CHAPTER 3 INTEGRATION

DBM 20023 – ENGINEERING MATHEMATICS 2

TABLE OF CONTENT

No. Topic Page

3.1 Explain Indefinite Integrals 3 - 14
15 – 22
3.2 Calculate Definite lntegrals 23 - 29
30 – 35
3.3 Apply lntegrals of Trigonometric 36 – 39
Function 40 – 48
49 – 63
3.4 Apply lntegrals of Reciprocal 64 – 87
Function

3.5 Solve lntegrals of Exponential
Function

3.6 Solve Integration by Parts

3.7 Solve Integration of Partial
Fraction

3.8 Apply the Techniques of
Integration

Page 1

DBM 20023 – ENGINEERING MATHEMATICS 2

3.0 INTEGRATION

Learning Outcomes:
At the end of this topic student should be able to:

 Perform indefinite integrals of algebraic functions
 Perform definite integrals
 Perform integral of trigonometric, reciprocal and exponential

functions
 Perform integration by part
 Perform integration of partial fraction
 Apply the techniques of integration to find area and volume

Page 2

DBM 20023 – ENGINEERING MATHEMATICS 2

Introduction

Definition: When 2 differentiated with respect to , the derivatives is 2 which

is ( 2) = 2 . Otherwise, given that the derivatives of a function is 2 , it is

clear that the function maybe 2.

If = ( ) , so = ′( ), then the integration of ∫ ′( ) = ( )


Integration is often introduced as the reverse process of differentiation.

3.1 Indefinite Integration of Algebraic
Functions

Definition: Indefinite Integrals is an integral expressed without limits, and so containing
an arbitrary constant.

Page 3

3.1.1 DBM 20023 – ENGINEERING MATHEMATICS 2

Indefinite Integration of Algebraic Functions

a) Integration of a Constant

If Then

= ∫ = +

Example: Example:

∫ 4 ∫ 9
2
∫ 4
= 4 + ∫ 9
2

= 9 +
2

Page 4

DBM 20023 – ENGINEERING MATHEMATICS 2
EXERCISE:
Find the following integrals:

a ) ∫ 0 b ) ∫ −15

[Ans: ] [Ans: −15 + ]

c ) ∫ 0.25 d ) ∫ 5
4

[Ans: 0.25 + ] [Ans: 5 + ]
4
Page 5

DBM 20023 – ENGINEERING MATHEMATICS 2

b)Integration of Algebraic Function

If Then
=
= ∫ = +1 +
+ 1

∫ = +1 +
+ 1

Example: Example:

∫ 6 ∫ 2 4

= 6+1 + 2 4+1
6+1 4+1
= +
7
= 7 + 2 5
5
= +

Page 6

DBM 20023 – ENGINEERING MATHEMATICS 2

Example: (use substitution method)

∫ 5 1 1
+

= 5 + 1

= 5 , =
5

∫ 1 1 = ∫ 1
5 + 5

= 1 ∫ 1
5

= 1 ln +
5

= 1 ln|5 + 1| +
5

Page 7

DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:

Find the following integrals: b ) ∫ 4 2
a ) ∫ 9 3

[Ans: 10 + ] [Ans: 4 3 + ]
10 9

c ) ∫ −3 d ) ∫ 1
√

[Ans: − 1 + ] [Ans: 2√ + ]
2 2

Page 8

DBM 20023 – ENGINEERING MATHEMATICS 2

e) ∫ 1 f) ∫ 1
(3−5 )4 5 +7

(use substitution method) (use substitution method)

[Ans: 15(3 1 5 )3 + ] [Ans: 1 ln|5 + 7| + ]
− 5

Page 9

DBM 20023 – ENGINEERING MATHEMATICS 2

c) Integration of a function involving addition and
subtraction

Addition of If Then
Integration = ( ) + ( ) ∫[ ( ) + ( )] = ∫ ( ) + ∫ ( )
= ( ) − ( ) ∫[ ( ) − ( )] = ∫ ( ) − ∫ ( )
Subtraction of
Integration

Example: Example:

∫(6 3 + 2) ∫(2 + 3)(3 − 1)

= 6 3+1 + 2 + = ∫(6 2 − 2 + 9 − 3)
3+1
= ∫(6 2 + 7 − 3)

= 6 4 + 2 + = 6 3 + 7 2 − 3 +
4 3 2

= 3 4 + 2 +
2

Page
10

DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:

Find the following integrals: b ) ∫(8 2 − 4 + 4)
a ) ∫( 2 + 6 − 1)

[Ans: 3 + 3 2 − + ] [Ans: 8 3 − 2 2 + 4 + ]
3 3

c ) ∫ 3 2−6 d ) ∫ (4 − 2)
3

[Ans: 3 − 2 + ] [Ans: 2 2 − 4 + ]
3 4

Page
11

DBM 20023 – ENGINEERING MATHEMATICS 2

d)Integration of Extended Power Rule

If Then
= ( + )
∫( + ) = ( + ) +1 +
( + 1)( )

Where and are constant and is an integer.

