MAT238 & MAT438 (Foundation of Applied Mathematics)

By Amirah Hana Mohamed Nor



Chapter 1 : Inverse Trigonometric Function

CHAPTER 1
INVERSE TRIGONOMETRIC FUNCTIONS

1.0 Introduction
1.0.1 Notation

If sin x = A, then x = sin−1A
If cos x = A, then x = cos−1A
If tan x = A, then x = tan−1A
If sec x = A, then x = sec−1A
If cosec x = A, then x = cosec−1A
If cot x = A, then x = cot−1A

1.0.2 Properties of Inverse Trigonometric Function

sin(sin−1x) = x, sin−1(sin x) = x

cos(cos−1x) = x, cos−1(cos x) = x

tan(tan−1x) = x, tan−1(tan x) = x

sec(sec−1x) = x, sec−1(sec x) = x

cosec(cosec−1x) = x, cosec−1(cosec x) = x

cot(cot−1x) = x, cot−1(cot x) = x

1.0.3 The graph of Inverse Trigonometric Function

© Amirah 2021 1

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

1.1 Evaluating Inverse Trigonometric Function using Triangle Method (without using
calculator)

Step 1 : let A = inverse trigonometric function
(or A = first inverse trigonometric function, and
B = second inverse trigonometric function

Step 2 : change inverse trigonometric function to ordinary trigonometric function

Step 3 : draw a right angle triangle and determine all the sides based on
trigonometric function obtained in Step 2.

Step 4 : find the relevant trigonometric ratios from a right angle triangle obtained in
Step 3

Step 5 : evaluate the given question (and refer useful formula)

Useful Formula © Amirah 2021
1. sin (A  B) = sin A cos B  cos A sin B
2. cos (A  B) = cos A cos B sin A sin B
3. tan (A  B) =

4. sin 2A = 2 sin A cos A
5. cos 2A = cos2 A − sin2 A
6. cos 2A = 2cos2 A − 1
7. cos 2A = 1 − 2sin2 A

8. tan 2A =

9. sec A =


10. cosec A =
11. cot A =

12. cos2 A + sin2 A = 1
13. 1 + tan2 A = sec2 A
14. cot2 A + 1 = cosec2 A

2

Simple Examples Ans : 4 Chapter 1 : Inverse Trigonometric Function
a) evaluate [ −1 (4)]
3 Try This !
3
Ans : 4
b) evaluate [ −1 (34)] 5

Solution :

© Amirah 2021 3

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

More Examples (from previous semester papers)

Example 1/page 111/ MAR 2013/ MAT238/ Q1a (7 marks)

Without using calculator, find the value of

i) [ + −1 1] (Ans : − √8)

23 3

ii) [2 −1 1] (Ans : − 2√8)

3 7

Solution :

i) Step 1 : let A = inverse trigonometric function
Let = −1 (1)

3

Step 2 : change inverse trigonometric function to ordinary trigonometric function

1 adjacent
= 3 hypotenuse

Step 3 : draw a right angle triangle and determine all the sides based on trigonometric
function obtained in Step 2.

3 √8 Using Pythagoras theorem :
ඥ32 − 12 = √9 − 1 = √8

1

Step 4 : find the relevant trigonometric ratios from a right angle triangle obtained in Step 3
√8

= 3

Step 5 : evaluate the given question (and refer useful formula)

+ −1 1 = + )
(2 (3)) (2


= 2 − 2

= (0) − (1)

= − Since we solve without using
calculator, so no need to write in
= − √8 #
3 decimal

4 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

• Notice that we have the same inverse trigonometric

ii) [2 −1 1] function −1 (1) from question (i).

3 3

Just let A=inverse-trigo (only) • So, we use step 1,2,3 from question (i)
because constant ‘2’ here means
we will use double-angle identity • So, we continue with step 4.

