Integral Calculus

Differrential Calculus
-

given the function where are to find for the differential or derivative.

Integral Calculus
-

given the derivative of a function where are to find for the required
function.

du

x 2 dx dx ( u2 ) =nua1 dx
d ( x3 ) =3 x 2 dx
1
2
2
d ( x ) = ( 3 ) x dx
3
x 2 dx
d

( 13 x +c )=x dx
2

1 3
2
x dx= x + c
3
*Two function having the same derivative differ only by constant.
y=x 35
'

y =3 x

y=x 3 +1000
y=3 x

Where: c = arbitrary constant.

Indefinite integral

f ( x ) dx

*Since the process of integration is not completely determine because of an

addition arbitrary constant.

PROPERTIES OF INDEFINITE INTEGRAL

dx=u+c

definite of integral

dw=w + z

dx= x+ z
*Indefinite of the variable of integration.
2.)

(dw+ dx+ dx)dx= dw+ dx+ dx

* if it contains some of term that can be integrate separately.

x5
(

Ex.

1 3
x +4 x 4 +4 )dx= x 5 dx 1 x3 dx+ 4 x 4 dx +11 dx
7
7

GENERAL POWER FORMULA OF INTEGRATIONS

d ( un )=nun1 du

from:

n n+ 1
d ( uu+1 ) =( n+1 ) un+11 du

u du=
n

un du=

n+1

d(u )
n+1

uu+ 1
+c
n+1

Examples:

un+ 1
x dx u du=
+c
n+1
2

u=x ; n=2
du=dx

x 2+1
x3
+c= +c
2+1
3

un+1
(43 x) dx u du= n+1 +c
5

u=43 x , n=5
du=3 dx

unwanted factor

Thus,

( 13 ) ( 43 x ) (3) dx
5

43 x

1

3
6

(43 x)
+c
18
2

( a2x 4 ) dx
=

( a42a 2 x 4 + x 8 ) dx
5

x
4
2 x
= a x2 a 5 + 9 +c

1+u
1+u
1+u2
( 2)

( 2) du
ln 3
+c
u ln 2
ln

3
4.)
=
=
1+u2
1
1

2
2
ln 3 (1+u)2
+c
=
6
Fundamental Integration Formulas
1.)General Power Formula

un du=

n+ 1

u
+c
n+1

2.)Logarithmic Function
a.)

du
u = ln u + c w/r: u>0

b.)

du
=ln (u ) +c w /r :u< 0
u

ln
c.) du = u w :u is not equal zero
u +c r
Ex.)
1
7
+
3
3 w6
=
w/r: u=3w-6 ; du=3dw

w+ 5
1.) 3 w6

(3)

( )
1
dw
3
= 3 dw+7 3 w6

1
7
w+
3
3

2.)

cos d
sin
2
5

ln (3w-6) + c
1
= 2 ln(sin2) + c

w/r: u= sin2,du=cos2 2 d
cos u
du w/r: u = sin u,du = cos u du
sin u

3. cot u du=

= ln sin u + c
5 xdx
1
4.) 1+ 2 x 2 = 5( 4 )

4 xdx
1+ 2 x 2 = 54 ln

3.)Exponential Function
a.)

eu

du=

e u+ c
4

b.)

a u du= lna(a)

+c

w/r: a>0

Ex.)
1
1. ( ) esin 3 x cos 3x dx
3
3

Let: u=sin 3x; du =(3)cos3xdx

1
e sin3 x +c
3
dx

2.)

105 x

=2

ln x dx

1
=( 5 )

1 105 x
5x
10
dx=
(
)+ c

5 ln10
( 5)

x
= 2x

+c
3

4.)

43 lnx dx=4 e lnx dx=4 x 3 dx

x4
4
= 4 4 + c= x + c

1+2 x

)+c

4.) Trigonometric Function

a.) cos u du=sin u du
b.) sin u du= -cos u du
2

c.) sec udu=tan u du

2
d.) csc udu=cot u du

e.) sec u tan u du=sec u+ c

f.) csc u cot u du=-csc u + c
g.) cot u du = ln u + c
h.) sec u du =ln(sec u + tan u) + c
i.) csc du = ln(csc u cot u) + c
5.) Inverse Trigonometric Function
a.)
b.)

du
u
2
2 = Arcsin
a
a u
du

a2 +u2

+c

1
u
= a Arctan a + c

Ex.
1.