Example: Example:

∫(2 − 3)5 ∫(6 + 1)−2

= (2 − 3)5+1 + = (6 + 1)−2+1 +
(5 + 1)(2) (−2 + 1)(6)

(2 − 3)6 = (6 + 1)−1 +
12 −6
= +

= − 1 1) +
6(6 +

Page
12

DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:

Find the following integrals: b ) ∫ 3(5 + 3)2
a ) ∫( + 2)3

[Ans: ( + 2)4 + ] [Ans: (5 +3)3 + ]
4 5

∫ −2 d ) ∫
(1+3 )2 (3 +5)4
c )

[Ans: 3(1 2 3 ) + ]
+ 9(3 +5)3
[Ans: − + ]

Page
13

e ) ∫(2 − )7 DBM 20023 – ENGINEERING MATHEMATICS 2
f ) ∫(3 − 6 )−4

[Ans: − (2 − )8 + ]
8
[Ans: 1 + ]
18(3 − 6 )3

g ) ∫(1 − )−21 h ) ∫ 8(3 − 2)3

[Ans: −2√(1 − ) + ] [Ans: 2(3 − 2)4 + ]
3
Page
14

DBM 20023 – ENGINEERING MATHEMATICS 2

3.2 Definite lntegrals

Learning Outcomes:
At the end of this topic student should be able to:

 Solve definite integrals

3.2.1 Definition

Definition: A Definite Integrals has start and end values; in other words there is an
interval ( to ).

I f ∫ ( ) = ( ) + , T h e n :



∫ ( ) = ( ) − ( )



Page
15

DBM 20023 – ENGINEERING MATHEMATICS 2

3.2.2 Properties of Definite Integral



a. ∫ ( ) = − ∫ ( )





b. ∫ ( ) = ∫ ( ) where is a constant





c. ∫ ( ) + ∫ ( ) = ∫ ( )





d. ∫[ ( ) + ( )] = ∫ ( ) + ∫ ( )



Example: Example:

∫02( + 4)2 ∫12( 4 + 5 )

= [( +34)3] 2 = [ 55 + 5 2 ] 2
0 2 1

= ((2+34)3) − ((0+34)3) = [((25)5 + 5(22)2) − ((15)5 + 5(21)2)]

= 72 − 64 = 82 − 27
3 5 10

= 152 = 137
3 10

Page
16

DBM 20023 – ENGINEERING MATHEMATICS 2

Example:

∫13( + 9)2 ( u s e s u b s t i t u t i o n m e t h o d )

= + 9

= 1 , =


Write the limit on using the substitution

= + 9 when:

= 1 , = 1 + 9 = 10 (lower limit)

= 3 , = 3 + 9 = 12 (upper limit)

So, we required:

12 [ 3 3]1102

∫ 2 =

10

= 1 [123 − 103]
3

= 782
3

Page
17

DBM 20023 – ENGINEERING MATHEMATICS 2

Example:

G i v e n t h a t ∫24 ( ) = 3, f i n d :
a ) ∫24 5 ( )
b ) ∫24 [7 ( ) − 1]

a ) ∫24 5 ( ) b ) ∫24 [7 ( ) − 1]
= 5 ∫24 ( )
= 5(3) = 7 ∫24 ( ) − ∫24 1
= 15
= 7(3) − [ ] 4
2

= 21 − (4 − 2)

= 19

Example:

Given that ∫12 ( ) = 3 ∫23 ( ) = 9 , find the following:
2 2

a ) ∫13 ( )

b ) ∫21 ( ) − ∫23 ( )

Page
18

DBM 20023 – ENGINEERING MATHEMATICS 2

a ) ∫13 ( ) b ) ∫21 ( ) − ∫23 ( )

= ∫12 ( ) + ∫23 ( ) = − ∫12 ( ) − ∫23 ( )

