4 - 1)' y = sin (2x • 3).

Theauxillaryequationis(n' - 1) 2 = (J; its roots are Li,- 1.-i

i,i-i,-i. Hence.
C.F.is c' (A, + A i x) + e' (l3 + B,x)
+ (0, +D,x)stnx ... (1)
P.1. = sin (2x • 3)= -, ^ —j-j-,sin(2z + 3)

sin (2r+3) .. (2)

Adding (1) and (2), we get the general solution.
 1

IJNFAR EQUAI1ONS 397

EX A MPLES X V flUC)
Solve

1. .-y=O. IC.P.19461

2. (i)--'-3 + 2y =O.

dx J dx 2 dx
d'y d3y
dx' dx dx' ,=
dx
(iv) (V + 1) 3 (0: + fly = 0.
3
dx dx3 ax
4, U) (0' - O)y = -c
(ii) D I - 1) y = sin 3x +1).

.4
dx' 5 2y 0.
dx' dx
6. (D' + D2 -. V I )y = sin'x.

7. - 3 -----.
d2 + 4y e''
dx'
dv
8. '-;i•- -L - + 4y = e- bin
9. (12' -. 302 + 40 - 2)y r' + cos
10. (0' - 40' + 302 + 40_ 4)y =
11. (0' * I ) y = 2 cos 2. x - I + e'
12. (0' + 202 + fly cosi
13. ( V - 1 '( D 2 )'y = '' + srt

14. - 2 LI
dxI + dx
15. 5 dy + 4y = 360in 1 cos -
38* INTEGRAL CALCULUS

ANSWERS
1. (I) y = Ac + - " (B sin + 431 + C cos2.43x)
60y = Ac S + Be - + C cos x + V sin x

2. (i)y=e(.4+Bx)+Ce'5. (ii)y=A+ Be" -s-Ce'

(iii) y =e ((A + Bx) cos x + (C • Dx) sin x!.

(iv) y=e (A 1- Bx + Cr 2) +D cos x + Esinx.

3. (i)y = Ac' +e/2 (B sin 43x i-Ccos-43x)-x +x'-6.

(ii) y = A + Br +Ce + j x - 35 x' +x.

4. (i)y=A+Be I +Ce' -4-.r(e'+e ).

(ii) y = c - / A cos x 1- B sin x ) + Ce

2 (

COS
s 73-0 (3x + I) -,sin(3x + 1).

5. y = (A 1 + Ax)e x + A3e-'

6. y=Ci' *(C 2 +C 3 x)e' +-s(n2r+-cos2x---.

7. y e 11 (C, • C 1 x)+ C 3 e' +-e'

8. y = C, e '' + e' ( Cl cos x + C 3 sin x )

COS
-
Ill e' ( ' x-.-3sin-x).

s
9. ye'(C1+C2 COS x+Cs in X)+xe' + -( cosx+3 sin x)

10. v (C, - C 2 x ) e' * C3e x + C i e - - + !r2e2

COS
fl. y r C ' j C, ax + C 2 ,in ax I • c' (C 3 cos ax

* C4 Sm ax I + ( cos x + e ), where a = I / '2 -

12. y(C,. C2 r) sin i+(C3+C4X) COS X-X 3 COS X.

1).

4 f I x 2 + - 1 sin x
 LINEAR EQUAI1ONS 389

ii. v = (C + C2 x)e' + C3 +e + 4 x 3 +2x.

15. y = C, cos x + C3 sin x + C3 cos 2x.
+ C4 sin 2x t sin 4x + sin 3x
18.14. Homogeneous Linear Equation.
An equation of the form

dv +
x'— + P, X"
dx"
+ Py =X,
* P. -, x ... (1)
dx
or symbolically, ( x" D" + P, X" -' D - +
+ P. -, x D+P,,)y=X, (2)
where P, , P2 ,..., P. are constants and X is a function of x
alone, is called a homogenecus hnar equation.
The substitution
x = e ,i.e., z log x
will transform the above equation into an equation with constant
coefficients , which has already been discussed in Art. 18.12. Here
the independent variable will be z
4y dz = _I dy ... (3)
Now, =_
dx dzdx xdz

dx? - dx \ x dz / x2 dx x dz 2 x

(4)
- x2 dz dzl

Similarly,
4y (5)
dz 3dz2
1(dy!)
=V

Let us write ö for with this notation (3) (4) , (5) can be
written as

dy (i,)
x
TZ =&y,
x 2= 8 (b - 1) y. (7)
dx
390 INTEGRAL CALCULIS

x 1 d' Y 8 (8- 1 )(8 - 2)y (8)

dx

x" =8 (5- 1)(8- 2)..(6- n + f ly (9)

dx
Note. This is sometimes called Cauchy equation.

18.15. Equation reducible to the Homogeneous Linear form.

An equation of the (arm

ax + b) P (ax + b)' -
dx dx'

...(1O)
dx
where P 1 , P2 P are constants and Xis a function of x alone
can be reduced to a linear equation with constant coefficients by
the substitution ax + b = z
Note. This is sometimes called Legendre eqvatwn

18.16. Illustrative Examples.

Ex. 1. Solve x' __.+3x1 -_ 2x + 2y = x2.
dx I dx2 dX
Put r =e, i . e., z = log x.
Then by Art. 18.14 the equation transforms into
18(6- I )( 6-2) + 36(6-1)- 26+21 y = e. (I)
where 8 = or, (6- 1) 2 (6 + fly =
the roots of the auxiliary equation are I , I ; -
The C. is y = ( C1 + Cj z)e I + Ci e la

And P. I. Is (8 - 1)2(8 e 2. =
+ 2)
the general solution of (1) is
1 , 2.
y =(C 1 + C2 z)e + C3e'
I-lance, the general solution of the given equation Is
y = (C + C 2 logz)x + C3 x- 1 +-x
 UNEAR EQUATIONS 391

Es. 2. Solve (x 2 D 2xD ) y = ze'

Put x = e i.e., z = log x.
by Art. 18.14 , the equation transforms into

(8(8-1) + 26}y = e e' ,

d
here 8=— , or,(8+8)y=ee"
dz

the roots of the auxiliary equation are 0. -

the C. F. is y = + C1ez

I ez eel
= 8(8 + 1)

-- ( 1 (8+1)

- 1 8+1

ee dz - e fe ll e dz. lBy Art. 18.6

= e'-e t( e -. 1) )=
I Sec Lx. 8 of Art. 187.

the general solution of (1) is

y=C, +C 2 e' +e' eel .
Hence, the general solution of the given equation is
y =C 1 + C2 x 1 +

EXAMPLES XVIII(D)
Solve the following equations

x.
1• dx -4x dx + 6y
x2 -

2. (x2D 1 + xD - fly = sin (logx) + xcos( log x).

3. x Li
dx2
+
dx +
= x4
92 INTEGRAL CALCULUS Lx. XVIII(D)

4. (x'D - 2) y = x 2 +

5. x' Li
dx'
+ 2x --- -2x
dx' dx
= 0.

6. (x' D' + xD - I )y = x'

7. x' L1+x1_yx.
dx' dx

8. (x + 2)'
dx' - 4(x + 2) 4dx -
y + = x.

9. (xD + 6x'D 3 + 9x'D 2 + 3x0 + I)y = 0.

10. x 1 i- 3x3 LL - 2x 2 . + 2xy = log x.

ANSWERS

1. C,x' + C,x' +
y =

2. y = C,x + C,x' --}sin(logx)

+ x (2 sin (log x) - cos, (log x))
3. y (C, + C, log x)x-' f%4

4. y = C,x 1 + C,x' + . x' Iogx - x'log X.

S. y=C,x'+C,x 1 + C,..
6. y= (C, +C,logx+C,(logx)')r+x'.
7. y = (C, + C,Iogx + C,(logx)' Jx +x(logr)3.
8. y C, (x+ 2)2 +C,(x + 2)' + . -(3x + 4).
9. y = (C, + C, Iogx)cos(logx)
+ (C, sC4 logx)sin(logx)
10.y = (C, + C,logx)x + C,x' +x - ' Iogx.
 CHAPTER XIX
APPLICATIONS

19.1. We have already considered in the preceding chapters

some applications of differential equations to geometrical probl-
ems. Here we shall have some other applications of differential
equations.
19.2. Orthogonal Trajectories.
If every member of a family of curves cuts the members of a
given family at right angles, each family is said to be a set of or-
thogonal trajectories of the other.
( A ) Rectangular Co-ordinate..
Supose we have one-parameter family of curves
f(x,y,c) = 0, ... (1)
c being the variable parameter.
Let us first form the differential equation of the family by dif-
ferentiation of (1) with respect to x and by elimination of c I See
Art. 15.2 1, and let the differential equation be

x.y,) dx
= o. ... (2)
If the two curves cut at right angles, and if i, y ' be the 1gles
which the tangents to the given curve and the trajectory at the com-
mon point of intersection,( say x, y ), make with the x-axis, we
have w - ' = - i , and, therefore, tan y = - cot w' . Since
tan i = it follows that the differential equation of the smystern of
trajectories is obtained by substituting
- j -s- i.e., - -
idv
/ dx
. dx
dy
dv.
for --
dx
in (2)
Thus, the differential equation of the system of orthogonal trajec-
res is

,,,_ 0. . (3)
394 INTFGRAL CALCULUS

Integrating (3) we shall get the equation in the ordinary form.

B) Polar Co-ordinates. -
Suppose the equation of a given one-parameter family of cur-
ves be
f(r,O, c) = 0 . . . (1)
and the corresponding differential equation, obtained by eliminat-
ing the arbitrary parameter c, be

F(r.8. = 0. ... (2)

J)
If 0, 0 'denote the angles which the tangents to the given curve
and the trajectory at the common point of intersection, ( say r, 8),
make with the radius vector to the common point, we have, as
befrre, tan Q = - cot 0 '
Since tan i' = r (dO/dr), it follows that the differential equation
of the system of orthogonal trajectories is obtained by substituting

- --ldr dO . dO dr.
- for r - i.e ., - r 2 - for - in (2)
rdO dr dr dO
Hence, the differential equation of the required system of orthogonal
trajectories is
a' r,O, -r' dO-= 0. .. (3)
F
dr
Integrating (3) we shall get the equation in the ordinary form.

19.3. Illustrative Examples.

F. 1. rind the orthogonal trajectories of the rectang4lar hyperbolas
zy =a I -
F)ifferentiating y = a with respect to x , we have the differential
equation of the Family of curves

x+y=Q (1)
dX
and hence, for the oithogonal trajectories, the differential equation Is

- = 0. or, xdx - ydy = 0

 AJ1'UCAllONS 395

Integrating this, we have xt - =c • the required equation of the

orthogonal trajectories. it represents a sy s tem of rectangular hyperbolas.
Es. 2. Find the orthogonal trajectories of the cardioides
r a (1 - cos 8 ) -
Since r = a (I - co' 9). .. log r = log a + log(] - cos 9).
Differentiating with respect to 0 • we get the differential equation of
the family of curve,
1 dr _ sin
r 40 I cosO
the differential equation of the system of orthogonal trajectories is
i( ±0) sin
r i.,' dr1 - cosO
4, 1 - cosO 49=0,
sin
4' sill O
or, - + - _______ 49=0
r I + cosO
IF
integrati.ig, log Coss = log C;
+
ic., r = c(I + coSO)
represents the required orthogonal trajectories.
Es. 3. Find the orthogonal trajectories of the system of curves
7 = 3' CO5 PI9

Since r' = a' cos nO, .-. n log r = n log a + log cos 'iO
Differentiating with respect to 9, ( and thereby eliminating a I we get
the differential equation of the family of curves
I dr sin nG
N = - P1
r 46
F
- cosprO
the differential equation of the system of orthogonal trajectorIes is

I I .1
dO\ sinnO
- • - j = -
r \ dri cos no

r sin no

integrating, log r - - log sin no = log c,

'I
 396 INTEGRAL CALCULUS

p.
i.e., log (ainnOP' - log C.

p. " = C" ,lnnO.

