INTEGRATION IN MATHEMATICS ENGINEERING 2(DBM20023)

DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3.0 Definition Of Integration

1) The process of obtaining from y ( a function of x ) is known as differentiation


and the reverse process of obtaining y from is called integration.


Differentiation process




Integration process

2) Integration is divided into two parts which are indefinite integral and definite
integral.

3.1 Indefinite Integrals
Integration of y with respect to x, is denoted by ∫ ( ) .

If = ( ),

The differentiation : = ′( )


The integration : ∫ ′( ) = +

Where c is an arbitrary constant.

Tips!

Note that a function is integrating with respect to the function given and it is not
compulsory to be integrated respect to x only. (See example given)

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Basic formulas in integration:

i)  a dx = ax + c ; where a is a constant. [Constant rule]
ii) axn dx = axn+1 + c ; where n is an integer, ≠ −1. [Power rule]

n +1

Example 1:

Determine the following integrals::

a)  3 x dx b)  −r3 dr c)  6 dk
5

d)  xdx e)  2 dx f)  4 3 dm
3x 4 16m

Solution:

a)  3 x dx = 3  x2  + c In indefinite integral, c must be stated. It is
5 5  2  known as an arbitrary constant where the
  value is yet to be determined.

= 3 x2 + c
10

b) − r3 dr = − r 4 + c
4

c)  6 dk = 6k + c The given function is integrating with respect to
k, therefore the integration will be in terms of k.
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

d)  1

xdx = x 2dx

3 2 Make sure that x is always in index form.
x
= 3 +c

2

= 2 3 + c

3 x2

2 dx = 2x −4 dx
3x 4 3
 e)

= 2x −3 + c
3 −3

= 2 +c
− 9x3

f)  3 dm = 3 1 dm
4 16m (16m ) 4

= 3 1 dm
11
(16) 4 m 4

= 3 m − 1 dm
4

2

3

= 3 m4 +c
23

4

3

= 2m 4 + c

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3.1.1 Determining the Integration of Algebraic Expressions Involving Operations
∫[ ( ) ± ( ) = ∫ ( ) ± ∫ ( )

Example 2:

Integrate the following equations: c)  (4 − x)(5 + 2 x) dx
x4
a) 4x3 +3x2 − 3x + 4 dx
5 d)   2 − 3 + x dx
 5x3
b)  ( x − 6)2dx

Solution:

a) 4x3 +3x2 − 3x + 4 dx = 4x4 + 3x3 − 3x2 + 4 x + c

5 4 3 25
= x4 + x3 − 3x2 + 4 x + c
25

b)  ( x − 6)2dx =  ( x2 −12x + 36)dx Integrate terms by terms in
addition or subtraction. If the
= x3 −12  x2  + 36x + c function given is in
3  2  multiplication, the function
  must be expanded.

= x3 − 6x2 + 36x + c
3

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

 c) (4 − x)(5 + 2x) dx = 20 + 8x − 5x − 2x2 dx Integrate terms by terms in
x4 x4 addition or subtraction. If the
function given is in division,
=  20 + 3x − 2x2  the function must be
 x4 x4 x4 dx simplified.
 
Tips! Operation
 ( )= 20x−4 + 3x−3 − 2x−2 dx

=  x−3  +  x−2  −  x−1  + c
20  −3  3 −2  2 −1 
  
  

= − 20 − 3 + 2 + c
3x3 2x2 x

 2 − d) 3 + x  dx =  2 − 3x −3 + x  dx
 5x3  5

= 2x − 3x−2 + x2 + c
(5)(−2) 2

= 2x + 3x−2 + x2 + c
10 2

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

EXERCISE 1 b) 3 (5 − x)(x + 6) dx
1:

Find the integral for the following:

a) 2x3 + 4x2 + 2x − 5 dx
3

c) 3k 2 +5 4 k −4 dk d)  x2 − x + 5 dx
k3 x

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

(4 − x)(3 − x) ( )f)1
e)  x5 dx  5x 10x3 −15 x + x dx

g) 3x2  x3 + 4x2 + 1  dx 2

4  3x   2z + z3

h) dz
z

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3.2 Definite Integrals of Algebraic Expressions

3.2.1 Fundamental Theorem of Calculus

b

If x=a to x=b for  f (x)dx , then it is written as  f (x)dx and it is called definite integral.
a

Fundamental Theorem of Calculus

b

 f (x)dx = F(b) − F(a)

a

3.2.2 Properties of Definite Integral

b

For the definite integral  f (x)dx , the constant a is known as the lower limit of integration while

a

b is known as the upper limit of the integration.

Properties of Definite Integrals

bb

a)  kf (x)dx = k  f (x)dx

aa

b bb

b)  f (x)  g(x)dx =  f (x)dx   g(x)dx

a aa

ba

c)  f (x)dx = − f (x)dx

ab

bc c

d)  f (x)dx +  f (x)dx =  f (x)dx

ab a

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3.2.3 Evaluate definite integrals using the properties of definite integrals
Example 3:

1) Evaluate each of the following definite integrals below.

3 0

a)  (3x + 5)dx b)  (x +1)( x − 3) dx
−1
1

c) 2 4x − x2 + 5x3 dx

1x

Solution:

a)3 (3x + 5)dx =  3x2 3
1  2 + 5x

1

 3(3) 2   3(1) 2 + 5(1)
2 2
= + 5(3) −

= 22

00

b)  (x +1)( x − 3) dx =  ( x2 − 3x + x − 3)dx
−1 −1

0

=  ( x2 − 2x − 3)dx
−1

=  x3 −  x2  − 0
 3 2 2  3x
 
  −1

=  x3 − x2 0
 3 − 3x

 −1

= 0 −  ( −1)3 − ( −1)2 − 

3 3 ( −1) 

= −5
3

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

  ( )c) 2 4x − x2 + 5x3 dx = 2 4 − x + 5x2 dx
1x
1

=  x2 + 5 x3 12
4x − 2 3


=  4(2) − 22 + 5 23   −  4(1) − 12 + 5 13  
2 3 2 3

= 85
6

48

2) Given that  g(x)dx = 6 and  g(x)dx = 10 , find the values of

24

44 4

a)  g(x)dx −  g(x)dx b)  ( g(x) − 2x) dx

28 2

28 8

c)  −5g(x)dx +  g(x)dx d) k if  (kx + g(x))dx = 9

44 4

8

e)  g(x) dx

2

Solution:

44

a)  g(x)dx −  g(x)dx = 6 − (−10)

28

= 16

4 44

b)  ( g(x) − 2x) dx =  g(x)dx −  (2x) dx

2 22

= 6 −  2x2  4
 2  2
 

= 6 − (4)2 − ( 2)2 


= 6 − (16 − 4)

= 6 −12

= −6

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

2 84 8

c)  −5g(x)dx +  g(x)dx = − −5g(x)dx +  g(x)dx

4 42 4

= −(−5)(6) +10

= 30 +10

= 40

8

d)  (kx + g(x))dx = 9

4

88

 (kx) dx +  g(x)dx = 9

44

 kx2 8 + 10 = 9
 2 
 4

k (8)2 k (4)2

− +10 = 9
22
32k − 8k = −1

24k = −1

k = −1
24

8

e)  g(x) dx

2

48

=  g(x) dx +  g(x) dx

24

= 6 +10

= 16

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

EXERCISE 2
1:

1. Evaluate each of the following;

(a)4 3x2 +x+ 1 dx
2 x2

(b) 2 2m3 − 2 dm
1 3m3

1 Page 140

(c) x (x + 3) (x − 3) dx
−2

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

(d) 3  2x − 1 2 dx
1  x

(e)−1 − 3x5 + 3x3 + 2dx
−3 x3

2 4 − x

(f) dx
1x

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

2. Given that ∫25 ( ) = 6, find the value of ;

5

a) 6 f (x) dx
2

b) 5  5 − f (x)dx

2

5

c) 2 − 3 f (x) dx

d) 5 f (x) + 6 dx
23

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3. Given that ∫13 ( ) = 5, ∫35 ( ) = 2 − 7 where m is a constant. If ∫15 ( ) = 4,
find the value of m.