Step 4 : find the relevant trigonometric ratios from a right angle triangle obtained in Step 3

√8
= 1 = √8

Step 5 : evaluate the given question (and refer useful formula)

(2 −1 1 = (2 )
(3))

2
= 1 − 2

2√8
= 1 − (√8)2

2√8
=1−8

= 2√8
−7
No need to write in decimal,
= − 2√8 # because we solve without using
7
calculator!

© Amirah 2021 5

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

Example 2/ page 119/ SEP 2013/ MAT238/ Q1a/ (6 marks)

Simplify [ −1 ( + 1) + −1 ] in terms of x. (Ans : √√ 2 2 −2+1+2 ++11)

Solution :

 Let = −1 ( ) opposite  Let = −1( ) opposite
adjacent  Let = = hypotenuse
+1
1
 Let =
1
+1 → =

 Right-angle-triangle

Using Pythagoras theorem :  Right-angle-triangle 1
+ 1
ඥ( + 1)2 + 2 = ඥ 2 + 2 + 1 + 2
= ඥ2 2 + 2 + 1 Using Pythagoras theorem : ඥ 2 − 1
ඥ( )2 − 12 = ඥ 2 − 1
 trigo-ratios (based on right-angle-triangle in
step )  trigo-ratios (based on right-angle-triangle in
step )
= 1
√2 2 + 2 + 1 =

+ 1 √ 2 − 1
= =

√2 2 + 2 + 1

 evaluate the given question (and refer useful formula)

( −1 ( 1) + −1( )) = ( + )
+

= +

√ 2 − 1 + 1 1
=∙ + ∙
√2 2 + 2 + 1 √2 2 + 2 + 1

√ 2 − 1 + + 1
=#

√2 2 + 2 + 1

6 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

Example 3/ MAR 2014/ MAT238/ Q1a (6 marks)

Use the Right Triangle method to find the value of [ −1 √2 + −1 √2]
3

ans : √14−√2
3√3

Solution :

Try This !

© Amirah 2021 7

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL) ans : − 1
2
Example 4a/ SEP 2014/ MAT238/ Q1a (4 marks)
Without using calculator, evaluate [2 −1 √23]
Solution :

Try This !

8 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

Example 4b/ SEP 2014/ MAT238/ Q1b (5 marks) ans : − 1
Simplify [ −1 2 + ] √1−4 2
Solution :

i) Step 1 : let A = inverse trigonometric function
Let = −1(2 )

Step 2 : change inverse trigonometric function to ordinary trigonometric function

2 opposite
= 2 = 1 hypotenuse

Step 3 : draw a right angle triangle and determine all the sides based on trigonometric
function obtained in Step 2.

1 2

Using Pythagoras theorem :
ඥ1 − 4 2 ඥ12 − (2 )2 = ඥ1 − 4 2

Step 4 : find the relevant trigonometric ratios from a right angle triangle obtained in Step 3

= √1 − 4 2 = ඥ1 − 4 2
1

Step 5 : evaluate the given question (and refer useful formula)

( −1(2 ) + ) = ( + )

1
= ( + )

1
= −

1
= (−1) − (0)

1
= −

1
= −

1
=− #

√1 − 4 2

© Amirah 2021 9

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

Example 5/ page 134/MAR 2015/ MAT238/ Q1a (5 marks) ans : 2√
Express [2 −1 (√ 2 + )] in terms of x.
Solution : −1

Try This !

10 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

Example 6/ SEP 2015/ MAT238/ Q1a (6 marks)
Simplify ( −1 + −1 2 )

Solution :  let A = first inverse trigonometric function
 let A = first inverse trigonometric function Let B = −1 2

Let A = sin−1 x

 change to ordinary trigonometry  change to ordinary trigonometry

 opposite  tan B = 2 = 2 opposite
= = 1 hypotenuse adjacent
1

 draw a right angle triangle and find  draw a right angle triangle and find
all the three sides all the three sides