19.4. Velocity and Acceleration of a moving particle.

If a particle be moving along a straight line, and if at any in-
stant t the position P of the particle be given by the distance s
measured along the path from a suitable fixed point A on it, then,
P denoting the velocity and f the acceleration of the particle at that
instant, we have
v = rate of displacement
= rate of change of s with respect to time
do
—..e,
and f = rate of change of velocity with respect to time
_dv ds
If, instead of moving in a straight line, the particle be moving
in any manner in a plane, the position of the particle at any instant
being given by the cartesian co-ordinates x, y, referred to a fixed
set of axes, the components of velocity and acceleration parallel to
those axes will similarly be given by
- dx
= rate of displacement parallel to x-axis -

= rate of displacement parallel to y-axis =

= rate of
f change of vi
= =

= rate oI change of =
f, = dt at) dt

The applications of these results are illustrated in the follow

ing illustrati ve examples.
 APPLICATIONS 391

19.5. Illustrative Examples.

Ex. I. A particle starling with velocity u moves in a straight line with a
uniform acceleration f . Find the velocity and distance travelled in any time
S denoting the distance travelled by the particle in time t, the accelera-
tion of the particle is given by the expression d 2 s I dt , and so, in this case.
d 1 s/di 2 =

integrating, Ls = ft s A , where A is the integration constant.

dt
Now, ds / at is the expression for the velocity v of the particle at time :,
and when t = 0,i.e., at start v = u - .. U 0 +A.

Hence, v = = ft + u - - - - (1)
di
Integrating (1), s = ft 2 + Ut + B,
where the integration constant B is found in this particular case from the
fact that s = 0 when t = 0, .. B = 0.
Hence. s =..ft 2 + ut = ut

Ex. 2. A particle is projected with a velocity u at an angle cx to the horizon.

Find the path.
Taking the starting point as origin, and taking the axes of co-ordinates
hori'ontal and vertical respectively, if x , y denote the co-ordinates of th'e
particle at any time I , since there is no mice and therefore no acceleration
in the iorizontal direction, and since the eertical acceleration in the horizon-
tal direction is always the same = g downwards, we have in this case

---. -
dt 2'dt2
dx - gt + B ... (1)
I l ence, Integrating, -= A dt
But, dx / dl , dy / dt represent the horizontal and the vertic corn-
porerits of velocity respectively, and these, at start when I = 0, are given
by u cos a and u sin a
.. 0 cos af l,u sin a=O+B ,
whereby the integration constants are obtained.
Thus, (U gives
dx
= ucosa, usina - gt.
- 7- =
398 INTEORAL CALCULUS

Integrating again, x = ut cos ct + C

and y = ui s ina..-..gt' + V.
Now, since xy=O when =0.we get from aboe C=D-L3.
Hence, x = ut cos a
and y = ia 5t h U - - gZ
Eliminating t, the path of the particle is gi ven by

y=xtana - - 12gU2 -
cos2a
which is evidently a parabola.

19.6. Miscellaneous Applications.

The examples below will illustrate some other applications of
different al equations.
Ex. 1. ' The population of a country increases at the rate proportional to the
number of inhabitants. If the population doubles in 30 year, in how many year will
it treble ?
Let r be the population in I year.
dx
solving, x = Ce
Let x=x, when I = 0 . :. C = x0 ; :. x x0 e
When x = 2x0 , 1 30; :. 2, = x0 e .. 2
When x3x, let S=T; ;. 3i0=xe; 3

30k = log, 2 T
log. 2 - O approirnateL,.
and kT = Iog,3 To

T 30 x 48 year approximately.
Ex. 2. After how many years will Rs. 100 placed at the rate of 5% con
Sinuously compounded, amount to Rs1000
Let x be the amount in t years.
dx = 5
x = kx say , where A
1 100
sols'ing.x = Ce5'
 APP[JCAI1ONS 399

When t O.,x , C 00. :. = 100e

When x 1000, let I = I ... 1000 = I00e
= Irl... kT = 10.
kT = log. 10 2.30 approximately.
T = (I / k) x 230 = 20 x 230 = 46 nearly.
the required time is 46 years nearly.
EXAMPLES XIX
Find the orthogonal trajectories of the following ía milies of cur-
ves
1. (I) y = mx (ii) y = ax'
(iii) x' + y' = 2ay . (iv) y' = 4ax
(v) ay = x 3(vi) x 1 -s- = a1
(vii) x 15 + y 213 = a'''. (viii) x 2 +y I +a 2 1 +2a4.
(ix) r• = a cos 0. (x) r 7 = a 2 cos 20
(xi) r (1 + cos 0 ) 2a . (xiii) r sin nO = a
2. (i) Show that the orthogonal trajectories of a system of con-
current straight lines form a system of concentric circles, and con-
versely.
I Take th point of concurrence as origin.
(ii) Sho.' that the orthogonal trajectories of the system of co-
axial circles
+ y 1 4 2Xx + c= 0
form another system of co-axial circles

where )and i are parameters and c is a given constant.

(iii) Show that the orthogonal trajectories of the system of

circles touching a given straight line at a given point form another
system of circles which pass through the given point and whose
centres lie on the given line.
3. (a ) Show that every member of the first set of curves cuts
orthogonally every member of the second
= x? +
(ii) die' y' + y * I
(i)

dx y7 + y + 1 o
dx x' + x + I
400 INTEGRAL CALCULUS Ex XIX

(b) Show that

(i) the family of parabolas y 2 = 4a ( x + a ) is sell-or-
thogonal.

(ii) the family of confocal conics

x1
a 7+
yt
= 1 ( being the parameter
+ b2+
is self-orthogonal.
4.(i) Find the curve in which the radius of curvature is propor-
tional to the arc measured from a fixed point, and identify it.
(ii) Find the curve for which the tangent at any point cuts off
from the co-ordinate axes intercepts whose sum is constant, and
..1entify it.
5. Find the Cartesian equation of a curve for which the tangent
is of constant length.
6. A particle is said to execute a Simple Harmonic Motion when
it moves on a straight line with its acceleration always directed
towards a fixed point in the line and proportional to the distance
from it in any position. If it starts from rest at a distance a from
the fixed point, find its velocity in any position, and the time for
that position. Deduce that the motion is oscillatory, and find the
periodic time.
7. A particle falls towards the earth, starting from rest at a
height h above the surface. If the attraction of the earth varies in-
versely as the squae of the distance from its centre, find the velo-
city of the particle on reaching the earth's surface, given a the
radius of the earth and g the value of the aceleration due to gravity
at the surface of the earth.
8. A particle falls in a vertical line under gravity ( supposed
constant ), and the force of air resistance to its motion is propor-
tional to its velocity. Show that its velocity cannot exceed a par-
ticular limit.
 APPUCA1IONS 401
Lx. XIX

9. A particle moves in an ellipse with an acceleration directed

towards its centre. Show that the acceleration is proportional to its
distance from the centre.
10. In a certain culture, the number of bacteria is increasing at
a rate proportional to the number present. If the number doubles
in 3 hour, how many may be expected at the end of 12 hour?
at the
11. Alter how many year will a sum of money, placed
rate of 5% continuOUSY compounded, double itself?
12. Radium disappears at a rate proportional to the amoint
present. If 5% of the original amount disappears in 50 year, how
much will remain at the end of 100 year?
13. A tank consists of 50 litre of fresh water. Two litre of brine
each containing 5 gram of dissolved salt are run into the tank per
minute; the mixture is kept uniform by stirring, and runs out at the
rate of one litre per minute If m gram of salt are present in the tank
after t minute, express tn in terms of t and f ; -.Ad the amount of
salt present after 10 minute.
and in-
14. The electric current 1 through a coil of resistance R
satisfies the equation RI + L ( dl I dt) = V. where V
ductance L
is the potential difference between the two ends of the coil. A poten-
tial difference V = a sin ot is applied to the coil from time t 0
to the time t = / , where a, 0 are positive constants. The cur-
rent is zero at L = 0 and V is zero after t =n/; find the Cf r
rent at any time both before and after t = n / al.
15. A horizontal beam of length 21 m, carrying a uniform load
of w kg per m of length, is freely supported at both ends, satis-
fying the differential equation
EI d__Y = wx2 .-w/x,
dx 2

y being the deflection at a distance x from one end. If y = 0 at

x • 0, and y1 = 0 at x =I , find the deflection at any point; also
find the maximum deflection.
-28
Inlegral Calculus maifl
 402 lNTEC;LL CALCULUS Ex . XIX

16. A horizontal beam of length 1

simply supported at its end
subject only to its own weight satisfies the equation

El d -'Y-
dx
where E, 1, to are constants. Civon
= y = 0 at x = 0 and at
x = I, express the deflction y in t:: s of x.
17. A harmonic oscillator consists of an inductance
L • a con-
denser of capacitance C and an e.m.f. E.
Find the charge q and
the current £ when E = Eo cos W and initial conditions are
q0 and i io at I = 0; i. q satisfying the equations

dqEo
dt2 COS(ot,,
+ LC L = dt

What hapens if o ?

ANSWERS
1. (1) r 1 + = . ( ii) x 2 + ny 2 = c (iii) x2 • y1 = 2cx
(iv) 2x 't • y 2 = c2 () 22 1 + 3y' = c2 .
NO y = cx
(vii) x 4/3 - y '3 =

(viii) y'I( I -y 2 )± x( ) -x1) + sn ^ y ± sin x= C

(ix) , = c sin 0. (x) r 2 = c2 sin 20.
(xi) r (1 - cos 0 ) = c . (xii) r cos no = C.

4. (i) Equiangular spiral. (ii) Parabola.

5. x =/a 2 _y 2 +fa( log (a_a2_y2)_ log (a+QiTj)}
if y =awhen x = 0

6. p '1-2
(a - x2 t I x
- - . when p is the acceleration at
a
a unit distance. Period
2n

a+h 10. 16 times the original number.

 APPLICAflONS
403
11. 14 year nearly. 12. L61 of the original anIour.t

13. St (i + 50
_._. ) gram 91 grant.

14. For t <! R sin wt - W L (COS wt - r ) }

IA ) i = L ?6)24R 2

awl
and for Rw RI
(a (i+tt)--r--

- 4x 3 + 81x)y 5w!'
= TTi
=

16. y -ix +

17 q (qo ._.)CosI
- 1-& LC +'o sin R.__!__ t
E0 C
+
f-- W I LC COS WI.

to Cos C)t_ qo .----'C

Eo C W
- I-&LC sin ox
Jf CO i.e.. frequency of e. m.
= natural frequency. Osdilaijo,
i.e., resonance will take place and the circuit will be destroyed. Before
destroying

q = qo cos WI + is E0
Slfl WI
+ I sin W;

= cos ,e - q o O3 g in cot +
3lfl WI + tCoSO)I
Eo(I )
 THE METHOD OF ISOCUNES

20.1. It is only In a limited number of cases that a differential

equation may be solved analytically by the preceding methods. In
many practical cases where the solution of a differential equation
is needed under given initial conditions, and the above me-thods
fail, a graphical method, the method of isoclines, is sometimes
adopted. We proceed to explain below this method in the case of
simple differential equations of the first order.
Let us consider an equation of the type
=f(x,y) ... (1)
dx
As already explained before, the general solution of this equa-
tion involves one arbitrary ccnstant of integration, and henL. rep-
resents a family of curves, and, in general, one member of the
family passes through a given point ( x, y).
Now, if in (1) we repace by m, we get an equation f (x, y)=rnt
which for any particular numerical vale of m represents a curve,
at every point of which the value of j, i.e., the slope of the tan-
gent line to the family of curves represented by the general solu-
tion of (1) is the same as that numerical value of m. This curve f
x , y ) = m is called an isoclinal or isocline. For different numeri-
cal values of n we get different isodinals, which may be graphi-
cally constructed on a graph paper. Through different points on
any one isocline, short parallel lines are drawn having their com-
mon slope equal to the particular value of m for that isocline.
Similar short parallel lines are drawn through points on other
isocllnals. If the tiumber of isoclines drawn be large, so that they
are sufficiently close to one another, the short lines will ultimately
join up and appear to form a series of curves which represent the
family of curves giving the general solution if (1), and a particular
number of the family passing through a given point represents the
particular solution wanted. All necessary information regarding
 METHOD OF ISOCLINES 405

the particular solution may now be obtained from the graph.

X,

rig. i
As an example, let .s :'nsider the differential equation
are
=x_y 2given.The iso.ik by m=x — y2,
Jx—
or Y' = x - tn, a series of equal parabolas shifted left or right from
= x ( which corresponds to m = 0 ), as shown in the figure.
The dotted curves represent graphically the solutions of the dif-
ferential equation.
 CUAPrER xxi
DOUBLE AND TRIPLE INTEGRALS
21.1. In Chapter VI we have discussed what is meant by the
definite integral of a function of a single variable with respect to
that variable taken between two prescribed bounds. We shall now
discuss briefly about the double intergation of a function of two in-
dependent variables taken over a two dimensional region and that
of a function of three variables taken over a three-dimensional
region.
21.2. Double integral over a rectangle.
First of all we confine our discussion of double integral of a
function of two variables over a finite rectangular region and then
we shall extend our idea to any finite region other than rectangles.
YI.n I-)

yl

Fig.l
Let f ( x, y) be a bounded function of two independent vari-
ables x and y defined over the rectangle ABCD, bounded by the
lines x = a, x = b , y = c, y = d. This rectangle will be denoted
by R[a,b;c.dl,or,simplyby R.
Let a=x0 <x1 <x2 < ... <x,1<xfr
and c=yo <y i <y <...<y 1 <y..=d.
 DOUBLE AND 1RWIB LECALS 407

We draw the lines x,x - x,...,x xj, ..,x X.

are parallel to Y" and the lines yy1, y
y • y, ., y y., -' which are parallel to OX to divide the
rectangle R into mn sub-rectangles.
Let us denote the sub-rectangle R I xj - , x1 ; y - , y, I by R,
and Its area by A ,1 We have A 4 - (x i x - )( y - Y i -'
Let mij and M 1 be the lower and upper bounds of
in R0, We next form the sums
f(x,y)

=E ZIN4A4
MqA q.

It Is evident that for every mode of sub-division of R into sub-

rectangles R ,m(b-a)(d-c)Ss S SS M(b-a)(d-c), where
M, mare the upper and lower bounds off( x, y) in R. Thus, we
can say that the two sets of upper and lower sums S and a are
bounded for all modes of division of R into a finite number of sub-
rectangles R,, as defined above.
The lower bound of the set of upper sums Is defined as the
upper integral of f ( x , y ) over R and Is denoted by I and the
upper bound of the set of lower sums Is defined as the lower In-
tegralof f(x,y)over R and is denoted by J.Wewrite:

f(x,y)dxdy, J .Jff(x,v)dxdy.

If I and I be equal, then the functlof ( x, y) I. said to be in-

tegrable over the rectangle and the common value denoted by

fJ ( / x, y ) dx dy or
55 f( x , y ) dA

integral of /( x, y ) over the rectangle R.

I. defined as the double
405 INThGRAL CALCULUS

Note. Norm of a division of a rectangle-

The norm of the sub-division D of a rectangle R is denoted byll D I I
or t and may be defined to be the greatest diagonal of sub-rectangle,,
i.e., II DII - max.4((xj - ;_) 1 + - yj- )2),thenadnumis
to be taken of all the diagonals of the sub-rectangles of R.
21.3. Condition of Integrability.
We state here, without proof, the necessary and sufficient con-
dition for the Integrability of a bounded function f ( x , y) over a
rectangle R. The condition is that to every positive number c, there
corrsponds a positive number 8, such that for every division of
R whole norm is c 8, the oscillation S - s is less than c.
21.4. Simple properties.
If f ( x, y ) and g ( x, y) are integrable functions over a rec-
tangle R, then the functions f ( x, y) ± g ( x , y) and cf ( x y),
where c is a constant, are also integrable over the same rectangle
R and
(I)
55 (f ( x,y )± X Y )) dxdY
-55 f(x,y)dxdy ±55
g(x,y)dxdy.
R R

(ii)554(xY )dxdY C55 f(x,y)dxdy.

(iii) Iff(x,y)!5 g(x,y)in

then jjf(x.y)dxdY 55Jg(xY )dx4Y .

(iv) If R = R 1 u Ra when R n R2 =

then JJa f(x,y)dxdy

55R
 DOUBLE AND TRIPLE INTEGRALS 409

Note 1. If R, and R, have a common region, as shown by the shaded

area In the adjoining figure, when we calculate

f(x,y)dxdy
+JJ f(x,y)dxdy

G
we are really integrating f ( x, y)
twice over the common region
EFDC so that the result (iv) is not
valid. We, therefore, introduce the R2
condition R, r R, = $, so that
R,and R2 have no common region.
Note 2. It is easily seen that B F
the results (i) (ii), (iii) and (iv) in MS . 2
§ 21.4 above remain valid for provided the functions are in-
double integrals over any finite region F
tegrable over the region F.
21.5. Calculation of a double integral. Equivalence of a
double integral with repeated integrals.
Theorem. If the double
ff R
f ( x , y ) dx dy exists, R being the

rectangle [ a , b;c,dland if the integral5 f(x,y)dx also

exists for all values of y in ( c, d ) , then the repeated integral

dy x, y ) dx exists and is equal to the double integral.

Proof. Let us divide the rectangle R [a, b ; c , d into rnn sub-

x - and y = y
I
rectangles by the lines x = x, , x = x2 ,..., x
y = y... - , where a = x0 .z x, < x2 < ... < X. = I,
Y = y .....
and C = y0 <y <Y2 < ......< y... = d . If rn,1 and
M,, be
the lower and upper bounds of f ( x, y ) in the sub-rectangle
R1, Iz - . I,
x, ; y, - 1 y, we have
!^f(x,y)c M,, . ... (1)
m,i
 410 INTEGRAL CA LCULUS

If
y remains fixed, then f( x, y ) can be regarded as a function
of one variable x only and then by using the mean-value theorem
of Integral Calculus we have
P
m 1 (x, -x, )5 Jf(x,y)dxsM 1, ( x - x 11 ), ... (2)
which holds for all Values of y in () - I , Yj ).
p
Let us now denote f( x, y )dx by g (y).
Jxi -
Since g(y)ls bounded ln(y11 .y, ),wehave
rn11 ( X . - )( y, - y, ) :5 Jo
Jo < M 1 xi - )( y • - y
i.e. , rn, 1 A , < 1 , J, 5 M,, A 1 , . (3)
where 10, 10 are respectively the lower and upper integrals of g (31)

Yi VI
in W, ,y, ),i.e.,!0 g(y)dy and Jo g(y)dy.
- Y -i

Now, taking summation with respect to I and j, we get

rn 1 A , 5fg(y)dy
±

d
) dy s n 7 M, A ,1,

I_I f-I

i.e., s5 5g(Y)dYJg(y)dy ig S, ... (4)

where s and S denote respectively the lower and upper sums for
the double integral of f ( x, y) over R.
 DOUBLE AND TRIPLE INTEGRALS 411

(4) can be expressed as

$ 5J dyf f(x,y)dx, f dyJf(xy)dx 15S.

r'
I :5
J
_c
dy
J f(x,y)dx 15 I
(5)
and I!;f dyf f(x,y)dx 5 I 1
Since the double integral exists, I = I and therefore

5 dy fi x ,y ) dx also exists and is equal to the double integral.

Hence, fdyff(xy)dx
=55 f OF, y)dxdy
('or. 1. irJf f(x,y)
R
dxdy exists and j f(x, y)dy exists, then also

arf f(x.y)dy exists and =fff(x.y)dydz.

f

Cor. 2. If the double integral exists, the two repeated integrals cannot
exist without being equal.

Thus, if the double integral exists, then the repeated integrals

5 drf f(x, Y)dY.f dYff(xY)dx both exist and they are equal.
 412 INTEGRAL CALCULUS

21.6. Double integration as a limit.

Let f ( x , y) be a continuous function in the rectangle
R I a, b ; c • d I . Let us divide R into sub-rectangles in the process
as described in § 21. 2. Let ( x, , y ) be any point of the sub-rec-
tangle R,, whose area is A,. Now, form the sum Z f( i,, y, ) . A,.
We can show that, as the norm of sub-division I D II - 0,
the limit of the above sum will be the double integral of f ( x , y)
over R,

i.e., Li -. f ( x, . y, ) A, = Jf f(x , y ) dx dy.

21.7. Geometrical interpretation of double integral.

Let us consider the double integral

Jj' f( x, y) dx dy, where

R istherectangle Ia.b;c,dJ.Let z= f(x,y) be the given func-

tion, which graphically represents a surface.
We divide R into rnn sub-rectangles by the lines x =
x x =x,,.., ,and y =y, y = Y1 .....y = y, - ,
where a=xo<x,<x<..<x=b,c=y0<y<y1<<yd
Let R,, be the rectangle I x, - , x, ; y, i . y. I whose area is A,,
Now, lines paraliel to the z-axis are drawn from points of R, 1 upto
the surface z = f( x , y ) to form a prism. Let V,, be the volume of
this prism and m, , M 1 the lower and upper bounds of f ( x , y)
in R,,. It is evident that
m, A,1 :5 V,, 15M,, A,,.

rn,,A,, V,, < EM,, 1•

-1 ,1 i1 ji
As the double integral
if f (x, y) dx dy exists, it is equal to

Li
E rn,1 A,, Li M,, A,,
 DOUBLE AND TRIPLE INThGRALS 413

= U V,

...5$f(xy)dxaY= LL being the volume

of the cylinder whose base is R, the generators of which are paral-

lel to the z-axis drawn from points of the sides of R upto the sur-
facez =f (x.y).
21.8. Double integral over any finite region.
We have already defined the double integral of a function over
a rectangle R. Now, we are going to define the double Integral of
a function over a given finite region E . As E is finite, we can con-
struct a rectangle which can enclose the given region £ . Let us
define a function g ( x, y) over R as follows
g(x,y) =f(x,y)forallpoirttsof F,
= 0 outside E.
The function f (x, y) is said to be integrable over £ if g (x, y) be
integrable over R . We have then
g(x,y )dxdy.
x ..y )dxdy
=55 It

Note. Let us take the z-axis perpendicular to the plane of E If lines

pac.aItel to the z-axis be drawn from points on the boundary of E upto the
surface a = f ( x, y), we get a cylinder. Geometrically f ( x, y) dx dy

represents the volume of the above cylinder. -

21.9. Evaluation of double integral.
Let E be the region bounded by the curves y = u (x), y v (x)
and the ordinates x=a,x=b.lf u(x),v(x)be continuous
functions and u ( x ) !; v ( x ) in (a, b ) and f ( x, y ) be a con-
tinuous function In E , then
,o(x)
55 f(x,y)dydx =J dxJ
£ a u(z)
414 DJTEGRAL CALCULUS

Y _y.fq_
Let R Ia,b;c,dj be
the rectangle which

jxb encloses the given region

E and let us thuine the
function g(x,y)In It as
follows
Y=C I
= f(x,y)
at all points of E,
Fiji = () outside E.

