CHAPTER

14
Indefinite Integra
a
I Definition (iv) a'd= log, a-+C= a log, e+c

If f') is derivative of f (x),then f (x) is perimitive or sin x dx=-cos x+c

anti-derivative or integration of f'). So, differentiation and (v)
integration are inverse to each other, i.e., cos x dx=sin x +c
(vi)
d
(sin x) = cos x, so integration of cosx is sin x. (vii)tan xdr = log I sec xl+tc =-log lcos x+r
d
(vi) cot x dx = log Isin xl+c
J
dx
(sin x +c) = cosx, so integration of cos x is sin x + c. sec xdx = log Ise. .r+ tan x l+c
(x)
d
f)+c]= F(r) ’ f()+c is perimitive of F(3). =-logl sec x-tan xltc = log tan
da
|4 2tc
» JF() dr= f(a) +c ()cosec xdx =- log lcosc x+ cot xl+c
Jisintegralsign and = log lcosec x- cot xltc: =log tan
|F()drmeans integrationof F(x) with respect to x, (xi) sec xtan x dx= sec xtc
where c is the constant of integration. (xi) cosec x cotxdx=-cosec x+ c
(xü)sec x dr = tan x+c
IBasic Theorems on Integration (xiv)cosec x dr=-cot x+c
If f(x) and g (x) are two functions of a variable x and kis
1
a constant, then (),adr=tan
x-a
(v) d= 20 log x+a
+c;a>a)
1 a+x
(ii)
d (xvi) 2
-log +c;(x<a)
d la-x
=dx = sin
(iv)

(is) dr =loglx+ t a ltc

ISome Standard Formulae
=sinh
tc; (n*-1)
n+1
1
(6o) Je.r= loglx+y+a te
(ü)dr= log, Ixl +c +a
Gi)e'dr =e'tc = cosh
indegral. One
Substitution
Integration 1
put In Integration
is Methods of (vm)f(ar
by I SolutionExample Solution"dxExamplequale s 2Solution Example (xxvi) (xvi) (orV) (cúY) (XxiI) ()
case
this the 0)
When 3
derivative e*Laf()+ |e e -Vr'+a
integrand 2 sin Ve-a (
=
cos
=+c eedr cos
we 11 1 + bx-a'
x/2 b) +b*Va' bx
of dr Va+b
dr dx d
2 +loglr+vr'
the is =f'()] = dx=sec-
= +c =V-a a
dx a'+b a
f(x) other the dx
to
equalso(ar+b) a+h -a'-logl
x+r-al+c Vr+a
equals to bx-costan bx-costan
=I product dx (a 2
i.e., a
to (a +
to ef(x) = cos sin
X+2x+c +c -cosh+c 2 2
brb
convert of bx+b +a'sinh+c 2 a'
two l+c
cos a
it +c sin a
factors bx)
into bx)
+c
a +c
such
standard
that
Solution
Obviously, if Solution Example the Example 4
directly. Solution
Example (ii)
Here, (ii) integrand Here, i.e.,
SolutionExample 7 The (iv) Jf) When Let
(b) Let 5 we log
6 Integral
put
following Some dr x' put
is integrand x=t
Let x
t
f) 1= =1 cosreduced
standard cos o(x). =ax+b=t of ’ equals
=sin Jcost r x)=t
three I=}cos cos3x function a xdr=dt dr is
equals to so di=d Indefinite
CHAPTER
integral
5xdr then and that a
’ equalsdt- log = forms forms dt f function
[f()] 3x convert of
=-sin (t)
dt= cos equals f(ar
are of the d. () d
n+1 cos2r) d 5rdr rc of
+c itegrals
very itform dr function
=dr (provided
n*-)+c b)
+ into
=
useful dr f dt
= standard (ar+ and

to
-o(ar b). in
that
write
+b) interal. case
iniegral
298 CHAPTER Indefinite Integral

Example 8 sec (v) Stundard substitution

d equals
Stan x Following standard subsitution will be usetul
Solution Let = tan x’ dt = sec rdr Integran

:
dt
=+ -2r+c= 2Wtanr tc J-ior va'

Example 9 +e
dr equals I=atan 9 or
I=asinhA
Solution Let e'+e=t
. - a or
’ (e'-e")dr = dt
dt
I=| =logt +c log(e' te ') +c or
Vatx
dx
(vIntegral of the JOrm a sin x+b cos x Vrta+) or Vxlat)
a-x
Put a=rcos0 and b=rsin 0. then r= Va+b and or =asin 9
Va-x
=tan -Ib
a
Vrta-x) or
dx 1
r sin (x+0) Scosec (x +0) dr |-a

