3.2 Integration using trigonometric substitution

MOOC MAT438/ UiTM

Integration Techniques

 Integration of Rational Function using Partial
Fractions

 Integration of Trigonometric Function
(Powers of sine cosine and secant tangent)

 Integration by Trigonometric Substitution
 Integration by Parts

MOOC MAT438/ UiTM

At the end of this session, the students should be able to
❶ Recognize terms in the given integrand which lend themselves to

trigonometric substitution, and decide which one to use:
for a term of the form 2 − 2, use =
for a term of the form 2 + 2, use =
for a term of the form 2 − 2, use =

❷ Apply an appropriate trigonometric substitution to a variety of integrals.

MOOC MAT438/ UiTM

• When we want to integrate functions with 'radicals', sometimes we

need to eliminate radicals using trigonometric substitution.
• There area three (3) types of radical function using trigonometric

substitution

Types of radicals Substitution Identity Related formula
 2 − 2 = 1 − 2 = 2

=

 2 + 2 = 1 + 2 = 2 = 2


 2 − 2 = 2 − 1 = 2
=
Radical radical symbol Where,
index radicand =
3 27

radical

MOOC MAT438/ UiTM

න 3 4 − 2 . Why we take
= 2 ??

Solution − = −

Using trigonometric substitution,

 = 2 ,  2 = 4 2 , 3 = 8 3
 = 2
2 + 2 = 1

 4 − 2 = 4 − 4 2 = 4 1 − 2 = 4 2 = 2
No more radical, but
Radical & no trigonometric function With trigonometric function

Therefore,

න 3 4 − 2 = න 8 3 ∙ 2 ∙ 2 Substitute  into the given question

= 32 න 3 2 Simplify = θ

= 32 න 2 2 Integrate powers of sine and cosine (odd) using u-subs.
= −

= 32 න 1 − 2 2 ∙ ∙ − From = θ
1 − 2 = 1 − 2 = −

2 + 2 = 1

MOOC MAT438/ UiTM

From = θ
1 − 2 = 1 − 2

න 3 4 − 2 = 32 න 1 − 2 2 ∙ ∙ − = 32 න 1 − 2 − 2 = 32 න 4 − 2


Integrate wrt u 5 3 + = 32 5 − 32 3 + 
= 32 5 − 3 5 3

 = 2 Relace  by x using triangle method
into 
2 x
= 2

− = 4 − 2
2

substitute into  5 4 − 2 3 32 4 − 2 5 4 4 − 2 3
8 +
Therefore, 4 − 2 32 2 + = 5 ∙ 32
2 −3 32 − 3 ∙ simplify
32
න 3 4 − 2 = 5

1 4 − 2 54 3
=5 −3
4 − 2 + #

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

න . Why we take
= 3 ??
+

Solution + = +

Using trigonometric substitution,

 = 3 ,  2 = 9 2

 = 3 2

 9 + 2 = 9 + 9 2 = 9 1 + 2 = 9 2 = 3

Radical & no trigonometric function No more radical, but
With trigonometric function
Therefore, Simplify
Substitute  into the given question
න 9 2 = 3 ∙ 3 2
+ න 3

= 3 න

Integrate trigonometric function wrt

= 3 +  න = +

MOOC MAT438/ UiTM


න = 3 + 
9 + 2

Relace  by x using triangle method

 = 3

+ x
= 3

= 3 → = 9 + 2 
3 9 + 2 3

substitute into 

9 + 2
න = 3∙ 3 +C simplify
9 + 2

= 9 + 2 + C #

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

− Why we take

න . = 1 ??
2
Solution
We need to factorize 4, because coefficient of 2 must one (1).

4 2 − 1 = 4 2 − 1 =2 2 − 12
4 2
Using trigonometric substitution,

 = 1 ,  2 = 1 2

2 4

 = 1

2

 4 2 − 1 = 4 ∙ 1 2 − 1 = 2 − 1 = 2 =
4
No more radical, but
Radical & no trigonometric function with trigonometric function

Therefore,

4 2 − 1 1 Substitute  into the given question
න = න ∙ 2
1
2

= න 2

= න 2 − 1 න 2 = +

MOOC MAT438/ UiTM

4 2 − 1 = න 2 − 1 න 2 = +
න

Integrate wrt = − + 

Relace  by x using triangle method

2x  = 1 into 

− 2
1
= 2 = 2 → = −1 2

1

= 1 , = 4 2 − 1 into 
2

Substitute and into 

4 2 − 1 = 4 2 − 1 − −1 2 + #
න

MOOC MAT438/ UiTM

4 2 − 1 Using integration by u-substitution
3 න = න ∙ 8
Can’t simplify!! = 4 2 − 1

= 8


= 8

11 = 1 න −21
3 න 4 2 − 1 = න ∙ 8 = 8 න 8

Can integrate
using power

rule !

1 11
3 න 4 2 − 1 = න ∙ 8 = 8 න = 8 න 2

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

න .

− 3 3 Why we take
− = 4 − 2 = 22 − 2 = 2 ??
Solution

Using trigonometric substitution,

 = 2 ,  2 = 4 2

 = 2

 4 − 2 = 4 − 4 2 = 4 1 − 2 = 4 2 = 2

Radical & no4t−ri g 2on3om=et2r i c f u n3ct=io8n 3 No more radical, but
with trigonometric function

Therefore, =න Substitute  into the given question
4 − 2 3

න 3
4 − 2 2

2
= න 8 3 ∙ 2

MOOC MAT438/ UiTM

2 =
න 3 = න 8 3 ∙ 2
4 − 2 2
= −
1
= 2 න 2
= −
Integration by u-subs

1
= 2 න 2 ∙ −

11
= − 2 න 2

= − 1 න −2
2
Integrate wrt

1 −1
= − 2 ∙ −1 +

1
= 2 +

MOOC MAT438/ UiTM

1 =
න 3 = 2 +

4 − 2 2 1
= 2 +

1  Relace  by x using triangle method
= 2 +
 = 2
substitute into 

12 2
=2∙ + = 2
4 − 2 x
− 4 − 2
1 = 2 → = 2 into 
= + # 4 − 2

4 − 2

MOOC MAT438/ UiTM

∶ 1 + Using integration by u-substitution
න 3 = න ∙ 4 − 2
4 − 2 2 3 −2 = 4 − 2
Can integrate
2 using power = −2

= 11 rule !
− න = −2
2 3

2

= − 1 න −23
2

1 −21
= − 2 ∙ −21 +
= −21 +

1
= +


1

= + #
4 − 2

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM