Example 1. Find the exact value of 58

20

1

25 16dx

x

FP2 Calculus 1 Inverse Trig functions

5 58 8

2 2250 0 16

1 1

25 16 4dx dx

x x

58

2 2405

1

4dx

x

1

2 2

1sin

xdx

aa x

5

81

0

1 4sin

4 5

x

24

Example 2. Express in the form

FP2 Calculus 1 Inverse Trig functions

72

5

2

1

5 4dx

x x

1

2 2

1sin

xdx

aa x

72

5

1 2sin

3

x

3

25 4x x 2( )a b x

Where a and b are positive constants. Hence find the exact value of

2 25 4 9 ( 2)x x x 72

5

2 2

1

3 ( 2)dx

x

Example 3. Express in partial fractions.

FP2 Calculus 1 Inverse Trig functions

1

2 2

1 1tan

xdx

a aa x

1

2 1

0

12 ln( 1) ln( 4) tan

2 2

xx x x

15 1 12 ln tan

2 2 2

3 2

2

2 5 11 13

( 1)( 4)

x x x

x x

Long division:3 2 2

2 2

2 5 11 13 3 3 52

( 1)( 4) ( 1)( 4)

x x x x x

x x x x

2

2 2

3 3 5 1 2 12 2

( 1)( 4) 1 4

x x x

x x x x

Partial fractions:

1

20

1 2 1(2 )

1 4

xdx

x x

1

2 20

1 2 1(2 )

1 4 4

xdx

x x x

Example 4. Given that , show that

FP2 Calculus 1 Inverse Trig functions

3

4 12

2 1(1 ) siny x x x

1

2

sin

(1 )

dy x x

dx x

14 3 1

20

2 sin (2 )

(1 4 )

x xdx

x

Hence or otherwise, evaluate

1 1

2 2

sin sin1 1

(1 ) (1 )

dy x x x x

dx x x

Use product rule

134

43 1

2 1

2 00

2 sin (2 ) 12 (1 4 ) sin (2 )

2(1 4 )

x xdx x x x

x

Example 5. Given that , derive the result

FP2 Calculus 1 Inverse Trig functions1tanz x 2

1

1

dz

dx x

1(tan ( ))in terms of , and .

d xy dyx y

dx dx

Hence express

1 1 1 1112tan tan tan ( )x y xy

Given that x and y satisfy the equation

21 2 2

2

tan 1 tantan s 1

1

1

ecdx

x z zdx

z x zdz

xdzdz

dx x

Prove that, when x = 1, 31

2

dy

dx

1

2 2

(tan ( ))

1

dyy xd xy dx

dx x y

1 1 1 11

12tan tan tan ( )x y xy

When x = 1 1 1 1 1112

1 1112

tan 1 tan tan ( )

2 tan 34

y y

y y

2 2 2 2

1 10

1 1 1

dyy xdy dx

x y dx x y

DifferentiateDifferentiate

1 1 1 1112tan tan tan ( )x y xy

Sub Sub xx and and y:y: 31

2

dy

dx

Example 6

1siny x(i) Given that derive the result 2

1.

(1 )

dy

dx x

(ii) Find 2(1 ).d

xdx

(iii) Using the above result, find1

1

0

sin x dx1 2

2 2

si(i) sin cos 1 sin

1 1

1 sin 1

ndx

y x y ydy

y

dy

dx y x

x

(ii)2

2(1 )

(1 )

d xx

dx x

(iii) 1

1

0

sin x dx1

2

sin 1

1

(1 )

dvu x

dxdu

v xdx x

1 11 1

20 0

sin sin(1 )

xx dx x x dx

x

11 2

0sin (1 1

2x x x

Example 71

xy

Given that show that

2 2

1 1

( 1) (1 )dx dy

x x y

Find 2

1

( 1)dx

x x

2 2

1 11dx

dy yx dx dy

y y

2

22 1 2

1 1 1( )

( 1)( 1) (1 )y

ydx dy dy

yx x y

2

1 1sin( )

( 1)dx

xx x