= 3 + 9 = − 3 − 9
2 2 2 2

=6 = −6

EXERCISE: b ) ∫12(3 2 + 4 − 2 )
Find the following integrals:
a ) ∫23(2 2 − 3 )

[Ans: 361] [Ans: 11]

Page
19

DBM 20023 – ENGINEERING MATHEMATICS 2

c ) ∫−11(2 + 3 ) d ) ∫01
(5 +1)3

[Ans: 6] [Ans: 772]
[Ans: −5]
e ) G i v e n t h a t ∫02 ( ) = −3. F i n d t h e f o l l o w i n g :

i) i) ∫02 3 ( ) i i ) i i ) ∫02 [3 ( ) + 2 ]
4

iii)

iv)

v)

vi)

vii)

viii)

ix)

[Ans: −49]

Page
20

DBM 20023 – ENGINEERING MATHEMATICS 2

x ) f ) F i n d t h e v a l u e o f t h e f o l l o w i n g i f ∫−12 ( ) = −1 and ∫23 ( ) = 5

i ) 2 [∫23 ( ) − ∫−12 ( ) ] xi) i i ) 6 ∫−12 ( ) + ∫23 ( )

[Ans: 12] [Ans: −1]

Page
21

DBM 20023 – ENGINEERING MATHEMATICS 2

iii) ∫01 2 xii) i v ) ∫13( + 13)3
(2 +5)

(use substitution method) (use substitution method)

[Ans: 0.336] [Ans: 6780]

Page
22

DBM 20023 – ENGINEERING MATHEMATICS 2

3.3 lntegrals of Trigonometric Function

Learning Outcomes:
At the end of this topic student should be able to:

 Solve integrals of Trigonometric functions

Integration of Trigonometric functions can be simplified as
below:

∫ = − +

∫ = +
∫ = +
∫ = − +

∫ = +

∫ = − +
∫
∫ = − +
∫

= +


= +


Page
23

Example: DBM 20023 – ENGINEERING MATHEMATICS 2

∫ 2 sin Example:

= 2 ∫ sin ∫ 8 2 (1 − 4 )
= 2 (−cos ) +
= −2 cos + = 8 ∫ 2 (1 − 4 )
= 8 [tan(−14−4 )] +
= −2 tan(1 − 4 ) +

Example: Example:

∫ cos (7 + 12)
6 (use substitution method)

∫ cos 3

0

= 7 + 12

= [sin33 ]0 6 = 7 , =
7

= 1 [sin ∫ cos(7 + 12) = ∫ cos ∙
3 7
3 ]06
1
= 1 [sin 3 ( 6 ) − sin 0] = 7 ∫ cos
3
1
= 1 [1 − 0] = 7 [sin ]+
3
1
= 1 = 7 sin[7 + 12] +
3

Page
24

DBM 20023 – ENGINEERING MATHEMATICS 2
EXERCISE:
Find the following integrals:

a) ∫ [4 cos 2 + 9 sin 1 ] b ) ∫[1 + cos 9 ]
3

1 : + sin 9 +
3 9
: 2 sin 2 − 27 cos +

Page
25

DBM 20023 – ENGINEERING MATHEMATICS 2

c ) ∫ sin (73 ) d) ∫ 2 3
5

: −3 cos (73 ) + : 1 tan 3 +
7 15

Page
26

DBM 20023 – ENGINEERING MATHEMATICS 2

i ) ∫02 [1 + sin ]

j ) ∫04 [ 2 ]

: +1 : 1
2

Page
27

DBM 20023 – ENGINEERING MATHEMATICS 2

k ) ∫ 3 cot(5 ) l ) ∫ 5

(use substitution method) (use substitution method)

: 3 ln |sin 5 | + : 1 6 +
5 6

Page
28

DBM 20023 – ENGINEERING MATHEMATICS 2

m ) ∫ sin(3 + 7) n) ∫0 ⁄2 cos(1 − )
(use substitution method)
(use substitution method)

[Ans: −cos(3 +7) + ] : 1.382
3

Page
29

DBM 20023 – ENGINEERING MATHEMATICS 2

3.4 lntegrals of Reciprocal Function

Learning Outcomes:
At the end of this topic student should be able to:

 Perform integrals of Reciprocal functions by using formulae and
substitution method.

Integration of Reciprocal functions.