Vr-a V

log tan or x(-a) or

Vx(I-a)
log tan at
Va +b I=acos 20
Vat

dx
Example 10 in+ cos x
equals
VB-x
or
A= acos'0+Bsin'
Solution Let l=r cos, 3=rsin e. then
Vr-axB-) (B>a)
r=v3' + =2 and tan =
Example 12 s
1- sinx
d equals
3
Solution I= rl+1- sinsinxA d
cos(x/2) + sin (x/2)
d
cos(x/2) - sin(x/2)
- [tan+
Example 11
dx
Vl+ sinx
equals -Jise-4 =2tan
Solution
d Example 13 dx
Vxta- equals
sin x/2+ cos a/2
Solution Letx = asin 0, hen
dx = 2asin cos 0 de
2asin -cos
=y2 logtan+ vasin 0- acos de =2|de =29 +c =2sin
 CHAPTER Indefinite Integral 299

(vii) (a) Integration by parts (vii) Integration of rational functions

If uand v are the differentiable functions of x. then (a) When denominator can be factorized (Using
partial fractions)
can
If denominator of a rational algebric function
be factorized. then its integral can easily be
Le., Integral of the product of two functions fractions. The
obtained by splitting it into partial
= first function x integral of second function following two standard integrals may be so
[derivative of first) x (Integral of second) ] obtained

How to choose Iand II function If two functions are of dx Llog X-a +c

2a xta
different types take that function as Iwhich comes first in the
word ILATE, where Istands for inverse circular function, L dx
Llog atx
tc
stands for logarithmic function, A stands for algebric functions, 2a a-x

Tstands for trigonometric and E for exponential functions.

dx
For the integration of logarithmic or inverse trigonometric functions Example 17 -2)
alone, take unity (1) as the second function.

Solution
Example 14 Evaluate r'e'dr equals
Solution / = r'e'd =re-f2r:e'á=re -2|xe'-f1e'd) Example 18 J dequals
On taking x as first function, we get 2x
=te'-2.xe' +2e' +c Solution x+ Da +2) dx
(b) If the integral is of the form e[f«)+ f'()] du , then dr = 4log (x + 2) - 2log (x + I) +c

by breaking this integral into two integrals, integrate one integral (r+2)°
= 2log
by parts and keep other integral as it is, by doing so, we get
[e'f()+f)]dr=e'f)+e (b) When denominator cannot be factorized
In this case integral may be in the form
dx pxtq dx
Example 15 Evaluatee' (sin x+cosx)dx, ar + bx +c ax' + bx +c
Solution /=|e' (sin.x+ cosx)dr For first integral we express its denominator in the
This is of the fornn form (x+ a)±ß and use the previous results.
fe'f()+ f' )]dr =e'f(a) + c=e'f() +c For second integral we express its numerator inthe
form Nr =A(derivative of Dr) + B and then we
Here, f(x) = sin r.. l= esinx+ c integrate it easily.
(c) If the integral is of the form |[x f'()+ f(r)] dx , then d

integrate one integral

Example 19 cquals
by breaking this integral into two integrals, du 2 x+1/2
so, we get
by parts and keep other integral as it is, by doing
tan
Solution eW2)' +3/4 J3 /3/2
JIx f() +f(a)] dr =xf (*) +c 2 2x+1

Example 16 Evaluate (x sec*x+ tan x) dx.

Example 20 J i dx equals
Solution Here I= (x sec* x+ tan x) dx
Solutiond-
+x+1 2x+I+1
a+
- [(u f'w +f)] dx , where f(x) =tan x (2r+1
I=xf(x) +c=x-tan x + c
300 CHAPTER Indefinite Integral
Integration of the Trigonometrical
(iN)
Ppowers ofa
ntegral of rational funetions containing only even (xi) (a)
d: dx Functions
atb sin' a cos x+b
Tofind integralof such functions, first we divide numerator
and denominator by x, then express Nr as dr tl/x) and Dr as dx dx
a function of (r+ l/). Following examples illustrate it. a cos +b sin (acos x +b sin x
multiply and divide
Example 21 -dx equals For their integration we by sec'x an
then put tan x = 1.
l+1/x d(r-1/x)
Solution dx
Example 25 Evaluate J3sin'x
tan