dx
9x 2

(let:

a =9, a=3, u =x ,u=x )

x
=Arcsin 3 +c
2

x +3 2 +( 6)

dx
2.)

dx
dx
x2 +6 x +15 = x2 +6 x +9+6 =
1
= 6

x +3
6
6
. 6 Arctan 6 . 6 + 6

6 arctan ( x+3 ) 6 +c
6

e2 x dx
3.) 44 e 4 x
2

2x

2x

2x

let: a =4, a=2 u =4 e ,u=2 e du=4 e dx

e 2 x dx( 4)

44 e 4 x

2x
1
2e
= 4 Arcsin 2 + c

1
2x
= 4 arcsin e +c

4.)

( 3 ) dy
dy
1
=
2
9 y + 6 y+ 3 3 ( 3 y +1 ) +2
2

1 1 2
3 y +1 2
{ .
.
+c
}Arctan
3 2 2
2 2

2 A
6

rctan

(3 y +1) 2
+c
2

6.)Integration By Parts

udv=uv vdu
Ex.)
=

cos 5 u sin2 u d

cos 4 u sin2 u cos 4 du

u
1sin

u
(
2

2
cos 2=
12 sin
sin
( 2u) cos u du
=
( 2 u+sin 4 u )

sin2 u cos u du2 sin 4 u cos u du+ sin5

w/r : u =sin u,du = cos u du

sin3 u 2 sin 5 u sin 7 u

+
=
3
5
7
2.)

+c

cot 4 x dx

xx
csc

( x1) dx
= csc 2 dx

2
cot x

2
Let: u =cot x,du = csc xdx

cot 3 x
csc 2 x dx + dx
=
3
cot 3 x
+ cotx+ x+ c
=
3
3.)

tan4 y sec2 y dy

Let: u= tan y,n=4;du=

sec 2 y dy

u cos u du

tan 5 y
=
5
4.)

+c

y ln y dy

dy
Let: u= ln y, du= y ,

y2
dv = ydy , v= y

y 2 ln y 1
ydy
=
2
2

5.)

y2
y2
+c
2 - 4

xe x dx ( let u=x , du=dx , v=e x )

x
x
= xe e dx

x
x
= xe e +c

Integration By Substitution
a.)Algebraic Substitution

e x 9

dx

>Change the variable of integration that seems to be causing trouble in the

process of integration by way of explicit substation.
yz
dydz
>Perform the integration using the appropriate method
>Return back to the original variable in expressing the final answer.
Ex.)
1.)

e x9

dx

2 udu
2
u +9
= u( )

= 2

u u+99
2
+9

(u2 +9)du
du
=2 (u 2+ 9) 18 u2+ 9

=2

du18

du
u2 +9

1
u
=2u-18( 3 arctan 3 + c

=2

e x 96 arctan e 39

2.)

1+ z

+c

dz

y
y (4 y)(1) dy
=

4
2
= 4 y dy4 y dy

4 y5
= 5

4 y3
- 3 +6s

5
4
4
= 5 (1 + z 2 - 3 (1+

2
3

z + c

b.)Trigonometric Substitution

 1+ z

dz

*if the integrand involves,

2
2
> a + x use : x = a tan

Ex.)
a

1.) dx

2+a tan

=a

2
sec d

a
cosd
= a3
2

2 2

1+ tan

3
2

sec

3
2

2 3

x
= a 2 a2 + x 2 + c

> a x us e : x=a sin

2
2
> x a use : x=a sec

Integration Of Rational Fraction

>quotient of two polynomials

>the numerator is of lower degree than the denominator

>the denominator is broken into real factors of partial fractions
*Distinct Linear Factors
*Repeated Linear Factors
*Quadratic Factors
DEFINITE INTEGRALS
f ( x ) dx definite integral withthe limits
b

a b
a

a=lower limits
b= upper limits
Ex.)

1.)