Reciprocal Function Integration

= ∫ = | | +


= ∫ = 1 | | +


= ∫ = | | +


= ∫ 1 = 1 | + | +
+ +


= ′( ) ∫ ′( ) = | ( )| +
( ) ( )

Page
30

DBM 20023 – ENGINEERING MATHEMATICS 2

Example: Example:

∫ ∫


∫


∫ = ∫


= | | + = | | +


Example: Example:

∫ ∫ 6 + 1
+ 3 2 +


( ) = 3 2 +

′( ) = 6 + 1

∫ ∫ 6 +1
+ 3 2+

= | + | + = ∫ ′( )
( )

= ln| ( )| +

= ln|3 2 + | +

Page
31

DBM 20023 – ENGINEERING MATHEMATICS 2

Example: Example:

∫ + 2 x dx
+
1 1 4x2

(use substitution u 1 4x2
method) du  8x
dx
u 3x2  x dx  du
du  6x 1 8x
dx
dx  du  2 x  du
6x 1 1 u 8x

  6x  1  du  1 2 1 du
u 6x  81u
1

  1  du  1 ln u2
u 8 1

 ln u  c  1 ln 2  ln1

8

 ln 3x2  x  c = 0.0866

Page
32

DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:

Find the following reciprocal integrals:

a ) ∫ 1 b) ∫ 1
6

[ : | | + ] [ : 1 | | + ]
6

c) ∫ 3 d) ∫ 1
2 5 +7

[ : 3 | | + ] [ : 1 |5 + 7| + ]
2 5

e) ∫ 4 f) ∫ 8 −3 ∙
2−3 4 2−3

[ : − 4 |2 − 3 | + ] [ : |4 2 − 3 | + ]
3

Page
33

DBM 20023 – ENGINEERING MATHEMATICS 2

g) ∫13 2 ∙ h ) ∫23 1 ∙
9 9−2

[ : 0.2441] [ : 0.2554]

i) ∫02 2 ∙ j) ∫−21 4 ∙
3 +1 2 2+1

[ : 1.2973] [ : 1.0986]

Page
34

DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:

Find the following reciprocal integrals by using substitution
method:

a)  2x dx b)  x 2x  5 1 dx
x2  2  5x 
3

 Ans : ln x2  3  c [ Ans : ln x2  5x 1  c ]

c)  x 1 dx d)  6x 3 dx
 4x x2 
2x 2  1

[ Ans : 1 ln 2x2  4x 1  c ] [ Ans : 3ln x2  3  c
4

Page
35

DBM 20023 – ENGINEERING MATHEMATICS 2

3.5 lntegrals of Exponential Function

Learning Outcomes:
At the end of this topic student should be able to:

 Perform integrals of indefinite Exponential functions by using
formulae and substitution method.

Integration of Exponential functions can be find using
formulae.

Exponential Function Integration
=
= ∫ = +

= ( + ) ∫ = 1 +


∫ ( + ) = 1 ( + ) +


Page
36

Example: DBM 20023 – ENGINEERING MATHEMATICS 2

∫ Example:

∫

∫ ∫
= ∫
= +
= +
Example:
Example:
∫
∫ 2−3

∫ ∫ 2−3

= + = ∫ 2−3


= − 1 2−3 +
3

Page
37

DBM 20023 – ENGINEERING MATHEMATICS 2

EXERCISE:

Find the following exponential integrals:

a ) ∫ ∙ b ) ∫ 4 ∙

[ : + ] [ : 4 + ]

c ) ∫ 2 5 ∙ d ) ∫ 6−3 ∙

[ : 2 5 + ] [ : − 1 6−3 + ]
5 3

e ) ∫ 2 (3 + 5 ) ∙ f) ∫ 3 +3 ∙
3

[ : 3 2 + 1 7 + ] [ : − −3 + ]
2 7

Page
38

DBM 20023 – ENGINEERING MATHEMATICS 2

g ) ex ; u 1 ex h ) 2e 2 x
1  e x dx dx
1 e2x

[ Ans :ln1 ex  c ] [ Ans : ln 1 e2x  c ]

Page
39

DBM 20023 – ENGINEERING MATHEMATICS 2

3.6 lntegration by Parts

Definition: The method of change of variable for integration

Introduction

Integration by parts is a technique for performing indefinite integration ∫ or definite
integration ∫ by expanding the differential of a product of functions ( , ) and
expressing the original integral in terms of a known integral ∫ . A single integration by parts

starts with :

( , ) = +

and integrate both sides,
∫ ( , ) = = ∫ + ∫

Rearranging gives

∫ = − ∫

Examples:
∫ 2 sin 3 , ∫ ln , ∫

Therefore the formula of part by part method is:

∫ = − ∫

Page
40

DBM 20023 – ENGINEERING MATHEMATICS 2
G u i d e l i n e s f o r t h e c h o i c e o f u a n d dv :

LIATEWe used word which is:

L – logarithmic functions
I – Inverse trigonometry functions
A – Algebraic functions
T – Trigonometry functions
E – Exponential functions

This word LIATE is arrangement to choose u which is a function to

derivatives first.