Now,
ff E
f(x,y)dydx =f5g(r.i)dydx

y ) dy
= J:dx 5 C g ( X,

5 u(x) v(x)

=5 dx [5 g(x.y)dy
+5
g(x,y)dy]
+5 V (z)

,.b v(x )
dx g (x, y) dy. (the other two integrals being Zero)
= Ja J v(z)

S viz)
dx f(x,y)dy.
= f
a fu (Z)

Note. If £ be the region bounded by the continuous curves

x =U(y),x = V(y) the straight lines y = c,y= 4 and f(x,y)bea
continuous function in E, then
,.V(p)

E
jJ f(x,y)dxdy
=J c
dyJ
thy)
f(x,y)dx.
 DOUBLE AND TRIPLE INThGRALS
415

21.10. Area of a region.

In the definition of double integral, If we put f(
x. y) = 1 'we
hwe the area A of the region bounded by the curves
Y = v ( x) • the straight lines x = a, x = y u (x),
b given by
b
A=
a
f v(z)
dydx.

21.11. Jaobian.
If i , ,..., u,, be n functions of n independent variables
- , x,. and have partial derivatives of the first order at
every point of the common domain in which the functions are
defined, then the Jacobian of u 1
, .. ...... ee,, with respect to x,,
,....,x,. is denoted by

u,..) (uu U,,

d( x1 , x ,...., x,, ) rer,
, ,•,•, X.
and defined to be the determinant
au,
ax, 5x2 ax,.
a, au, au,
a2

ax, Tx, ar,,

Cor. If x - r cos 0, y r sin 0, then

L. ax
ps.. = coso - r sin 8
r,o) = ar aq =r

sin r cos 0

21.12. Change of variable in a double integral.

Sometimes double integrals can easily be evaluated by chang-
ing the independent variables by suitable transformations.
416 ThTEGRAL CALCULUS

Let x= •(&,t).y =v(E,) .. (1)

be two functions of k, il defined in a region £' of the - q plane
bounded by a curve C'. Moreover we assum e that
(i) the above two functions 0 and w have continuous first
order partial derivatives at all points of E' and C'
(ii) the equations (1) transform the region E' bounded by C,
into a region £ of the ry plane bounded by a curve C in such a
way that there exists a one-one correspondence between £, E' and
C. C';

(iii) the Jacobian does not change sign at any point

of E, but it may vanish at some points of C', then

f(xY)dxdY =JJf1$(.1)Wh1)hl jL ddi.
JJ

I Proof of the theorem is beyond the scope of this elementary treaties I

21.13. Application of double integral.
(a) Mass of a plate.
Let a plate be bounded by the curve C and let the mass per unit
area (i.e., the density) at the point (x , y ) be given by p = fx , y).
Divide the plate into elementary areas by lines parallel to the axes
of co-ordinates. Let 8A be one of these elementary areas with (x, y)
its centre of mass. The mass M of the plane area is given by

M =L t X p8A
JJE

(b) Centre of mass of a thin plate.

Let us divide the plate E into elementary areas by lines paral-
lel to the axes of co-ordinates. Let 8A be onef these elementary
areas with ( x , y) its centre of mass, then the co-ordinates of the
centre of mass of the plate are given by
 DOUBLE AND TRIPLE INTEGRALS 417

- SAL2O £p5A
Ix p&it 55 pxdfl 55
=
JfpdxdY

pydA 55
55i
y= Lf

55 pdA 55
A 0 pM
p dx dy
where p = f (x, y)is the density of thc plate at (tx,y)
(c) Centre of pressure of a plane lamina.
IF a plane lamina be Immersed in a liquid, the point at which
the resultant pressure acts is called the Centre of Pressure of the
lamina.
Let a lamina be immersed vertically in a liquid. Let us take the
axes of co-ordinates in the plane of the lamina, the x-axis horizon-
tal and the y-axis vertical. Let 'is divide the lamina E into cicmen
tary areas by lines drawn parallel to the axes. Let 8A be one of
these elementary areas and (x, y) be its cntroid. Let p = f( x, y)
be the pressure at ( x, y).
Total pressure on the lamina
= LtLp.6A =5JPdA =5L

If (, ) be the centre of pressure of the lamina, by taking mo-

ments about the axes of co-ordinates, we get

55 pxdA 55
x=
8
Lt ZpxM
Ep SA -
A -, 0
_____ •________
55 pdA ,jf p dy
E £
dx

JJPYdA
y= Lt
55 pydxdy
E E
8A-,o Ep8A
pdA 51 pdxdy
Inlegral Calculus (main) -29
55£
411 ThrlTc.RAI. CALCULUS

(d) Moments and product of inertia of a lamina.

If r ........r, be the distances, from fixed line, of par-
ticles of masses m 1 , m 2 .....m,. respectively, then mr 2 is defined
as the Moment of Inertia of the systerr " particlesabout the line.
Let E be the plane lamina. Take the axes in the plane of the
lamina. Divide E into elementary areas by lines drawn parallel to
the axes. Let 54 be such an area whose centroid is (x, y). Let
p
= f(x,y)bcthcdeflSitYat(1,Y).
I, Moment of inertia about the x-axis
... JJ py2dA dxdy;
AA
Lt
0
L p&f l.y
=55 py 2

I, = Moment of inertia about the y-axis

= 6A
L 0 EpSA.x2
=55 px' dA =55 px'
F = Product of inertia with respect to x - and y-axes

= &Ao
Zp8A .x .y =55 pxydA =55 pxydxdy.
21.14. Illustrative Examples.
.R /2 .fl
Ex. 1. Evaluate cos (x + y) dx dy.
Jo o

We have I
=j 0
dy
5 Cos (x +
0
y)dx

= J: /2 dy [ sin( X+

x /2
+ y) -sin (0+ y )]dy
J
 DOUBLE AND TRIPLE INTEGRALS 419

*/2 */2
(- siny - sin y)dy [2 cos y]

= 2 cos - 2 cos O = -2.

1-y2
Ex. 2. EeaIuateJ
5 1)' +y2 ] dxdy.

Here, 1 = s: dy J' [(x_ I P y2 ] dx

.1 I-y7
= dy [(X1)3+y2x]
Jo

-
p1
- Jo
dy +y2(1- y1)
-
< _)
3 j
1
-4
f+
I -
r_ + . ls+X
[ 21 3 5 3] 0 iri1

£x.3. Show that

j0
.1
dx
p I
I
J0 (x +
- dy 5 1 1
aYj (x +- y)3d1•

l I -y
Jf 0
x
We have d. dy
(x+y)3

.1 I
=J0dxj0
(' +

.1 .1
axJ [ 2x
=1 (x+y)3(x+y)hJ'
J
20 INTEGRAL CALCULUS

= f ',. [ _ Zr
2x + y) x

•1 ,
dx_, ' -!
=j 0 ______ +

dx
=
r 1 1' 1
+
1
= 2
J0

Again, f dy I __...__Y_..
(x +
dx
D

=J0aJ3
d5
+
rTdk

.L.. L 1
jx + y i(x +y>'j

f'
L y+I
_
1
^_._L-+._11
(y+l) y y11

C I t
J_j._j.._^_.jpdY I1JQ
r I 1
I 1' —
2

Thus, the two given integrals air unequal;

i.e., the result Is proved.
Note. This resul not unexpected as th function ((x - y)l( x + y) 3)
is discontinuous at (0,0). 1 Proud
over the triangle formed by the
Lx. 4. Evaluate IJ 'I( 4.r 2 — y 2 ) dx dy
O X y X. (C. H. 1967 1
straight lines y 1,
 DOUBLE AND TRIPLE INTEGRAl 421

The given double integral can be expressed as

•1 .X ______
I dxl IF _y'dy
Jo Jo 'a

=
Jo
dx
[yiL
L
_____________

2 +4
I

1
+2x' stn' I )a
=5 0
11

5 (-r1 + 7x1

-
13r' nx'l "
TI = '
0
2 + y' )1 dl dy , She region of
Es. S. Evaluate II 2g ' — 2a (x + y) - (x
C H 1962
i ntegration being the circle x' + Y' + 2a ( x + y ) = 20' . I
i-lore R ,the region of integration, is x 7 + y' + 2a ( x + y ) = 2a'

ie..(x +a)' + (y+a)' =4a7.

Now, use the transformation x + a= X y + a = Y,

i.e. z= X -a,y= Y - a.
lax ax _ I 0! =
JWcy) ax
=Il-I ay
lo
ax a) I

=55 14a 7 -(r • a)' - (y

-z - - Y' ) ax ay. where the new region '

the circh X = -Ia'

Ft nI v use the polar 1 ran 1nrmation
X.Y \
= rco=9.Y =r ,n9 :. I =
422 INTEGRAL CALCULUS

2a
I =I I (4m _,)rdrdO
e.o ,-o
a
,21, ,23
do (442 - ,)rdr
=
Jo Jo

=Fo1 1 4a 2 -Y
I
29 r f 1.

.Ti0

=2,(2a2.4a2_40)=8IW4.

Note. If the region of Integration R be the complete circle x 2 + y 2 = a2

and we use the polar transformation x = r co. 0, y = r sin 0, the limits of
, will be 0 to a and those of 0 will be 0 to 2i.
If R be the positive quadrant of the above circle,, will vary from 0 to
a and 8 from 0 to x/2
U R be the upper half of this circle, rwill vary from 0 to a and 0 from
0 toe.

EX. 6. EudluateJf (i - !. - Ygj ) dxdy, where R consists of points

in the positive quadrant of the ellipse+ I. (C . 1-1. '63, '72 1

Use the transformation x = aX, y = bY.
ax ax
ax ay a0
J X,Y) y _i 0 b =ab.
ax a
R transforms to R' which is the positive quadrant of the circle
X 2 + Y, =
I -Y2 )dXd.
J5R. 51
Finally use the transformation X = r cos 0, Y r sin 8.
I X .Y \
 DOUBLE AND 11UPLE INThORALS 423

,12
I ab (1 - r ),i,dO
e-o rO
,f2 .1
=abl ae (1 -r2)rdr
Jo Jo

r' i 1 1 irab
ab[8] IT
=.b-!(I
_) =--
=

Ex. 7. Evaluatef e' 2yco+y2 ) dxdy. (0 SixSc)

J
Here the region of integration is the posttfrye quadrant. Use the trans-
. .
formation x = r cos 0 y = r sin A.

,I2 .-
= e- • 2,2 ,Co.5O) r drd9
Jo .o ro

i.1/2 i•
(I ,o2I),2
sI rdr
=
f
'o
do
Jo

(f2 -
•c0aa.2e)'2 ]
J dH [
L - 2(1 + cos a sin 2A)

• IC / 2
- I cos
• a sin 20)
dO
I do
= 2 Sx/?
0
cos'O + sin 1 8 s-2 cos asin0 cos O

= I Jsec2ede
tan 1 0 + 2 cos a tan 0 + I
0
424 INTEGRAL CALCULUS

- U-
- 0
7F_+
dz
2zcosa + (where z =tanel

- dz
- 2J (z+ cos a)2+ sin 2a

- 1 1 Z +cosal 1 1*
- 2stna [tan- , ina ,J0 = 2sinaI,_tcot(1)

= I a
2 sin a COt'cotcj=-
2 sin a -
Ex. B. The density at the point (I y) of a lamina bounded by the
circle x 2 + y 2 - 2ax = 0 is p = x - rind its mass -

M = mass of the dx dy,

lamina = ffR p

Rbcing the circle x I + y2- 2ax = 0

= ff R x dx dy.
Use the polar transformation x = r cos a,y = 'sine. Then Xy =
r, and
the equation of the circle becomes r = 2a cos S
*/2 .2acoa
I r cos erdrde
=J 8-*/2 J,=0

*/2 21 CO5 0
cos$.dO.[!_]
=1 -* /2 0

*/2
8a'
3

cos 'O d9=!._3

16a3J 0* /2
 DOUBLE AND TRIPLE INTEGRALS 425

Lx. 9. A plane laming of uniform surface density is bounded by the upper w!

of the cardioide r = a (1 + cos 8) and the initial line. Find the co-ordinates of
the centre of mass.
Let (x . y) be the centre of mass.
pxdrdy
-
X= R being the upper half of the carJioide

I JI R pdxdy r = a(I + cos O)

r cos 8rdrd9
ff R
Pr
rdrd8
R

cos odoJ r dr
J

f f
doa(l.co.0)
a
- ,0

f
x g(. •rc0)

[1 cos ede

- x

f L 0[-r2

I
Jo
•a' cos 0(I+ co g 0)dO

a' (I + cos 9)' do

Jo

( cos 20 (1 + cos 2 )3 2d0

J
- - 2a o O =20

fI
Jo
(1 + co g 20 2dê
426 INTEGRAL CALCULUS

(2 cos Q - 1)8cos'$d$
2a o fI
- 3 .X/2
8 co s ' d
J0

F 7531* 531*1
2a j 2 864 226422J2a3 a
- T 531* 342
6422
ydxdy
-ffR p
pdxdy
= 55R
=
ff it

ff it ,drd8
a(I .co.0)
sin OdO
50 [-i]
- * (l+co.0)
l[n]

2a 'olg
I (1 + cos O)sin
3 OdO

-
(1 + Cos B)1d8
fo
J

+cosO)4
29
4cos 4 dO
fo
 DOUBLE AND TRIPLE INTEGRALS 427

2a 4 re
a -
3 */2 ]
41 cos4.2d
Jo

2a1 1 16i
a --
3231* 9*
422

co -ordinates of the centre of mass are ).

Ex. 10. A semi-circular lamina of radius a is immersed vertically in a liquid,
the pressure of which varies as the depth with the bounding diameter in the sur-
face. Find the centre of pressure of the lemma.
Let us take the centre of the lamina as the origin, the bounding diameter
as x- axls and the vertical radius say-axis.
p pressure at the point (x.y) =ky.Let(x,y)betbeceiitreofpres-
sure.

= 55R
,R:x2 +y2^a2 y;t 0
JJ p dx dy

a
IL xy dx
Iputrarcoo 9, Y = r sin 01

IL
a' rn2el*
sl
II o
r.in9co,0drd0
L
TT10
=
.1 .a
=
a =0.
,2.inedrde [—co.e1
JJ 3 Jo
fjjtpydx°jY
a
flit
JJPdXdY
flit
428 INTEGRAl. CALCULUS

c & r 3 sin' 9 drdO

Jo
r2 sin 9drdO
0 0

_Cos ej
- [

I a4
2 4 3ita
- a3
2.3
- 16

0, 3na
the centre of pressure is (
)

EXAMPLES XXI(A)
1. Evaluate

(I) +y)2dydx.

i•4 a.1
xy(x - y)dydx.
(ii)J0 J 0

sin(x + y)dxdy.
(iii)J Cl J0
tog. 2 1
ye'Y dxdy.
(iv)5 5
 Ex. X X I(A ) DOUBLE AND TRIPLE INTEGRALS 429

w/2
e2 cos(y - x)dydx.
(v) J 0 J

(vi) f o
f o Y/' dydx
5J,+ dy dx

1 -Jr

5 j'
(viii) (x2 + y 2 )dydx.

- Y _________________
- x 2 - y 2 dxdy.
(ix) J0 j0

(x) I5
Jo
2 'J2x- r 22
x dy dx. (xi)
5 53
dy
4
dx

f Cos O x a(i •co.0)

(xii) r sin 9 dr d8. (xiii) rdr A
JoJo J0J0
(xiv) II xy dx dy over the positive quadrant of the circle
x + Y2=

(xv) IJ( x2 + y 2 )dx d y over the region in the positive quad-

rant for which x +y 15
(xvi) ffxdxdy over the ellipse b B x l + ay' = I
(xvii) 11 xy ( x + y ) dx dy over the area bounded by y = x
and y = x.
2. Prove, by evaluating the repeated integrals, that
430 114TEGRAL CALCULUS Ex. XXI(A)

3. Evaluate, by using suitable transformations:

(I) If x 1 + y 2 )dx dy over the region enclosed by the tri -
angle having its vertices at(O,O),( 1 ,O),( 1,1).
I C. II. 1965

(ii) + y2dydx.
50 5 0
(iii)11x2y2 dxdyextended over the region x ^! O..y ^! 0,
x 1 +y 2 !5 1. IC. It. 1969j
(iv)Ifx2y1dxdy over the circle x' +y'!5 1.
I C. H. 1964 1
(v) 11 4( 44 2 - x2 - y I ) dx dy taken over the upper half
of the circle x' + y' - 2ax = 0 . I C. 11. 1966
(vi)fI 12- 2(x + y)-(x'+y')l dxdy,the region of
integration being the circle x' + y' + 2 ( x +v) 2.
(vii) If xy (I' + y 2 ) /2 dx dy over the positive quadrant
of the circle x' +y' =a',(n +3> 0.
" a
f (i x 1 dxdy
(viii) i
j j v(x' + y' )
0 y

(ix) If sin
X Y >dx dy over the region to the first
it

quadrant bounded by = O,y = xand 12 +y' =iO

(x) If r' sin 8 dr dO over the upper half of the circle

= 2acosO.
(xi) If (x+y)'dxdy over the ellipse {(x'/a') + (y 2 /b 2 )1 = L
I C. H. 1977J
(xii) fJx'y dx dy over the positive quadrant of the ellipse
(x2/a2+y h /b 2 )=1. IC.I-l.1971J
 Ex. X xi (A ) DOUBLE AND TRIPLE INTEGRALS 431

4 2 b 2 - bx'- a2y2
(xiil) JJ(a2 b 7 + b 7 x 2 + a2y2) dxdy, the field of
integration being the positive quadrant of the ellipse
a1 b2

dxdy over the triangle with vertices

(xiv)J5 + +
(O,O),(2,O),(I,'5). IC.H.19701
dxdy
taken over one loop of the
(xv)Jf (I + + 2 )

lcmvtiscate(x 2 2 = x1
+ y2 ) yt IC. H. 29741
(xvi)Jlxdxdy over the region r ? a ( 1 + Cos O).

e - ( - 2 - 0 ) dxdy. /
(xvii)s: s:
x1 dx dy
(xviii)J -- (•1 +'(x +

Je_(. .2.+2,l) dydx.

(xix)50 0

(xx) II zy dx dy over the region bounded by the parabolas

Y2 = 4x, y2 = 8x. x 1 = 4y, x1 = 8y.
(Put 1.. , - =j
x y
4. Find, by double intcration, the area of the revlon bounded
by the curves:
(i) y 2 4x,y =16x,x = 1.x = 16 IS the positive
quadrant;
(ii) y7 4ax;x' =4ay;
432 It'IIEGRAL CALCULUS Ex. X X I(A )

(iii) x 1 + y 2 100,x 1 +y 2 = 64 .,y =13x, -/3y x

In the positive quadrant;
12 + 1!
(iv) the ellipse 1 and Its auxiliary circle;

(v) r = a (1 + cos 0), the initial line and the line 0 tc/3;
(vi) y 2 = 8X, y 2 16x, zy = 25, xy 16 in the posi-
tive quadrant.
L For the thin plates bounded by the following curves find
the mass, the centre of mass and the moments of inertia about the
axes, p being the density at (x , y)
(i) x ^! O,y 0, 1 2 + y'!5 1,p = kry;

(ii) the parabola y 2 = 4ax and its latus rectum, where

p = constant;
(iii) y 0,x 2 + y 1 15 a 1 .p = 12 + y2
12 2
(iv) positive quadrant of the ellipse - + = 1 . where

p = kx;
(v) trlanglewhoseverticesare(0,0),(1,13).(2,0),
where p = constant;
(vi) upper half of the circle x 2 + y' = 2ax, where

P ='[x + y2
6. (I) A quadrant of an ellipse of semi-axes a and b (a > b)
is just immersed vertically with the semi - major axis in the surface
of a liquid in which the pressure varies as the depth. Find the centre
Of pressure. (C. H. 1962 1
(ii) Find the position of the centre of pressure of a quadrant
of a circle of radius a which Is just immersed vertically, with one
edge in the sufaceof a liquid, the pressure of which varies as the
square of the depth. i C. ii. 1964 1
 DOUBLE AND 1RIPLE D(TEGRALS 411

ANSWERS
1. (1) .-. (ii) a. (iii) 2 (iv) }
(v) I (vi) I. (vii) log 2
3 Ii
(ix)- .
6 (x) ji.. (xi) log -. (xii)
(xiii) i n a l . (xiv) 1.44. (xv) I. (xvi) 0
(xvii) 3 (i) .j. (jj)isIs. (iii) -x.
is
(iv) 24 . (v) 93 O R - 4). (vi) Sis a4
(vii) 2 (n + 4)'
(viii) .- log( 42 + 1). (Lx)!... (x) 2a 3

(xi) nab (a'4 +b 2 ) b. •..( gab

(xlii) -- (is - 2

(xiv) "3 Ian I I . (xv) )3-( g - 2 (xvi) 10 isa3

is . ... is is
(xvii) - ( X V III) (xix) . (xx) 192
4. (a) 84. (ii)
(Vi
_a (iii) 3is. (iv) isa (a - b
(v)(4g+9'.'3); )31og2.

(l1) 1 pa i ,( .3 a,O),..pa
37
I .pa
• . . isa
(iii)
I / Sa \ isa' Isa

ka 2 b/3 gct 3b ka'b' 2ka41,

(iv) -i- k-,- .-),---- 15
4 3 P43P(i -) 2'6
16a3 ( 6a 9a \ 256.
(vi) --- —-, 512a

6. (I) On the minor axis and at a depth 3b

(ii) On the vertical radius, at a depth 32a

.3,5

Inlegral Calculus (main) -30

434 INTEGRAL CALCULUS

21.15. Triple integral over a rectangular parallelopiped.

Let I ( x • y . z) be a bounded functionof three independent
variables z , y • z over the rectangular parallelopiped bounded by
the planes x a,x = b.y =c,y d.z = e.z =f. This region
will be denoted by R I a, I'; c, d; I 1 or simply by R.
Let a=x0 <z 1 <x2< ... <X.,i<xm=b,
c=y 0 <y 1 <y , .c...<y 7<y .=d,
e=z0 <z <z< ... <z,<z,=f.
We nowdraw the planes x x 1 , x ........ x =x., which
are parallel to Y Z plane, planes y = y1 , y = y2 .... . y = y.. -'
which are parallel to ZX plane and planes z = ;, z
z= ., which are parallel to X Y plane to divide R into mnp
sub-regions, each of which being a rectangular parallelopiped.
Let us denote the sub-region
R Ix , xe ; y, - y , ; Zk. , z k I by R,,& and its volume
(x - - 1 )( y, - y , )( Z - Z - I) by
Let in qk and M 1z be the lower and upper bounds off (x y, z)
in R, . We now form the sums

Mijk U-jk
=

P ii m
S Mijk Vi,è

k -i j-i i-I

It is clear that for every mode of sub-division of R into a finite

'umber of sub-regions
in ( b - a )(d- c)(f -e) S s S S M(b - a)(d-c)(f -e),
where in, M arc the lower and upper bounds o(f( r, y, z) In R.
rhus, the two sets of upper and lower sums S. s are bounded.
 DOUBLE AND IRIPLE INTEGRALS

The lower bound of the set of upper sums is defined as the

upper Integral of f( x, y, z ) over R and is denoted by I and the
upper bound of the set of lower sums Is defined as the lower in-
tegral off 0, y, z) over R and is denoted by land we write

If I and J be equal, then the function x, y, z) is said to be in-

tegrable over R and the common value denoted by
f(x .y. z)dxdydz orJfJ f(x,y, z)dVisdefined
MR
as the triple integral of f ( x , y, z)over R
21.16. Condition of integrability.
The necessary and sufficient condition of integrability of a
bounded function f ( x , y, z ) over R is that to every positive num-
ber £, there corresponds a positive number 8, such that for every
division of R whose norm is < & • the oscillation S - $ Is less
than t.
I Proof is omitted.
21.17. Calculation of triple integral. Equivalence of a
double integral with repeated integrals.