1 ඥ1 + 4 2 2

A B
ඥ1 − 2 1

 find the relevant trigo-ratios  find the relevant trigo-ratios
sin A = x
cos A = 1− x 2 2
→ =

√1 + 4 2
1

→ =
√1 + 4 2

Therefore,

[ −1 + −1 2 ] = ( + )

= −

= ඥ1 − 2 ∙ 1 2
− ∙
√1 + 4 2 √1 + 4 2

√1 − 2 − 2 2
=#

√1 + 4 2

© Amirah 2021 11

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

1.2 Derivative of Inverse Trigonometric Functions

1. ( −1 )= 1 2. ( −1 )= −1
− √1 − 2
√1 2

3. ( −1 )= 1 4. ( −1 )= 1 −1
+ 2 + 2
1

5. ( −1 )= 1 6. ( −1 )= −1 1
√ 2 √ 2 −
− 1

Simple Examples 2 2
Differentiate with respect to x. =
a) y = sin−1(2x)
b) y = tan−1(3x2) √1 − 4 2
c) y = cos−1( )

2

d) y = sin−1(e4x)
e) y = sec−1(ln x2)

solution :
a) y = sin−1(2x)

= 1



ඥ1 − (2 )2

( −1 ) = √1 1 2
− (2 ) = 2

b) y = tan−1(3x2) 6
= 1 6 = 1 + 9 4
1 + (3 2)2

( −1 ) = 1 (3 2) = 6
+ 2
1

12 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

c) y = cos−1( )

2

1 1 11 = 11
= √1 − ( 2 )2 2 ∙
= ∙ 2 (√4 − 2) 2
√4 − 2 2
4

−1 1 1
( −1 ) √1 − 2 (2 ) = 2
=

21 1
= √4 − 2 ∙ 2 =#

√4 − 2

d) y = sin−1(e4x)

= 1 4 4 4 4
=#

√1 − 8
ඥ1 − ( 4 )2
(4 ) = 4
( −1 ) = 1 ( ) =
−
√1 2

e) y = sec−1(ln x2) 1 2
= 1 2 2 =#

1 2ඥ( 2)2 − 1
( ) =
2ඥ( 2)2 − 1

( 2) = 2

1
( −1 ) = √ 2 − 1

© Amirah 2021 13

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

More Examples (from previous semester papers)

 page 26/ OCT 2007/ MAT238/ Q1b (5 marks)

Differentiate = √ −1 2 + −1 2 with respect to x.

ans : ඥ( −1 1 + −1 2 + 2
2 )(1−4 2) 1+ 4

Solution :

= = −1 2

= ඥ −12 = ( −12 )12 = −1 2 ′ = ′ = 1 + 1 ∙ 2
( 2)2

Using generalized power rule Using product rule
(combine with inverse-trigo rule) (combine with exponential rule and
inverse-trigo rule)

′ = 1 ( −12 )−21 ∙ ඥ1 1 ∙ 2 ′ = ′ + ′
2 − (2 )2

11 1 ′ = −1 2 + ∙ 1 ∙ 2
′ = 2 ∙ √ −12 ∙ √1 − 4 2 ∙ 2 1 + ( 2)2

1 ′ = −1 2 + 2
′ = 1 + 4

ඥ( −12 )(1 − 4 2)

= ඥ −12 + −1 2
= +

Differentiate wrt x:

= ′ + ′


→ = 1 + −1 2 + 2 #
ඥ( −12 )(1 − 4 2) 1 + 4

14 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

 page 22/ APR 2007/ MAT238/ Q1b (6 marks)

Find the derivative of the function = −1 ( 2 ) ans : 2 −2 2 2
√4 2− 2 2
2

Solution :

= −1 2
( 2 )

Differentiate wrt x: = 2 = 2
′ = 22 ∙ 2 ′ = 2
−1 ′ − ′ ′ = 2 22
= √1 − ( 2 2 )2 ∙
2