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3.3 Integral of Reciprocal Function
3.3.1 Perform Integrals of Indefinite Reciprocal Function

Formula of integration of reciprocal function:

) ∫ = | | +


) ∫ = × | + | +
+

Example 4:

Integrate the following:

a)  1 dx b)  3 dx
5x 5x

c)  dx 3 d)  8 + 2 dt
5x + −16t

Solution:

a)  1 dx = 1  1 dx b)  3 dx = 3  1 dx
5x 5 x 5x 5 x

= 1 ln x + c = 3 ln x + c
5 5

c)  dx 3 =  1 3 dx d)  8 + dt = 8 1 + dt
5x + 5x + −16t 2 −16t 2

= 1 ln 5x + 3 + c
5

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

= 8 1 ln −16t + 2 + c
−16

= − 1 ln −16t + 2 + c
2

3.3.2 Evaluate integrals of definite reciprocal function

Example 5:

Evaluate the value for each function.

a) −3 1 dx 1 2 dx
−4 x
b) 0 4 + 2x

c) 2 3 + 4x dx 5 3 dt
1x
d) 1 2t − 3

Solution:

−3 1  12 1
dx
b)
a) 1 dx

dx = 2
−4 x 0 4 + 2x 0 4 + 2x

 = ln x −3 Ignore = 2 1 ln 4 + 2x 1
−4 negative  2  0

= ln − 3− ln − 4  =ln 4 + 2x 1
= ln 3− ln 4  0

= 1.0986 −1.3863 = ln 4 + 2(1) − ln 4 + 2(0) 
= −0.2877
= ln 6 − ln 4 

= 1.7918 −1.3863
= 0.4055

c) 2 3 + 4x dx 3 5 1
1x
dt = 3
 5 1 2t − 3 dt

d)
1 2t − 3

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

=  | x | + 4x2  2 3 1 5
3 ln 2  −1 2 1
  =  ln 2t − 3

 = 2
3ln | x | +2x 2 −1

   = 3ln 2 + 2(2)2 − 3ln 1 + 2(1)2 =  3 ln 2(5) − 3  −  3 ln 2(1) − 3 
 2   2 
= 10.0794 − 2

= 8.0794 =  3 ln 7  −  3 ln − 1 
 2   2

= 2.9189 − 0

= 2.9189

EXERCISE 3

1:

1. Integrate the following.

a)  dx b)  1 dx
− 3x 2x + 5

c)  2 4 dx d)  6 dx
4x + 6 − 24x

e)  3 − (x 7 2) dx
(x −1) +

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

2. Evaluate the value for each function.

2 2 dx
−1 2x + 3
a)

b) 3 3 dx Page 147
1 x−4
JMSK, PUO

DBM2013 Engineering Mathematics 2 Chapter 3 Integration

−1 4 dx

c)
−2 2x − 4

2 4 dt

d) 1 3t + 3

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

3.4 Integral of Exponent Function
Formula of integration of exponential functions:

1) e x dx = e x + c
2) eax+b dx = 1  eax+b + c

a
= eax+b + c

a

FiEndx:ample 6:  ( )b) e2x e3x + ex dx c)  4e x − e5x + 2e3x dx
ex
a) e5−3x dx

Solution:

a ) e5−3xdx = e5−3x + c
−3

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

 ( )  ( )b ) e2x e3x + ex dx = e5x + e3x dx

= e5x + e3x + c
53

 4e x − e5x + 2e 3 x dx (4 − e4x + 2e2x ) dx
 c ) ex =
Simplify terms by terms

= 4x − e4x + 2e2x + c
42

= 4x − e4x + e2x + c
4

EXERCISE 4
3441:

Integrate the following equation:

a) e10xdx b) e3t−2dt

c) e−2xdx 1+ et

d) et dt

e) e3x + e−x f) ex (e2x − e3x ) dx
dx
e2x Page 150
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

g) e5 x  e2x − 3  dx
7 e3x

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

h) (e4x − e−4x )2 dx

3.5 Integral of Trigonometric Function
Formula integration of trigonometric function:

1) ∫ sin( + ) = − 1 + ) × cos( + ) +
(

2) ∫ cos( + ) = 1 + ) × sin ( + ) +
(

3) ∫ sec2( + ) = 1 + ) × tan ( + ) +
(

3.5.1 Perform Integrals of Indefinite Trigonometric Function Page 152
Example 7:

Find:
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

a) sin (5 − 6x) dx b)  cos  3x +   dx
 4 

c)  sec 2  x  dx d) 8 sin 2x dx
 2  3

Solution:

a)  sin (5 − 6x) dx = − cos (5 − 6x) + c sin 3x +  
−6  4 
b)  cos 3x +   dx = +c
= cos (5 − 6x) + c  4  3
6

tan x  d) 8 sin 2x dx = 8  sin 2x dx
3 3
c)  sec2  x  dx =  2  + c
2 1
8  − cos
2 = 3  2 2x  + c