Page
41

Example: DBM 20023 – ENGINEERING MATHEMATICS 2

I n t e g r a t e ∫ Example:

I n t e g r a t e ∫(2 − 5)

L e t = a n d = L e t = 2 − 5 a n d =
∫ = ∫
= 1 ∫ = ∫ = 2 =


= 2

= = So using

So using ∫ = − ∫
∫ = − ∫
∫(2 − 5) = (2 − 5) − ∫ 2
∫ = − ∫ = (2 − 5) − 2 ∫
= − + = (2 − 5) − 2 +
= ( − ) +
Example:
So using
I n t e g r a t e ∫ cos
∫ = − ∫
L e t = = cos
∫ = ∫ cos ∫ = sin − ∫ sin
= 1
= sin − (− cos ) +
= + +

= = sin

Page
42

DBM 20023 – ENGINEERING MATHEMATICS 2

Example:

I n t e g r a t e ∫ 4 ln

L e t = ln = 4

= 1 ∫ = ∫ 4


= 1 . = 5
5

So using

∫ = − ∫

∫ 4 ln = ln [ 5 5] − ∫ 5 . 1
5

= 5 ln − ∫ 4
5 5

= 5 ln − 1 ∫ 4
5 5

= 5 ln − 1 [ 5 5] +
5 5

= − +


Page
43

DBM 20023 – ENGINEERING MATHEMATICS 2
EXERCISE:
Find the following integrals using Integration by Parts:

a ) ∫ 2 ln

[Ans: 3 ln − 3 + ]
3 9

b ) ∫ 2

[Ans: tan + ln| | + ]

Page
44

DBM 20023 – ENGINEERING MATHEMATICS 2

1 . ∫ ln( )

[Ans: ln − + ]

2 . sin

[Ans: − cos + sin + ]

Page
45

DBM 20023 – ENGINEERING MATHEMATICS 2

3 . ∫ 2

[Ans: 2 − 2 + ]
2 4

4 . ∫(2 + 3)

[Ans: (2 + 3) − 2 + ]

Page
46

DBM 20023 – ENGINEERING MATHEMATICS 2

5 . ∫ 2 cos

[Ans: 2 sin + 2 cos − 2 sin + ]

h )  ln x2 dx

   Ans : x ln x2  2ln x  2  c

Page
47

DBM 20023 – ENGINEERING MATHEMATICS 2

i)  2 ln x dx
x2

 2 ln x  2  c
 x x

j ) ∫ 3 2

3 2 3 [ 2 2 2 2
2 2 2 2 4
[Ans: − − − ] + ]

Page
48

DBM 20023 – ENGINEERING MATHEMATICS 2

3.7 Integration of Partial Fraction
Introduction

This technique is used to integrate rational functions which cannot be solved using integration
techniques learnt so far.

Integration of Proper Fraction

Any algebraic fraction is said to be proper rational algebraic if degree of numerator is less than
the degree of denominator.

( ) < ( ) → ( )
( )

To integrate a proper rational function, express the function in partial fraction form.

Examples:

1. x x 1  A  B (Linear factor)
x 1 x3
1x  3

2. x 1  A B  C (Repeated linear factor)
x 1 x2
x 1x  22 x  22

Page
49

DBM 20023 – ENGINEERING MATHEMATICS 2

3.7.1 LINEAR FACTOR

When the denominator has any linear factor factor of the form ( − ), then there are exists a

partial fraction of the form
( − )

EXERCISE:

Find the indefinite integral:

1. ∫ −5 +11
2+ −2

Step 1: The denominator factorizes as ( + 2)( − 1)

Step 2: Rewrite into form below:

∫ −5 +11
( +2)( −1)

Step 3: −5 +11 resolve into partial fraction
( +2)( −1)

Partial Fraction

−5 +11 = +
( +2)( −1) +2 −1

−5 + 11 = ( − 1) + ( + 2)

When = 1,

−5(1) + 11 = (1 − 1) + (1 + 2)

6 = 3

=

Page
50