−1 4 22 − 2 2
= ∙
22 (2 )2
√1 − 4 2

−1 4 22 − 2 2
= ∙
√4 2 − 22 4 2
4 2

−1 4 22 − 2 2
= (√4 2 −2 22 ) ∙
4 2

−2 4 22 − 2 2
= √4 2 − 22 ∙
2 ∙ 2

−1 4 22 − 2 2
= √4 2 − 22 ∙
2

−(4 22 − 2 2 )
= 2 √4 2 − 22

−2(2 22 − 2 )
= 2 √4 2 − 22

−(2 22 − 2 )
= √4 2 − 22

2 − 2 22
= √4 2 − 22 #

© Amirah 2021 15

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

 page 18/ OCT 2006/ MAT238/ Q1b (6 marks) ans : 3 3 (1+4 2) −1 2
Find if = 3 −1(2 ) 1+4 2−2 3



Solution : = 3 = −1(2 )

= 3 −1(2 ) ′ = 3 3 ′ = 1 + 1 ∙ 2
(2 )2

Differentiate implicitly wrt x: ′ 2
+ 4 2
= 1

′ + ′
=

3 3 −1(2 ) + 2 3
= 1 + 4 2

2 3 3 3 −1(2 )
− 1 + 4 2 =

2 3 3 3 −1(2 )
(1 − 1 + 4 2) =

1 + 4 2 − 2 3 3 3 −1(2 )
( 1 + 4 2 ) =

3 3 (1 + 4 2) −1(2 )
= #
1 + 4 2 − 2 3

16 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

 page 99/ SEP 2011/ MAT238/ Q1b (5 marks) ans : = 10 −1 5 + 4
Differentiate = ( −1 5 )2 + 4 −1 √2 with respect to x. 1+25 2 √2 −4 2

Solution :

= ( −15 )2 = 4 −1√2

Using generalized power rule Using product rule
(combine with inverse-trigo rule) (combine with exponential rule and
inverse-trigo rule)
1
′ = 2( −15 )1 ∙ 1 + (5 )2 ∙ 5 1 1 = √2 = (2 )12
∙
10 −1(5 ) ′ = 4 ∙ √1 − (√2 )2 √2
′ = 1 + 25 2
′ = 1 (2 )−21 ∙ 2
2
4
′ =
√2 √1 − 2 ′ = (2 )−21

′ = 4 ′ = 1
√2
ඥ2 (1 − 2 )

′ = 4

√2 − 4 2

= ( −15 )2 + 4 −1√2

= +

Differentiate wrt x:

= ′ + ′


10 −1(5 ) 4
→ = 1 + 25 2 +#
√2 − 4 2

© Amirah 2021 17

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

 MAR 2013/ MAT238/ Q1b (4 marks)

Find if = −1(3√ ) • ( ) + −1 2



ans : = + −1(3√ ) − 2
32√1− 2 √1−4 2
3
3

Solution :

= −1(3√ ) ∙ = 3√ = 1

3

Using product rule ′ = 1 −23
(combine with inverse-trigo rule and ln 3

rule) = −1( 3√ ) = −1 1 =
′ = ′ + ′
( 3 )

′ = + −1 ( 3√ ) ′ = 1 1 −32 ′ = 1
∙ 3
12
2 √1 − 2 √1 −
( 3)
3 3 3
1
′ =

= −12 3 23√1 − 2

3

Using inverse-trigo rule

′ = −1
∙2
ඥ1 − (2 )2

′ = −2
√1 − 4 2

= −1(3√ ) ∙ + −12

= +

Differentiate wrt x:

= ′ + ′


−1(3√ ) −2
= + +
√1 − 4 2
3 23√1 − 2

3

−1(3√ ) 2
→ = + −#
√1 − 4 2
3 32√1 − 2

3

18 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

 SEP 2014/ MAT238/ Q1c (5 marks) ans : = 2 −1 (4) − 8
Find the derivative of if = 2 −1 (4) √ 2−16



Solution :

Try This !