= 2 tan x  + c
2 = − 4 cos 2x + c
3

3.5.2 Evaluate Integrals of Definite Trigonometric Function

Example 8: 600 
Evaluate each of the
Evaluate each of integrals below: b) 5 cos 5x dx c)  sin (4t − 3) dt
400 4


a) sin (6t) dt
0.25

Solution:

a) ∫0 . 25 sin (6 )

= [−cos6(6 )] Π Use mode radian(R)
0.25 because the limit is in

= [−cos6(6 )] − [− cos(66(0.25 ))] radian
= −0.167

JMSK, PUO TIPS!!!! Page 153
π=180°

DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Or Use mode degree (D)
because the limit is in

degree
∫ sin(6 )
Page 154
0.25

180°

= ∫ sin (6 )

45°

= [−cos6(6 )] 180°
45°

= [− cos(66(180°))] − [− cos(66(45°))]

= −0.167

b) ∫4600°° 5 cos 5

= [5 sin 5 ]600
5
400

= [sin 5 ]640000

= [sin 5( 60°)] − [sin 5(40°)]

= [sin 300°] − [sin 200°]
= −0.8660 − (−0.3420)
= −0.5240



c)  sin (4t − 3) dt
4

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

= [sin (4 − 3)]
4

4

= (sin (4( ) − 3)) − (sin(4(0.245 ) − 3))
4

= −0.0705

EXERCISE 5 b)  6 cos (2x +1) dx
3441:

1. Integrate each of the following:

a)  4 sin (4x + 4) dx

c) 5 − sin (2 − 3x) dx d)  2 sec2 x dx
5

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

e)  sin 3x + cos x dx f) 3 sin 3x +   dx
2 2  3 

g) 1+ 2x3 − sin (2x) dx h) e2x + sec2 2x dx
4 4

2. Evaluate each of the following: b)  sec2 3t dt

a) 50 2cos d
20 4

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

 30

c) 3 3 + cos 4 d d) sin (2x + 3) dx
 20

4

3.6 Solve Integration Problem by Substitution
The expressions of the form ( + ) can be integrated by using substitution method.

Consider  (2x − 3)5dx

Steps Illustrations

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

1. Find a suitable substitution u= (2x -3)
2. Differentiate u with respect to x
3. Determine the suitable differentiation. du = 2
4. Substitue the function with u. dx
5. Integrate u.
6. Substitute again = (2 − 3). du = dx
2

 u5 du
2

u6 + c
62

y = (2x − 3)6 + c
12

For linear substitution such as  (2x +1)3 dx and  2 dx , we can solve these questions

3(8 − x)4

by using a formula:

 (ax + b)ndx = (ax + )b n+1 +c

a( n + 1)

Example 9: c)  (2x)(5 + x2 )4 dx e) 8cos xesin xdx
Solve the integral below:
Page 158
a)  (4 + 2x)3dx

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

b)  (6 2 x )4 dx d) 4x3 sec2 x4dx ( )1
+2
f) x x2 + 2 3 dx

0

Solution:

a)  (4 + 2x)3dx

By using substitution method,

u = 4+ 2x  (4 + 2x)3dx =  u3  du 
du = 2 ; du = dx  2 
dx 2
= 1  u 3 (du )
2

= 1  u4  + c
2 4

= 1 (4 + 2x)4 + c

8

By using formula,

 (4 + 2x)3dx = (4 + 2x)4 + c

4(2)