Theorc,'. If the triple integral f( x , y z ) dx dy dz

ff5
exists over R 1 a, Li; c , d ; e ,f I and ifhe double integral

55 s f(x,y,z)dxdyalsoexjstsforallvalues of in
then the repeated integral 1dz5 [55 S
f( x , y, z ) dx dy] exists
and is equal to the triple inLgral.
Proof. Similar to that of § 21.5 and left as an exercise to
the student.
Cor. I( f(z,y,z)be continuous over R, we have
436 INTEGRAL CALCULUS

f(X.y,Z)dXJYJZ J14Z5dYJgf(X'Y
555
we can change the order of integration to suit our convenience.

21.18. Triple Integral over any finite region.

Let E be a finite region bounded by any surface. We can con-
struct a rectangular parallelopiped R enclosing E completely. Let
us define a function g ( x , y, a ) over R as follows:
g(x.y,x) = f(x,y,z)atallpointsof E.
0 outside £.
The function f ( x, y , a) is said to be integrable over E if
g (x , y, a ) be integrable over R . Then we have

55$ E
f(x,y,z)dxdydz
= 55$
20.19. Evaluation of triple integral.
Let E be the region bounded by the sufac.es z - u C x y),
z v(x ,y ) ; y = q, ( x), y = , (x) ; r =a ,x = Li. If f( x , a)
be a continuous function in E • then
b • i. ( t) v(r.y)
f(x.yz)dxdydz.5dxj dyf f(xy,z)dz.
J5f F •(x) w(x,y)
Proof. Similar to that of j 21-9. I
21.20. Change of variable in a triple integral.
Let x(,1 , ç),y (,,ç),z= w ,) .. (1)
be three functions of k,ij, , defined in a region £ ' of the E,
space bounded by a surface 5'. Moreover, we assume that

(i) u v, w possess continuous partial derivatives of the i irst

order at each point of E 'and S',
 DOUBLE AND TRIPLE INThGRALS 437

(ii) the equations (1) transform the region E' bounded by 5'
into a region E of the xyz space bounded by the surface S In such
a way that there exists a one-one correspondence between E, E'
and 55',
a(.)
x ,y x
(iii) Jacobian does not change sign atany point of
a (.
E', but it may vanish at some points of S', then
f (x ,y ,z )dx dy dz
111JE;
J

= 4f If j (u(,i .v ( i.Q,w(,.ti ,C))

a(x,y J
d d'q dç.

21.21. Applications of triple Integral.

(a) Mass of a solid.
Let a body Ebe bounded by the surface S, and p=f(x,y,z)
be the mass per unit volume ( i.e., the density at the point (x, y, z).
We can show as in § 21.13 (a) that the mass of the body is given by

M
= 1511: p
(b) Centre of mass of a body.
Let p =f(x,y, z ) bethedensityat (x y, z ) ofa solid body
E. If ( x , y , z ) be the centre of mass of the body, then

55J p xdxdydz '51E p ydxdydz

= 555 p dxdydz = fJj p dxdydz
zdxdydz

=r fff dxdydz
48 INTEGRAL CALCULUS

(c) Moment of inertia of a body.

L.t p = f(x,y.z)be the density at (x,y,z)ofa solid body
E, Then 1,, 1,, 12 its moments of inertia about ox, oy, oz are given
by

= 55J p(y + z° )dxdydz,

=555 p(z° + x2)dxdydz,
= 555 E p(x° + y 2 )dxdydz.

21.22. Illustrative Examples.

pa -x
E:. 1. Evaluate
JOJ0
I (y + : 2 ) dz dy dx
I C. 11. '76 I

a 'g -x 2 b
We have I =
Jo
dx j dy f (y 2 +z )dz
0

.2 b

= dx dy [y l
J O
z + _. ] 0
0

f 11

(by' +)dy
= dxj
o

[2

dx [ +
=
Jo a

I
-Jo [ti(a 2 2) &' (a° -x° ) 2/2 ] a
- I.
3
 DOUBLE AND TRIPLE INTEGRALS 439

(ba3cos39b3acOs9)
Ix = sin OJ
JC

=5 2 (ç_'cos
2
b' cos2e)ae
- ---
o ,
0

- a 4 b 31 it a 2 b 3 1,t - ita 2 b(3a 2 +0)

3 22 - 48
Ex. 2. Evaluate III (x + y + z + 1)' dx dy dz over the region defined by
X? O,y^ O za 0, x +y + z!^ I
The region of integration is the tetrahedron as shown in the ligre.

Fig.4
In this region z varies from 0 to 1-x-y,y varies from O tol - x
and x varies from 0 to 1.
I. tI-x-y
I = f dx
..J dj (x+y+z+I)4dz
0 0

I I-x
axj
=$ dy 1
[x + y +z
+
1)5]

rtx
I
I
i , l
dx
J0
132 - (x + y + 1)5Jdy

0
440 INTEGRAL CALCULUS

64 (1.x)''
- Jdx
J[32
0 6

..1
= 32( x 32 x + 1 (1 '+
_. r)' J
[
0

1 r 32.1' 32 128 11 = 351 117

.--=--.

Er. 3. Find the mass of a solid in She form of the positive octant of the
ellipsoid x z2
= I ,the density at (x,y. z)being xyz
* +
M = Mass of the solid:ffJ ryz dr dy dz. where E is the positive
octant of the ellipsoid.
Put r =sX,y = bY,z = cZ.
_- a 00 =abc.
0b0
00
abc XYZ abc dX d't dZ where E' is the positive octant
M =: EI
of the sphere X2 +Y 5 + V =
= alb 20 XYZdXdYdZ.
Put X =rin6cos. Y =rsInOsin, Z =rcosO.
a(X.YZ)_ sinOcos rcos9 cos Q -. rsin8sin
- sin8sin r cos 9sin rsinO cos $
cosO - r sin 9 0
= ,2 sin 9.
,/2
M =a2b2c2 f2
J J 0
r'sin'Ocos9sin4cos 2 sinOd,d3d,
0

1 /2
= a2b7c5 ['6'
- I Ocos9dO sin$ cos d
1 oJo Jo
 DOUBLE AND TRIPLE TNThGRALS 441

X/2
- ab'c1 [ 11n4-0
0
[2] 0

= 1a 2 b2c 2 j
Note. Generally in the case of a spherical region X 1 + y 2 + z 2 = a 2
we use the transformation x=rsinOcos$.y=rsinOsin,z = ,cose.
(I) limltsforr,O, are 0,a;0,ir;0,2x for the whole sphere;
(ii) these are O,a;O,x/2;0,2x for the upper hemisphere;
(iii) they are 0,1; 0, it/2 ; O, x/2 for the positive octant
EXAMPLES XXHB)
1. Evaluate:
2.
(i)J J J (x + y + z)dxdydz.
00 0

,1x
(ii)J II
0
J
y
I O
xdzdxdy.

Iz 1z.
(iii)JJ J e"Ydzdydx.
000

2 z x'13
(iv)f I dydxdz.
Jo o Io 2 + 2

dxdydz
(v)
fff(l + x + y + z)2
extended over the tetrahedron

bounded by the planes = O,y = 0,z = 0,x + x = 1.

F C. ii. '70 I
NO JJJ x 2 dx dy dz extended over the volume of the ball
x2+ y 7 + 2 1 5 a 2C. 1-1. '69)
442 INTEGRAL CALCULUS Er XXI(B)

2. Evaluate by using suitable transformations

(j) JJJ ( x + y I + z' ) xyz dx dy dz taken through the
the sphere x 2 + y 2 + zS 1. IC.II.'641
dxdydz _
over the sphere
(•i)JJJ 2 + y' + (z -
r 1 + y' + zt C.!i.731

(iii)
fri.
'\
I; —X1
+ 2 +
y1 -z2 dxdy dz over the positive
+
-

octant oI the sphere 1 2 +y 2 +z5 1.

(iv)fff\/- + dxdydz taken over the ellipsoid

a2 b2 +

a2 b2
+—^1.
C2

(v) hi (ax 2 + by + cz 2 )dxdydz over the sphere

1 2 + Z 5R'.
3. For the solid bodies bounded by the following surfaces find
the mass, Centre of mass and the moments of intertia about the axes,
p being the density at ( x, y, z
(I) x ? O.y ^ O,z ^! 0 .2 2+ y 2 + z 2 sa 2 ;p = kxyz.
(ii) x ? O,y 2t 0.z 2! 0,x + y z!5 1 ;p = constant.
(iii) z ? 0 • x 2 + y 2 + z 2 a 2 p = constant
x 1 ,,2 22
(iv) x e 0.y ^ 0,z 2: 0,- +- + —2 1;p = constant.
4. A mass M of gas is diffused through all space. If the den-
sity of the gas at (x, y, z) be e . 7 I , show that M 2 =

ANSWERS
L (i) 18i (ii) 35
(iii) 8 ( •" - 6e 2l • ge l 3)
2 (v)j log (2%
./e 5 ) (vi)
(iv) —
3
 DOUBLE AND TRIPLE INTEGRALS 443

2. (1)0. 00(2 - log 3) (iii) I1(R( ! - g!

2 4 2

4,(a + b + c)R
(iv) rcabc (v) - 15
(16a 16a 16a)ka ka' ka'
3. (I)ka'
48- 35 35 35 '96' 96 '96
I I I
(ii) I p,(1, !,1);

4lIa S
npa' , (0,0,+a); -- jgPjP.jP
na 4 4Ra2

' 3a I
(iv) 3b3c
iabcp; ;.wabcp(b? + c').

30 ,abcp(c' + a ! ), 30 ,abcp( c 2 .s.b 2 )

 Miscellaneous Examples II
lnt.egrate the following ( Ex. Ito Ex. 21 ):-

cos z + Sin X
1. i cos2x log dx.
.1 COSX - sin X

2. Jx log ix + .l(x 2 + a 2 )Jdx.

xdx
•
j J(x+ a) + \( x + 1')

cotx
x + ])dx.
4.
I (1 - sin x)( sec

tan P 2+ cos x
5
(i)f I + sin x
dx
J cosx(1 + cosx) dx
(ii)

6. (I) 5 '( sec x - I) dx (ii) 5 '( cot x ) dx

f 2x1 + 3x + 7
7. dx.
jI (x + 3)(x 2 + 2x + 5)
tan xdx
8.
f (a+ btan2x) ,a>b.

dx
9.
f + I)
X( X"

1O.$e:irx cos x cos 2x cos 4xd.