© Amirah 2021 19

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

 MAR 2015/ MAT238/ Q1b (5 marks) ans : = 3 + 6 3 −1 √ − 3 2 3
Differentiate = 2 3 −1 √ − 3 √ (1+ )

Solution :

Try This !

20 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

1.3 Integration of Inverse Trigonometric Functions
Appendix (given in Final Examination)

1. ∫ 1 2 = −1 + 14. ∫ 2 = −1 +
√ 2 − ( ) √ 2 − ( )

2. ∫ 2 1 2 = 1 −1 + 15. ∫ = 1 −1 +
+ ( ) 2 + 2 ( )

3. ∫ 1 2 = 1 −1 + 16. ∫ 2 = 1 −1 +
√ 2 − ( ) √ 2 − ( )

Where =
= ( )

 MAR 2014/ MAT238/ Q1cii (4 marks) ans : √2 −1(√2 ) +
Evaluate

2
∫

√1 − 2 2

Solution :

21
∫ = 2 ∫
√1 − 2 2 = 1 Where =
√(1)2 − (√2 )2 = ( )
= √2
= 2∫ 1
∙ = √2
√ 2 − 2 √2

21 =
=∫
√2 √ 2 − 2 √2

= √2 −1 + 2
( ) = √2

√2

−1 √2 Because,
(1)
= √2 +

2
2 (√2) √2√2
= = = √2
√2 √2 √2
= √2 −1(√2 ) + #

© Amirah 2021 21

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

 SEP 2014/ MAT238/ Q2c (5 marks)

Solve the following integral

2
∫ 4 + 3 2

ans : 1 −1 (√3 ) +

√3 2

Solution :

21
∫ = 2 ∫
4 + 3 2 (2)2 2 = 2 Where =
+ = ( )
(√3 ) = √3

1 = √3
= 2 ∫ 2 + 2 ∙ √3

21 =
= √3 ∫ 2 + 2
√3

= 2 ∙ 1 −1 +
√3 ( )

= 2 ∙ 1 −1 √3 ) +
√3 2 (2

= 1 −1 (√23 ) + #
√3

22 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

 MAR 2013/ MAT238/ Q1c (4 marks) Ans : √5 −1 ( −5) +
Evaluate the integral
√5
5
∫ ( − 5)2 + 5

Solution :

51
∫ = 5 ∫
( − 5)2 + 5 5)2 2 = √5 Where =
( − + = − 5 = ( )
(√5)
= 1
1 =
= 5 ∫ 2 + 2

= 5 ∙ 1 −1 +
( )

=5∙ 1 −1 − 5 +
( )
√5 √5
5
= 5 −1 − 5 + = √5
( )
√5 √5 √5
Because,

= √5 −1 − 5 + # 5 (√5)2 √5√5
( )
√5 = = = √5
√5 √5 √5

© Amirah 2021 23

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

 MAR 2015/ MAT238/ Q1c (5 marks) Ans : 2 −1 ( 2 ) +
Use the substitution u = e2x to solve
33
4 2
∫ 4 + 9

Solution :

4 2 2
∫ 4 + 9 = 4 ∫ ( 2 )2 + (3)2
= 3 Where =
= ( )
= 2
2
= 4 ∫ 2 + 2 ∙ 2 2 = 2 2


1
= 2 ∫ 2 + 2 = 2 2

= 2 ∙ 1 −1 +
( )
( 2 )2 = 4
= 2 ∙ 1 −1 2 + Because,
3 (3)
( ) =
2 2
= 3 −1 (3 ) + #

24 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

 Evaluate
2

∫
ඥ4 − ( )2

Solution :

2 1
∫ =2∫ Where =
ඥ4 − ( )2 ඥ(2)2 − ( )2 = 2 = ( )
=
= 2∫ 1 ∙ 1
=
√ 2 − 2 =

= 2∫ 1

√ 2 − 2

= 2 −1 +
( )