= (4 + 2x)4 + c

8

b)  (6 2 x )4 dx
+2

By using substitution method,

u = 6 + 2x  (6 2 dx =  2  du 
u4  2 
+ 2x)4

du = 2 ; du = dx = u−4du
dx 2

= u −3 + c = 3(6 −1 + c
−3
+ 2x)3

By using formula,

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

 (6 2 x )4 dx = 2 (6 + 2 )−4 dx
+2
x

= 2 (6 + 2x)−3 + c = 3(6 −1 + c
(2)(−3)
+ 2x)3

c)  (2x)(5 + x2 )4 dx ( )(2x) 5 + x2 4 dx = ( 2x )u4  du  = u4du
 2x 
By using substitution method,
u = 5+ x2 = u5 +c
5
du = 2x
dx ( )= 5 + x2 5 + c
du = dx 5
2x

d) 4x3 sec2 x4dx  4x3 sec2 x4dx = 4x3 sec2 u du
4x3
By using substitution method, =  sec2 udu
= tanu + c
u = x4 = tan x4 + c

du = 4x3
dx

du = dx
4x3

e) 8cos xesin xdx  8cos xesin xdx = 8cos xeu du
cos x
By using substitution method,
u = sin x =  8 cos xe u du
cos x
du = cos x
dx =  8eu du
du = dx
cos x = 8eu + c = 8esin x + c

( )1 Page 160

f) x x2 + 2 3 dx

0

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

By using substitution method,  ( ) 1 x x2 + 2 3 dx = 1 xu 3 du
u = x2 + 2
0 0 2x
du = 2x ; du = dx
dx 2x

= 1 1 u3 du

20

=  1 u 4 1
8 0

1( )= x2 4 1
8 +2  0

 = 1 (1 + 2)4 − (2)4
8

= 1 (81 −16)

8

= 65
8

TIPS!

It is recommended to use substitution method in
multiplication and division which cannot be
simplified to a single function.

EXERCISE 6 Page 161
3441:

1. Integrate each of the following below by using substitution method:

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

( )a) 5x2 5x3 + 5 5 dx b) 39xx32++55xdx

c)  2x x2 − 3 dx d)  3 (2 1 3)4 dx
x+

e) ex(ex − 2 )3dx f) 3x2 ex3 dx

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

g) sin x cos2 x dx h)  2 tan (4z + 2)dz

2. Evaluate the following integrals: Page 163
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

 ( )a) 2 3x2 x3 − 3 2 dx
0

b) 0 2t 2 1− 4t 3 dt
−2

c) 6 3 dx
1
(3 + x)4

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

2 3k 2 + 4

d) 1 k 3 + 4k dk

3.7 Integral of Inverse Trigonometric Function Page 165
3.7.1 Formula of the Integrals of Inverse Trigonometric Function
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INTEGRATION OF INVERSE TRIGONOMETRIC FUNCTIONS

 1 du = sin −1 u + c a  1 du = 1 sec−1 u + c
a2 −u2 u u2 − a2 a a

 − 1 du = cos−1 u + c  − 1 du = 1 cos ec−1 u + c a a
a2 −u2 a u u2 −a2

 a2 1 u2 du = 1 tan−1 u + c − a2 1 u2 du = 1 cot −1 u + c
+ a a + a a

Note that, to use these formulas, we have to consider the denominator of the function.
If the denominator of the function has only two terms (means that one term is the constant and
another term is the function in term of x ), we can directly choose the value of 2 and 2.
*We have to select the constant as a2 and the function in term of x as u 2 .

If the denominator is quadratic function, we have to factorize the function first. So, we can use
completing the square method.

Completing the square method Page 166

If ax2 + bx + c = 0 , where a = 1,
Then, ax2 + bx +  b 2 + c −  b 2 = 0

2 2

3.7.2 Perform Integrals of Indefinite Inverse Trigonometric Function
Example 10:

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Integrate the following functions. 2.  dx 3.  1
9x2 − 5 dx
1. −1 dx x
5 + 4x − x2
9− 4x2

Solution: 2)  dx
9x2 − 5
1) −1 dx x

9− 4x2 Let a2 = 5
Let a 2 = 9
a= 5
a= 9=3

u2 = 4x2 u2 = 9x2

u = 4x2 = 2x u = 9x2 → x=u
du = 2 u = 3x 3
dx dx = du
dx = du
3
2

  −1 dx = −1  du 
9 − 4x2 a2 −u2  2 

= 1 −1 du 1 dx = 1  du 
2 a2 −u2 x 9x2 −5 u2 −a2  3 
  u

= 1 cos −1 u + c 3
2a

= 1 cos −1 2x + c = 1 dx
23
u u2 −a2

= 1 sec−1 u + c
aa

= 1 sec−1 3x + c
55

3.  1 dx
5 + 4x − x2

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 Completing the square : 5 + 4x − x2 = − x2 − 4x − 5