1 2 1
( x 3 x + + 11)dx
2
3
4

2++11 x 42
4
3
2
=[ x 1 x + 1 x
4 2 3 3

=[

1
(2)
4

TRANSFORMATION BY TRIGONOMETRIC FORMULAS

Type I:

sin m y cos n xdx either n or m is a positive add integer

If n is add
n1
u
sin cos xsinxdx reduce
power formla
Use =

sin x=1sin x

If m is add

cos n1 sinm xcosxdx reduce

power formla
Use =

cos 2 x=1sin 2 x

tann xdx cotn xdx where n is aninteger

Type II:
2

use=tan x=sec x1
2

cot x=csc x 1

Type III:

tanm x sec n xdx cot m x csc n xdx where m is a positive even integer

use=sec 2 x=1 tan 2 x

csc 2 x=1+ cot 2 x
Reduce to power formula
Type IV:

Use =

sin m x cosn xdx where n or m are botheven

1
sin 2 x= (1cos 2 x )
2

1
cos 2 x= ( 1+cos 2 x )
2
sin 2 x=2 sinx cos x

cos 2 x=cos 2 xsin 2 x

2 cos2 x 1
12 sin 2 x

INTEGRATION BY SUBSTITUTION
A Algebraic Substitution
B Trigonometric Substitution
ALGEBRAIC SUBSTITUTION
change the variable of integration that seems to be causing trouble in
the process of integration by way of explicit substitution.
yz

zx

dydzdzdxdwdy

wy

> Perform the integration using the appropriate method.

> Perform back to the original variable in expressing the final answer.
NOTE: If the expressing contain an exponent equals to
exponent in the substitution of the variable.

Example:
1

e xg dx

u e x g

2 udu
u2 + g
u[]

u2=e x g

(u 2+ gg) du
= 2
u2 + g

u + g=e

(u 2+ g)du
du
= 2 ( u2+ g) 18 u2+ g

2udu=e x dx

2 du18
2

= u u

du
2
u +g

dx=

a =g

= uu

= 2u-18

1
u
( ) arctan +3
3
3

a=3

2 e x g6 arctan e 3g +3
x

Answer =

TRIGONOMETRIC SUBSTITUTION
If the integral invovles,

a x x=asin

2udu
x
e

1
2 , include the

 a2x 2
x
=sin
a

a2x 2
cos=
a

cot=

a2x 2
a

If the integrated involves,

csc=

a2 + x 2 x=atan

a2x 2
a

sin2
=

= 2sin cos
x
a
2 2 2 2 2
a +x a +x

2 ax
= a2 + x 2
If the integrated involves,
x
=sec
a
x
=cos
a
sin

x 2 +a2
x

Example:
1 In the integral

we may write

a2x 2 x=asec

so that the integral becomes

2 In the integral

we may use

INTEGRATION BY RATIONAL FRACTIONS

Quotient of two polynomials.
Two degree of numerator must be of lower degree the degree of the

denominator.
If the denominator contains a quadratic expression according to this
2
form: a x + bx+ c
b2=4 ac

a discriminant to determine behavior of the roots.

b24 ac <0

If

roots are imaginary and unequal.

That count for factored without distinct lines factored.

Can also be integrated by still.
3 FORMS of PARTIAL FRACTION
A Distinct Linear Factors
B Repeated Factors
C Quadratic Factors

a n> leads to power formula

b n> leads to power formula

If n=1 leads tologarithmic

du
= u =luu+ c

A
a x +bx +c arctangent

A
(a x + bx+ c)

du

a2 +u2

1
2

arctan

u
a

+c

n 1 lead by trigonometric substitution.

By completing the square

DISTINCT LINEAR FACTORS
dx

Adx

Bdx

Cdx

(xa)(x b)( xc ) = xa + xb + xc
=

dx
dx
dx
+
+
xa xb xc

Logarithmic Integration Formula

Various Method of Solving the Constant: ABC

a Taking the original zero of the denominator as the assumption for the
values of the variable.
b Equate the coeff. of some power of x.

c Using the formula:

A=

f ( a)
g( a)

f(a)=function of the numerator, g(a)=other factors of the

denominator
Example:
1

(3 y 24 y6)dy
( y 1)( y 3)( y+ 1) Distinct Linear factor
dy
dy
dy
a
+
b
+
c

=
y1
y3
y +1
2

(3 y 4 y 6)dy
= ( y1)( y3)( y +1)

2
= 3 y 4 y6

y1 + y 3 + y +1

= (y-1)(y-3)(y+1)

= A(y-2)(y+1)+B(y-1)(y+1)+C(y-1)(y-2) 1.

y=1, y=2, y=-1

say y = 1:
3(1)-4(1)-6 = A(1-2) (-1+1)
-7 - -2a A =

7
2

say y = 2:

say y = -1:

3(2)-4(2)-6 = B(2-1) (2+1)

3(-1)-4(-1)-6 = C(-1-1) (-1-2)

12-14 = 3B B =

2
3

3+4-6 = C (-2)(-3)
1=6C C =

Using method B:
2
= 3 y 4 y6 = A(y-2)(y+1)+B(y-1)(y+1)+C(y-1)(y-2)

1
6

2
= 3 y 4 y6

y2

2
2
2
= A( y y-2)+B( y 1 )+C( y 3 y +4 )

: 3 = A+B+C 1

y : -4 = -A 3C 2
C : -6 = -2A B +2C 3
Equate: 1 and 2

Equate: 4 and 2

3= A+ B+C
6=2 AB+2 C
3=A 3 C

3=A +3 C
4=A3C
7=2 A

A =

7
2

From 4:
From 1:

3= A+3 C
7
3= +3 C
2
7
3 C=3+
2

3C =

3C =

3= A+ B+C
7
1
3= B+
2
6
3 7 1
B=
1 2 6

6+7
2
1
2

C=

1
6

18211
6

4
6

2
3

Plane Areas
*enclosed by the given boundaries (a curve or a line)

A= yi x
i=1

Y 1 x +Y 2 x + Y n x
x 0
yvariable

dA=ydx
a

ydx= f ( x ) dx
o

A=

Examples:
1 Find the area bounded by the parabola
y 2 =4ax

w/v(0,0)

Soln:

12=4a

; a=3

LR=12

; 2a=6

dA=Area of Rectangle (A=LxW/Thickness)

dA= ydx
dA=(3-x)dy

y
=12x x= 12

dA=2
6

y 2 =12x of the line x=3

=12(3)

( 3x ) dy
0

y2
3
dy
dA=2
12
0

=36

dA=2

+ 6

Y=

[ ( )]
1 y3
3 y
2 3

[()

36

dA=2

1
(6)3
36

dA=2 [ (86) ]
y=k

horizontal

x=k

vertical

dA=24s.u

ans.

x=3

contd
dA=ydx
3

A=2

ydx
0
3

A=2

2 3 x dx
0

A=4

A=4

3 x dx
0

3 3
2

[]
x
3
2

[ ]
3

8 3
2
A= 3 (3)
1

8 2 2
A= 3 (3) (3)

8 2
A= 3 (3)
A=24s.u ans.

2 Find the area bounded by the curve

x 2=3 y

and the line x-y=0.

y y 2 ) dx
dA= ( H
3

A=

( y H y 2 ) dx
0

A=

A=

x2
dx
3

3 3

x 1x

2 3 3

1 2 1 3
A= 2 ( 3 ) 9 ( 3 )
9
A= 2 -3
3
A= 2 s.
x 2=3 y
x 2=3 x
x=3

x 2=3 y
y=

x2
3

Therefore:
y=3

SOLIDS OF REVOLUTION
A solid is generated when an area is revolved about any reference axis
or line.
CIRCULAR DISK

CYLINDRICAL SHELL

dv=2 y xdy
b

V =2 yxdy
0

dv=2 ( b y ) xdy
b

V =2 ( b y ) xdy
0

CIRCULAR RING

y2
y

dv=

y2

V =
0

Examples:
1 Find the volume of a sphere.

(V = 34 r )
3

[ ]

V =2 r 3

dv= x dy

V =( 2 ) ( x 2 ) dy

V =2

r
0

[ ]
2 3
r
3

4 3
V = r cu .units
3

V =2 ( r 2 y 2 ) dy

3 r

y
V =2 r y
3
2

2 Find the volume of revolving the area bounded by the curve

and the line

y=3 x

r3
3

about the x-axis, y=axis & the line

y 2=6 x

y=2 .

To get the point of

intersection
2

y=3 x y =6 x
2

y =6 x
y 2=6
y
x= xR
3

( 3y )

y=3 x

y ( y )=2 y
2

y =6 x
y=0

x=0

y=2

y 2
x= =
3 3

x=

y
xL
6

dV =2 y ( xRxL ) (dy )
2

V =2 y ( xRxL ) dy
0

y y
V =2 y
dy
3 6
0

[ ( )( )]

1 y3 1
V =2

3 3 6

( 2 ) 3 ( 2 )4
V =2

9
24

y4
4

]
V =2

V=

V=

6448
72

( 16 )
36

4
cu .units
9