= 2 −1 + #
(2)

 Evaluate
3 3

∫ 8 + 9
Solution :

3 3 3
∫ 8 + 9 = 3 ∫ ( 4)2 + (3)2 = 3 Where =
= ( )
= 4
3
= 3 ∫ 2 + 2 ∙ 4 3 = 4 3


31
= 4 ∫ 2 + 2 = 4 3

3 ∙ 1 −1 +
=4 ( )

3 ∙ 1 −1 4 ) +
=4 3 (3

= 1 −1 4 + #
4 (3)

© Amirah 2021 25

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

 Evaluate
5 5

∫
√25 − 2 5

Solution :

5 5 5 5 Where =
∫ = ∫
√25 − 2 5 ඥ(5)2 − ( 5 )2 = ( )

= 5

5 5 = 5
= ∫ √ 2 − 2 ∙ 5 5 5
= ( 5 5 ) ∙ 5
11
= 5 ∫ √ 2 − 2
= 5 5 5

= 1 −1 +
5 ( )

= 1 −1 5 + #
5 (5)

 Evaluate


3 3
∫
√1 − 2 3
9

Solution : Where =
= ( )
3 3
∫ = ∫ = 1
√1 − 2 3 ඥ(1)2 − ( 3 )2 = 3

3 = (− 3 ) ∙ 3
= ∫ √ 2 − 2 ∙ −3 3

11 = −3 3
= − 3 ∫ √ 2 − 2

= 1 −1 +
−3 ( )

= 1 −1 3 +
−3 (1)

= 1 −1( 3 ) + #
−3

26 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

Continue…

Hence,

3 1 1
− 2 [− 3 3 −3[
∫ 3 = −1( = −1( 3 )] 3
3 )]
√1 3 9
9 9

= − 1 [ −1 ( (3 ∙ − −1 ( (3 ∙
3 3)) 9))]

= − 1 [ −1( ( )) − −1 (
3 (3))]

= − 1 [ −1(−1) − −1 1
3 (2)]

2 Set your calculator
= 0.6981 # ( 9 ) in mode ‘radian’

 Evaluate


∫
ඥ25 − ( + 2)2

Solution :

1
∫ =∫
ඥ25 − ( + 2)2 ඥ(5)2 − ( + 2)2 = 5 Where =
= + 2 = ( )

1 = 1
= ∫ =

√ 2 − 2

= −1 +
( )

= −1 + 2 + #
( 5 )

© Amirah 2021 27

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

1.4 Integration of Inverse Trigonometric Functions involving Completing the Square Method

Procedure completing the square method,
 arrange based on the highest power of x, such that

2 + +

 coefficient of x2 must positive 1. If coefficient of x2 not equal to positive 1, then factorize

coefficient of x as follow

[ 2 + +
]

 put + ( )2 − ( )2 after the term containing x, where the expression inside ( ) must be 1 •coefficient

2

of x = 1 • = (as follow)

2 2

[ 2 + + 2 − 2 +
(2 ) (2 ) ]

 write the first three terms into the form of ( )2 and simplify the next terms

Or, using formula,

2 + + = [( + 2 − ( 2 + ]

) )
2 2

When we use completing the square method?
When:
i) numerator is a constant, and
ii) denominator is in the form of quadratic expression with the

term containing ‘x’

28 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

Example 1
Evaluate


∫

√21 − 4 − 2

Solution :

Using completing the square,

21 − 4 − 2 = − 2 − 4 + 21   Re-arrange : 2 + +

= −( 2 + 4 − 21)   coefficient of x2 must positive 1

(factorize coefficient of x2 if necessary)

= −( 2 + 4 + (2)2 − (2)2 − 21) 

= −(( + 2)2 − 25)   put + ( )2 − ( )2 after the term containing x,
= 25 − ( + 2)2
where the expression inside ( ) must be 1 •coef x
2

Hence,  write the first three terms into the form of ( )2
and simplify the remaining terms


∫ =∫
√21 − 4 − 2 ඥ25 − ( + 2)2

Complete this integral.