=  − 4x +  − 4  2 − 5 −  − 4  2 
−x2 
 2  2  

 = − x2 − 4x + (− 2)2 − 5 − (− 2)2

 = − (x − 2)2 − 9

= −(x − 2)2 + 9

 1 x2 dx =  1 dx
5+ 4x −
− (x − 2)2 + 9

Let a2 = 9 u 2 = (x − 2)2
a= 9 u = (x − 2)2
a=3 u= x−2
dx = du

 1 x2 dx =  1 dx
5+ 4x −
− (x − 2)2 + 9

= 1 du

−u2 + a2
= sin −1 u + c

a
= sin −1 x − 2 + c

3

EXERCISE 7 Page 168
3441:

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Integrate the following functions. 2.  − 3 dx
25 − 5x2
1

1. dx
1− 4x2

3. − ex dx 4. sec2 x dx

49 + e2x tan2 x +131

 − dx 6.  4
dx
5.
36 − 9(4 − x)2
x 4x2 −13
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

7. cos x dx

cos 2 x + 1

8. x2 −1 + 3 dx
+ 2x

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

9. 4x2 1 + 13 dx
+ 8x

3.8 Integration by part
In integration we are able to integrate all the function given except such as the following;

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

 xe3xdx ,  x2 sinh 5xdx ,  ln x dx ,  sin-1x dx

The next formula will enable us to evaluate not only these, but other types of integrals.

 u dv = uv −  v du

Choose u, dv in such a way that:
1. u is easy to differentiate.
2. dv is easy to integrate.

Sometimes it is necessary to integrate by parts more than once.

Tips!!!
Choose u as a function that can differentiate and dv
will be the function that you need to integrate.
We can choose u by using acronym, I LATE.
I-Inverse trigonometry function
L-logarithmic function
A-Algebraic function
T-Trigonometric function
E-Exponential function

Example 11: Page 172
Evaluate  xe4xdx

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Solution:

Both of the functions are possible choices for dv but we choose e4x as dv as it is the most
complicated part. Thus, let us display this expression as follows:

 dv =  e4xdx u=x
du = dx
v = e4x
4

Substitute these expressions in by parts formula,

= xe 4x dxxe4x−e 4x dx
4 4

= xe 4x − e 4x + c
4 16

Example 12: Page 173

Evaluate x2e2xdx

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

Solution:

Sometimes it is necessary to use more than once by parts integral until we may integrate single
function as illustrated below.

 dv =  e2xdx u = x2
du = 2xdx
v = e2x
2

= x2e2xdx x2  e2x  −  e2x  2x dx
2 2

= x 2e2x − e2x x dx
2

We must again integrate by parts since the expression of the integration is still in multiplication
form.

dv = e2x dx u=x
v = e2x du = dx

2

 x 2e2x dx = x 2e2x − e2x x dx
2

x2e2x (x) e2x e2x 
2 2 2 dx
= −  −


= x 2e2x − xe 2x + e2x + c
2 24

EXERCISE 8 Page 174
3441:

Determine each of the following integrals:

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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

a) xe−9xdx b)  x sinx dx

c)  ln x dx d)  4 ln x dx

ln x f)  sin−1 xdx
e)  x 2 dx
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g)  3x3 cos x dx Page 176

h) ∫ 2 3
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

i) ∫ sin

3.9 Integration involving partial fraction Page 177

Recall...
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DBM2013 Engineering Mathematics 2 Chapter 3 Integration

There are two types of fractions; proper fraction and improper fraction.

If we are given a single fraction and we can break it down into sum of easier fraction, so is called
partial fraction.

Types of fraction Types of partial fraction Partial Fraction

Proper fraction 1. Denominator with Linear A+B
Factor, ( x + a )( x + b ) x+a x+b
(numerator <
denominator)

2. Denominator with x A a + (x B
Repeated Linear Factor, +
+ a)2
a. (x + a)2

b. (x + a)3 A + (x B + (x C
+a
x + a)2 + a)3

If the denominator is a quadratic function, ax2 + bx + c or cubic function ax3 + bx2 + cx + d , we
need to change the function into a linear or repeated linear factor.

3.9.1 Solve integrating partial fraction mathematical problem Page 178

Example 13:
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