secxdx
Solution 2
secx+3 tanx
(dividing Nr and Dr by cos
Example 22 du equals
secx dx
dx + 1/x)
Solutiond= J,(r+l/-2 Ltan-'(2tan x) +c
=4tan'x
-log +/)-5 dx
dx dx
2/2+l/) + 2 (b) asin x +b cos r
a sin x +b a cos x+b
-V2r+1 +c
242 log '+V2r+1 dx
asin x+b cos Xtc
(x) Integration of irrational functions
If any one term in Nr and Dr is irrational, then it is made For such types of integration first we express them in tems
rational by suitable substitution. Also, if integral is of the form of tan x/2 by replacing
dx
Nar' +bx +c du, then we integrate it by 2 tan x/2 1- tan x/2
Nax +bx+c Sin x= and cOS X =
1+ tan /2 1+ tan x/2
expressing ax +bx+c=(1+a) +B.
and then put tan x/2 =t.
Also, for integrals of the fornm
dx
a'r' + b' Example 26 5+4cosx equals
Vax +bx+c
= dr, Ja'r+b)War' +bx +c dx
Solution - dx seex/2 dx
firstwe express a'x+b in the form
d
5+4 1- tan /2) 9+ tanx/2
a'x +b' = A (ax 1+ tan x/2 )
dx
and then proceed as usual with standard forms. =2.-tan -i/tan x/2 |+c
3 3
dx
Example 23 equals
(c) [P Sin xtq cos xd Psinx
J'+2x asin x+b cos x a sin x+b cos x
dx,

dx
Solution =cosh'(x + I) +c q cos x
asin x+b cos x dx,
Example 24 (Nr'+2x dr equals For their integration we first express Nr as folloWs
Solution N(x+I' -1 du Nr=A(Dr) + B(derivative of Dr)
Then, integral = Ax+Blog (Dr) + C
 CHAPTER Indefinite Integral 301

sin x + cOS X
Example 2 2sin x+ 3cosx-dx equals Example 28 dx equals
Solution Let sinx + cos X= A(2sinx+3cos x) + B
(2cos-3sin x)
Solution Here, dx
2A-3B =1 1
3A +2B=1
A=,B=
13 13
J=log(1-e")+c
51 (2sin x+ 3cos x) +c
:l=-log
13
Example 29 (Ve'-1d1 equals

ISome Integrals of Different Expressions of e* Solution Here I=[ve'-1 dr

e
ae" dx dx
-dx (put e =t] Ve-1
b+ ce
Let e-l=', then e'dr =21 dt
(i) dx
dt = 21 -2tan'()+c
(multiplying and divide by e and put e =t)
= 2(ve'-1 -tan-'e'-1)+c
(üi) -dx (multiply and divide by e')

(iv) -dx f'

fx)
fom IStandard Types of Integration
1
1. dx
() dr (multiply and divide by e ) Ja +bx+c)
2. - dx
(vi) da (integrand = tanhx) Vax + bx+c

3. + bx +c dx

(vii) dx (integrand = coth x) Make the coefficient of is unity. After making perfect
-1 square.
1
pxtq
(vifi)
(e'+ej2
dx (integrand =sech'x) 4.
+ bx+c
dx

1
(x) d (integrand=cosechx) 5.
pxtq dx
Vax+bx +c
1
dx
6. (px+ q)var² +bx +c dr
(multiply and divide by e and put e =) d
Put (px tq) = A(-(ax? +bx+c)+B
dx
(xi) -dx (multiply and divide by e ) p(x)
7. dx
ax +bI+c
(xiü) dx (multiply and divide by V2e-2) pr) is a polynomial of second or higher degree in
x(n 2 2)
W2e*-1 Dividing Nr by Dr, the expression is expressed as
(integrand =(|te')/ite') p) f)
follows
ax'. + bx+c + bx+c
(xiv) jve'-i dr (integrand =(e'-1)/Ne-1)
where f ) is linear.
+1
(xv) le+a dx (integrand =(e' +a)/e-a) 8. -dx
J+kx +1
302 CHAPTER Indefinite Integral
-dr
17. bsin
dr
(k ’ constant) and Nr and Dr dividing by x. 18. a+bcosx
r 1
10. dr
(asinx+ bcosx)²
19.
1
*= + I) +(r -1)) 20.
(asinr+ bsin r COSNt C- Cosx)
dr

Nr and Dr divided by cosx and put tan x

and Dr divided by sinx and put cotr=| =t or N
+kr'+1
1
d
I=}(+)--) 21.
nsin x+ bcos x)
(asin
Polar form a= r cos , b=rsin0
12. du
22. -dx
1
l=a'+a')
2a -(r' -a')) Put sinx=
2 tan x/2
1+tanx/2
1
13. d
(ar+ b)Wcr +d 23. d
Put (cr + d) = r
1-tan x/2
Put cosx = and put tan x/2 =|
14. | d 1+ tan x/2
(px+ q)vax +br+c
Put (px + q) =
1 24. | (a sin x+bcosr+c)
dx

Psinx
25. dx
15. dr a sin x+b cos x
(ar' +bx +cpx+ q q sin x
26. dx
Put (2x + q) =r asinx+bcos x
16. 27.
Psin x+qcos x+ r dx
dx
(Ar+ B)ycr +D asin x+bcosx+c

Put (r = 1/) d
Nr= A(Dr) + B (Dr) +C
dx