You can Refer page
27, no. 

© Amirah 2021 29

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

Example 2 ans : 1 −1 ( −3) +
Evaluate
42

∫ 2 2 − 12 + 26

Solution :

Try This !

30 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

Example 3 ans : 5 −1 (√2 ( + 1)) +
Evaluate
√14 √7
5
∫ 2 2 + 4 + 9

Solution :

Using completing the square,

2 2 + 4 + 9 = 2 [ 2 + 2 + 9
2]

= 2 [ 2 + 2 + (1)2 − (1)2 + 9
2]

= 2 [( + 1)2 7
+ 2]

= 2 [( + 1)2 + √7 2
( )]
√2

Hence,

51 Where =
∫ 2 2 + 4 + 9 = 5 ∫ 2 = ( )
2 [( + 1)2 + (√7) ]
√2 √7
=
= 51
∫ √2
2 2 = + 1
( + 1)2 +
(√7) = 1
√2 =
Note :
51
= 2 ∫ 2 + 2

√ √ = (√ ) 5 1
2 ( )
=

= ∙ −1 +

5 ∙ 1 = 5 ∙ √2 5 1 + 1
2 (√√72) 2 (√7) (√7)
2 √7 = ∙ −1 +

Note :

= 5 ∙ √2 √2 ( √2 ) + 1 = ( + 1) ÷ (√7)
(√7) √2
√2√2 √7 5 (√2( + 1)
= −1 ) + √2
= 5∙1
√14 √7 = ( + 1) ∙ (√2)
√2 √7 1 √7

=5 = 5 −1 (√2 ( + 1)) + # = √2 ( + 1)
√14 √7 √7
√14

© Amirah 2021 31

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

Example 4
Evaluate


∫

√6 − 2

Solution :
Using completing the square,
− 2 + 6 = −( 2 − 6 )
= −( 2 − 6 + (−3)2 − (−3)2)
= −(( − 3)2 − 9)
= 9 − ( − 3)2

Hence,

1
∫ = ∫
√6 − 2 ඥ9 − ( − 3)2

1 = 3 Where =
= ∫ = − 3 = ( )

ඥ(3)2 − ( − 3)2 = 1
=
1
= ∫

√ 2 − 2

= −1 +
( )

= −1 − 3 + #
( 3 )

32 © Amirah 2021

Chapter 1 : Inverse Trigonometric Function

Example 5

Evaluate

∫ 4

2 2 − 6 + 13

ans : 1 −1 ( −3) +

22

Solution :
Using completing the square,

2 − 6 + 13 = 2 − 6 + (−3)2 − (−3)2 + 13

= ( − 3)2 + 4

therefore,

1
∫ 2 − 6 + 13 = ∫ ( − 3)2 + 4

1 Where =
= ∫ ( − 3)2 + 4 = ( )

1 = 2
= ∫ ( − 3)2 + (2)2

1 = − 3
= ∫ 2 + 2
= 1
= 1 −1 + =
( )

= 1 −1 − 3 +
2 ( 2 )

Hence,

∫ 4 = 1 −1 − 34
[2 ( 2 )]
2 2 − 6 + 13
2

= 1 [ −1 − 34
2 ( 2 )]

2

= 1 [ −1 4 − 3 − −1 2 − 3
2 ( 2 ) ( 2 )]

= 1 [ −1 1 − −1 −1
2 (2) ( 2 )]

= 0.4636 # Set your calculator
in mode ‘radian’

© Amirah 2021 33

MAT238/ MAT438 : Foundation of Applied Mathematics (Notes & Workbook for ODL)

Example 6 ans : 4 −1 (√3( −1)) +
Evaluate
√3 √10
4
∫

√7 + 6 − 3 2

Solution :

Try This !

END OF CHAPTER 1
34 © Amirah 2021