J cosh x + sinh x sin x di

I+ Cos x

'. x2+Sx+7
12 e' dx.
(x+3)2
dx
13.
J xlogxlog(logx)
 Misc. II MISCELLANEOUS EXAMPLES II

14.xdx
f
(x - a)(x b)(x - c)

15
• J xdx
(12 + a 2 )( x 2 + b 1 )( x + c 1 )

x'dx
16
• f X2 — I
________
X +I xl - I
17
• x4x2+I
ax . (ii)J - dx .
(i) j

f dx x2 dx
18. ()
J x 4 -x 2 + 1 f
2x1+3x + 3
19 2x + 2)dX.
J0 1+ 1)(x+

cos 8d0
20
I s- 2sinO +2sin 2 O + 2sin 3 8+ sin40
0

dx.
21f + cos . x
0

22. Show that

x <-. log (I -x) < I —f--- (0 < x < U. IC .H. 196.31

--x

-dx
23. If n > I, then 0.5
<J < 0.524.
I C. II. 1963, '66 I
24. Show that log ( m / n ) = log n - log n from the defini-
tion of log x as a definite integral, i.e. . from
I
dt
log x=
f -
446 INTEGRAL CALCULUS Misc .11

25. Find the area of the region included between two car-
dioldes r =a(1+ cos O),r=a(I-cosO),a>O. (C.H.1967J
26. Find the area of the loop of the curve
a sin 3O ,y= a sin 3O a>O.
x=----sin cosO
27. Find the area of the loop of the curve r cos 0 a cos 28.
25. (I) The area between the curves y 2 4ax and x 2 = 4ay
(a> 0) revolves about the axis of x. If V be the volume of the
solid thus formed, then show that 5V = 96ita
(ii) lithe curve r = 2a cos 0 revolves about the initial line,
then show that the area of the surface of revolution is 4ita2
29.If the area lying within the cardioide r = 2a ( I + cos 0) and
without the parabola r (I + cos 0) = 2a revolves about the initial
line, then show that the volume generated is 18 Ira'.
30.Find the area common to the circle r = a and the cardioide
r = a( I + cos 0 ).
31. Show that the area included between one of the branches
of the curve x 2 y 2 = a 2 ( x 2 + y 2 ) and its asymptotes is equal to
one-fourth of the square formed b y the asymptotes (a > 0)
32. The distances of the vertices A B, C of a scalene triangle
of area 5, from a fixed fine MN, are x ,x 2 , The line MN does
not cut the triangle ABC Find the volume generated by the revolu-
tion of the triangle ABC
ANSWERS
ir
1.- I sin 2z log c05 - log cos 2x
2L C0SX - SiflX

2. -logfr + Ix 1 +a 2 ) _! '/( x 2 +a)+— sinh

3.(x L
x + a)' 2 -( + a)3'1
[
(r + b) 12 + - ( x + b)312
 MISCELI.ANEOLJS EXAMPLES If 447
4. log tan r 4 1 sec' r + tan 1z
S. (I) I1+inx
- log
I
- sin x 2 1 + sin x
I

(ii) 2 log (sec x + tan x) - tan - x

6. (i) - 2 cosh ' ( 12 cos f xi.
(ii) I lOgs I +'12 -tan x + tan
- '1(2 2 tan z
tan xi + tan x + '-2I t3fl -, (-'1tan x)
7. 2 log (x + 3) -jtan ((x + 1)12).
('1(a - b)
S h x)
- '1( a - b) sir, - 4b
- n-1 (x + 1).
9. log log
10. .8 in (8x - tan ' 8) II. cosh x tan
(65)
12. e rx+3 +2
13. log (log ( log x))
14. +
(a - b)(a -c)
Iog(x - a).
a
- (a2 b2 )( 0 2 - ç2 ) tan1 J
16. 12 log( X 4 I) --log(x' + x 4 + 1)
24
tan-' (2r 4 ; 1)
+
-
17. (I) tan1 x (ii) log 2 -+ 73
'13.x + 1
.x
- I
18. (I) ! tan-' ______ - I x1 -'13.x +1
log Ty- + 43.
- x+
___
(ii) tan ., - +_ log x' 'l3.x + I
x x -73.x + 1
19. . i + 2 log 2 - tan ' 2. 20. - (log 2 + I) . 21.
25. (}!_ 2). 26. (3 43 27. 2a 1 (1 ..j).
30. (x - 2)s2 .32.41t(r, + x 2 + r ).
 INDEX
Acceleration, 396 Centroid, 289 . 296
Applications, 393 circular art, 296
Area between two curves parabolic lamina, 297
cartesian, 230 quadrant of an ellipse, 298
polar, 234 solid hemisphere., 299
Areas of closed curves, 245 Chainette, 312
Areas of plane curves Change of variables, 13
cartesian, 225 Cissoid of Diocles, 317
polar, 234 Clairaut's equation. 356, 358
Area of Complementary function, 369, 384
cardloide, 236 Complete primitive, 327
dssoid, 241 Condition for integrbiIity, tOO
cycloid, 228 Constant of integration, 3
ellipse, 227, 231 Convergent integral, 151
folium of Descartes, 237 Cycloid
parabola, 228 vertex downwa j s, 311
loop, cartesian eqn., 229 vertex upwards, 310
loop, polar eqn., 236
Astroid, 314 Darboux's theorem, 100
i
Auxiliary equation. 363 Defin te integrals, 2, 92, 94,113,
equal roots, 364 as the limit of a sum, 92
pair of complex roots. 365 general prtperties, 128
real and distinct roots, 364 geometr.al interpretation, 110
lower limit, 2, 93
Bernouli's equation. 351 upper limit, 2., 93
Beta function, 191, 200 Differential equation
Binomial differentials, 197 definitions 34
By parts integration. 38 degree, 325
exact, 341
Card ioide, 319 first degree, 331
Catenary. 312 first order, 331
Cauchy's equation, 390 formation, 325
 INDEX 449

geometrical interpretation, 326 Gamma function, 200

homogeneous, 336 Geometrical interpretation
nth order, 384 definite integral, 1110
order, 325 differential eqn., 324
ordinary, 324 General laws of integration 4
partial, 324 General properties of
resolvable into (actor,, 354 definite intergals, 128
second order, 363 General solution, 327
solution, 326, Generalized definition, 93
solvable for x, y, 356
Delta function, 152 Homogeneous equation, 336
Divergent integral. 151 special form, 337
Double integral, 406 Hyperbolic (unction, 24,66
application, 416 Hypocyloid, 314

Elementary rules of integration, 4 Improper integrals, 150

Equations of second order, 363 convergent, divergent,
Special type, 376 oscillatory, 151
Equiangular spiral, 318
Inequalities and limits, 136
Eulerian integral, 191 Inferior limit, 93
Evaluation o(defin,te integral, 1113 Infinite range, 150
Evolutes of parabola, ellipse, 315 Integrability, 100
Exact equation, 341
necessary and sufficient
Exponential curves, 316 condition, 11(X)
Exponential function, 136 eir.ble function, 100
lnteg.. '-e- Of
First mean value theorem ' .' t.t and co-tangent, 65
of integrat calculus, I(;) '4 Tntegratwg i,rs .3.42
First principle, 94
Folium of lescartes .116
..e, .. 2. 91. 113
Fundamental integrals, 5 Fulerian, 191
Fundamental theorem, 1 1 mproper, 150

Integral Calculus (maIn -31

 45') L"ll-GKAI. . U.S

indeto'iite, 1, 2 L.-ngth of plane curve froi

r'firute, 150 cartesian. 257
nvolving one parameter, 181 pairwfric, ?Y,*
Riemann. 10) .J'd, f7
Integration, 2 polar. 261
as the limit of a s'Jrn, 91 - ,jscate, 7,20, 321
by parts, 38 acon, 320
by Successive reduct i on, 131 Limits. .3
from first principle, 94 L;ne integral, 247
of infinite scries, 159 Linear equation, 350, 33
of power series, 159 Logarithmic curve, 316
of rational fraction. 79 Loga'ithmic spiral, 38
Intrinsic equation of Lower integral, 100
cardloide, 268
catenary. 267 Method of substitution
cycloid, 268 definite integral, 115
Intrinsic equation to a curve from ndefinite integral. 18
W cartesian eqn., 265 Miscellaneous application, 398
pedal uln., 267 Moment of in"rtia, 289, 300
polar eqn., 266 circular plate, 303
Irrational functions, 165 elliptic lamina, 302
Isodine, 404 retant'. lat l amina, 302
sphere, 304
ja . -obia- 415 hn unifr'rn l rod, 301

'egenr. equtinrs 3O C ,..,rnc i-known curves. 310

.'gth of air 01 Urt1 --, na! actories, 393
Ic, 262 —In
,3. .. eq. 394
•li 53
Ic.ip, 260 c' 'u,' th *uren , 283
pLral.011, 239 f'araL-311c rule, 249
 INDEX

Particular intera!, 369, 3M Special trigonometric functions, 58

meths, 320 Spiral of Archimedes, 319
i'erfvd di(fe.'n1ial, 341 'tandard rnehods of nk"gra t' 7
rnttive5 and integrals, 107 Standard integrals, 23, 41, 42, 58
Priripal value, 151, 155 Slrophoid, 317
Probabllky cireq , 317 Subst i t' ti in definite

integral, 115
Aad'us of gyrn!on, 300 Superior limits, 93
Rational fractions, 79 urfaceres, 275 . 2.84
Rcti fica hon, 257 S,.rnrnation of series., 118
Reduction frni.l, 141, 181 Syrnboiical oprealion, 17t'
double parameter, 139 Symbolical o 1icr1'tura, 369, 370
single parameter, 181
special devkec, 195 Tractriv, 313
Rie,nann integral, 1XI Trial solution, 363
Rose petal, 32 Triple i1egrsls. 434
application, 437
Series represented by definite
integrals, 118 Upper integral, 100
Separation of variables, 331 Upper limit, Z 93
Sign of an area, 243
Simpson's rule, 2,47 Velocity, 396
Sine spi,al, 3Z2 Volumes. 274, 276
Volume and sur ace-area
Singular solution, 327, 358, 359 cardioide, 280
Solids of revolution, 274, 281 Cycloid, 279
volume, 274, 280 parabola. 22
Solution of a differential paraboloid, 2.78
equation. 326
Some -ell-known curves, 310 Witch of Agriesi, 318
 UNIVERSITY PAPERS
B. A./B. Sc. (Pass)
CALCUTTA UNIVERSITY
1994

2 (a) Evaluatef - --- -- dx. L) Prove thatf sin' dx -

srnx+ cos x

(c) Find the area of the region bounded by the parabola y 2 - 4x

and its latus rectum.

(d) If B(rn n) -f r (I - - dx (pr, n >

show Ihat R(.. .j ) - it

(e)F.valuatejj y dx dy over thc region in the positive quadrant

for which x + y 1.

3. (a) Show that v- + B satisfies the differential equation

4 2 v I2dt'
----0 A B are constants
dr 2r dr
(b) Find the differential equation of all parab,)Ias having their
axes paralle l to y-axis.

(c) Show that the curve in which the polar subtarigent is

froportional to the length of radius vector is C - e , where k is
constant and C is arbitrary.

(d) Show that the curvature is zero at every point on a straight

line.

(e) Solve +I-c-

f
dx
cos 3x + cos 4 t dx,
9. (a) Evaluate ()
1 Zcos3x

(ii)
I %r sXz
dx;
 UNVERSITYPAPER 453

(b) Evaluate Ii !
H fl-p

10. (a) Ill. cosx dx, n is a positive integer,

.5

show that I, I2 and hence find 5 sin' xdr

(?) Show that the total area of the ellipse a2

! j,2 - I is nab.

(c) Assuming I (i + I) ii (n is positive integer), and i()_

find the value of

f [-TnO dO x -°r

ir: of the curve

11. (a) Find the length o
_n(Cos 0+) sin t))
y.a(siflfl - OCOSOL from 0-0 to 0-O
(I') The circle x + - a' revolves round x-axis Find the surface
area and the volume of the whole sphere generated.

(r) Show that

5 a(xx:
dx-- log (+ 1).

12..(a) Find the value OfJf .'ty(x2+y)dxdy

where L00sY' )
(h) Find the intrinsic equation of the catenary y - c cos h
 451 INTEGRAL CALCULUS

(c) Obtain an approximate value of log 2 by calculating by

x
J
Simpson's rule dividing the interval into two equal parts.

dO
13. (a) Show that ifI PO +gO -, 0 and if 0 - uand -0 whent -0

hen
0. It co3vl t.
(b) Solve

(i) x2(xdT+ydy)+Zy(rdy-ydx)..0,
(ii) (x 2 +y 2 +4)rdx +(x -y2+9)ydy-0;

(in) dx + logy- (log Y)2

(it') xy (t _y) -

14. (a) Find the curve for which the sum of the r ciprocals of the
radius vetor ad the polar '.uhtargcnt is constant

(b) Solve. () -2 - O iu

d2 ' 'iw d'q

(ia) r - - x - - tug r , ( III)+ - a
dx dt dx
z) (r) - I y -2, Lv 3 - I ad v - - I W 1 CU -2
BURL) WAN uNIvrRsrfl
1
7. (a) L vatuate the following
dx
(:ij -----;--' (zJ dr
(4x-
C xi- $iflx
(in)
f dx, rz
 ftV flY PAPERS 455

(b) Evaluate 5 x2 dx from the definition of Definite Integral as

lirni of a sum.
dx -
(c) Show that . acos 1 xs b2'in2x
[a> 1'> 0 I
A. (a) State and prore the Fundamental Theorem of Integral Calcukis.
a
(b)f fIx)dx- ff(o - x)dx. Prove it, and hence show tia
bgtanxdx - 0.

(c) Evaluate, when possible

f dx
—j----

y 2 - 4x - 4- 0 and
9. (a) Calculate the area bounded by the curves
y 2 s4x-4 .'. O. (figure necessary).
(b) Find the perimeter of the hypo-cycloid

(c) By the method of Integration. Determine the volume of a sphere

of radius r.
10. (a) Show t!t, a p lane curve for which the length of the normal at
a point of the curve is equal to that of the radius vector at that point is
either a circle or a recangu1ar hyperbola
(v) Solve the owing:
(i) (x 2 + y 2 )dy xydx,
() (2x- v+ l)dx+ (2y- x- I )dy 0,
456 INThGRAL CALCULUS

+ x)c X 3ecy , (iv ()+ 2x_3x2_

- dx
(v) (xycosxy+ sinxy)dx+ x 2 coszyay. 0.

11. Solve the following:

(i) L2L_ 4y.. r3e1•,

dx 2 dx

+ y- sin2x when x- Oy- 0 and

(ii) dx—2dx 0,

dy
(iii) x d y + y - sin logx 2,
s .-

(iv) d2y +'ty- 2sinxcosx,

dx

(v) 0 when x- 0, y- 4andwhenx- y- 0.

VIDYASAGAR UNIVERSITY
1993

1. (b) (i) if f(x)- Ix- fl, evaluate f

0
f (x x

a a

(ii) Prove that f(x) ' IX _f f(u - x)dx.

0 0

(iii) Evaluate
f f
a.0 y-O
sin (x + ) dx Iy.
 UNIVERSIT\ FAI'ERS 457

(iv) A point P(x, y) moves in xy- plane satisfying

0 x /2, 0 s y s sin x Find the area of the regioMraced out by P.

(v) Use method of integration to evaluate

5
1 5 +25 +35 + ...+ n
lim l
n-•

(c) (i) Examine whetheris an integrating factor of the dil-

ferential equation y(l + xy)dx- xi'y - 0.

(ii) Obtain the differential equation whose general solution is

ax+ by+ c- 0, a, b, care arbitrary .nstanLc.

(i' Find tie solution of the equation

2c + c 2y- 0, cisaconstant.
Jr2 dx

7. (a) Evaluate

. dx I CO X
(' ; (H)J
(x2 + 1) (x'+ x+ 1) .)x+

dx [eT.-r dx
(ui C 'x
1+ 3e+ 2 7

1) ( ii + 2) it 3) (2u)
(b) Evaluate urn /' 1

. (a) Evaluate log in x dx

438 INTEGRAL(.ALCULUS

(t)Fva!u.'te: xy(x 2 + ?)
j
P
f dxdy, where

R :10 S x s a, 0 s hJ.

.2
9. (a) Show that :
.. dx
--= x
j sinx
f r
sir.x
d_

dx
(b) Evaluate: f ------
x(^ 1)

.4

, here n is a positive integer,

(c) If I , - f
0
tan" x

I
then show that I(,. I) + 1 (n- l) -
?1

10. () Find the area of the region Included between t he cardioiues

r-a(I+ csO) and r- n(1- cosO).

(b) Find the length of the arc of the curve x- c0sinO,

0
cos 0 measured from 0 - 0 to 0 -

II. (n) curve is such that the segment cut oil on the *,-axis by the
normal at any po int ( x, y ) on it is equal to the distance of the point of
contactformthe origin !'tbecurc the point( 0, 1) ,find
its equation
dy
(I .) k'e: pn / - ; (V3, + 1) + x 0, p
-dx

+ str,- x3c052y.
dx

oIve: - +ly- 2.e

V
 t NIVERSITY PAPERS 459

- x
(c) Suh:
f x 2x + y- 3

Solve
(j) : 4y- sin x, given that

Y- I and 1, when x- 0.
(ç) If the gradient of a curve at any point (x, y) is 2 + x and it
passes through the orgin, then find the eqution of the curve.

KALYANI UNIVERSITY

1992
4. () Sr- w that the interaI

dx (u> 0) is convergent i f n> I

') Ie ..t the cosivergtnce

Jr
('). (z)
j
U I-

X -

t . ') !)efcw Gariuna function F ( r) Evaluate F 1).

D' t. e Rta 'unction 13( ,,: u)

Prove hi t 13 n)-B( in 5. 1 n ) + B ( in U + I
460 INTEGRAL CALCULUS

(c) Eva Iu1

() f 4
in x cos x dx, 6 (ii)f x 5'2 (1 x) 3 dx,

(iii) f0 c 4' x 312 dx.

NORTH BENGAL UN!VERSflY

1989

6. integrate
sin x dx
(i'
/ JI e dr, (vi) j_j
(x+ ),

1' .2dx
fdx
2
x + 6x+ 8
(iv)
J a4 - x4

(v) J dx
4+ 3coSx

7. (a) State and prove "Fundamental Theorem" of Integral Calculus.

(b) Evaluate

22) 2
. — . [ 1) 6n-
 UNIVERSITfl' PAPERS 461
(c) Evaluate
K Jr
-

dx
(i) fo2 cos x d.,,, (i
o
71 COS
2 x+ b2 S i n 2x
K
dx
(m)fo 1-2a cos x+a

$. (a) Find the whole area of a 2 2-y a2 ? - x4.

(b) Find the perimeter of r- a( 1 + cos 0

(c) Show that the volume generated by the rotation of the conic
x- a cos O, y- h sin 0 about the line x- 2a is 4a2b.

9. (a) Find the equation of the curve for which the cartesiansuhnormal
at any point is constant.

(19 F:nd a curve whose subtangent is of constant length a.

(c) Explain the method of solving the linear first order differential
equation

- Q- Py, where Pand Qare functions ofx.

dx
(d) Solve;

(r) (i+ y2 )dx- I

(tan- y- r)dy_ 0,

(ii)
4+ _ y2 , (w) d'y- 2
dx 2dx
+ Sy- lOsinx,

(ii) i-4+ gO- a, when 0-u. 1-0,

dt
0.
 INTEG1'J CMfULUS

TRIPURA UNIVH(.9FY
1992

1.(b)() Show thatf e f(x) + f'tx) I dx - e J(x)

.: (, if j) be an function
(ii) Show 11

- 2fJx) (jx, it fly) be an even function.

cosxix :1
(:r) Show tha! I' sin.x+ cosx 4
0
q
(iv) Find the area er.d'ed by the elIipse'%x - a C's
y - 1' cos tpand the a y es in the first quadrant.
(v) Define definite integral as
the linii of a sum.

(c) (1) Solve the equation

j dx + x di - 0
(v 1 + y Z - i
ax + bx
(z)E liminate the co-MM5 a, I' 1om ii -
(I) 2 L.
(iii)Soive 5f)+ 4) y
:tn of conc rb:
() Find the dif1erefltl euOf l f
cirek; having cen'res at fie origin
( j Red uc +
Ll—
-iiUo tlu' ax
+
dx x x

Q are functions of x alone or constants

7. (a) Integrate the fullowing

()f dx;

f sinx+ 2 cos x dx
(aia)j 2 sin x + cos x
 tiNt V ERStTY PA J'FRS

(b) Obtain the reduction formula f

f c,s cos ox IX and
h e nce I.nd cv,i!uiff COS 3X C0s5x Jx
th rjan'.cnjal I K.orum of!nt2gai Ckuus,

(h) Ivaivat' t de(inite intgrat C dx by the rnettwd of

lirn't cIa s,yi

2
(e) If 1, r"snxdx and ,>

l)1,,_
14 'tl.n-
(-J
- I

f
2 2 2)
'. (o) show that sm '0 cos "0 10 -
- (m Pi + 2
I

Where i n - 1, ii -1

(b) FvaJuiteff dx dy over the triangle fonred b y the

straight lines '.1 0, x - 1, y- x
10. a) Find the r-tation between S ind y for the parabula 3/ ia
with its vcrtcx as the fixed point. (S .vmLx)ls having usual mcaai r i gs .)

(h) Find the Vo!ulne generated by revolving the upper ha1f of the
IP of x(x 1 + - v about the xaj.
11.(a) Soi',.-e 1k' ( lowirg
I dv
dr
li * II ax- (2x 2q+ 1) dy- t);
+,
•.x 1-x

1 e pp • - y(x + y). 1' - dy

 U. (a) Solve the following; where D-
dx
(1) 2 4D+
3)y- 2e3;
(n) (DD- 2)y- sin 2x;
(f)2_ 4E) s- 4)y - x 2 , given x - 0, y - ' Dy - I

(h) A particle starting with velocity u, moves in a straight line

.vith a uniform accctcrationf Find the v e locity and distance travelled in
my time.
KALYANI UNIVERSITY
1992

4. (a) Show that the integral

J - dx (a > U) is (nnvergnt II
ID

(b) Irsi -the ccrwrrgenre

(b(
Ii)
.3 I fl
\ - H +)

CY
(Lx'

5. (u) E)e1inc Gamma Cunc1-uifl Fix). Evaluate r iii.

(b) [)I1nc 13eta runion 5 (t)i. r)
Prove that B(rn.r)=L'*I n-B(rn.n+l).

(c) 1r,Iiiatr

(I) JSin 4 X COS6 cix. (iij J I-x) cix

(iii) J /dX.