Further Pure 2 Calculus

Chapter Assessment 1. Find the exact value of each of the following:

(i) /4 2

0sin 2 dx x

π

∫ [5]

(ii) /6 3

0cos dx x

π

∫ [5]

(iii) 1.5

21.5

1 d9 2

xx− −

∫ [5]

(iv) 7.5

22.5

1 d4 75

xx +∫ [5]

2. (a) Find the exact value of ∫ −

1

0 d

²341 x

x. [5]

(b) Solve the differential equation ²41dd yxy

+= , given that y = 0 when x = 4.

Express y in terms of x. [7]

3. (a) Find the exact value of 0.4

20.4

1 d25 4

xx− +∫ . [5]

(b) (i) Differentiate arcsin2x⎛ ⎞

⎜ ⎟⎝ ⎠

with respect to x (where 0 < x < 2),

simplifying your answer as much as possible. [2]

(ii) Using integration by parts, show that 3

0arcsin d 1

2 3x x π⎛ ⎞ = −⎜ ⎟

⎝ ⎠∫ .

[6]

4. Use the substitution 3 2 tanx θ= to show that ( )

32

2

0 2

1 1d83 4

xx

=+

∫ . [6]

5. (i) Sketch the graph of ( )arccos 2y = x

). [3]

(ii) Differentiate (arccos 2x with respect to x. [2]

(iii) Use integration by parts to find ( )arccos 2 dx x∫ . [4] Total 60

© MEI, 07/06/05 1/5

Further Pure 2 Calculus

Solutions to Chapter Assessment 1. (i) ( )

[ ]

π π

π

π π

π

= −

= −

⎛ ⎞= × − −⎜ ⎟⎝ ⎠

=

∫ ∫/4 / 42 1

20 0

/ 41 12 8 0

sin 2 d 1 cos 4 d

sin 4

1 1sin 0

2 4 8

8

x x x

x x

x

(ii) ( )

( )

( )

π π

π

π

= −

= −

= −⎡ ⎤⎣ ⎦= − −

= − =

∫ ∫∫

/6 / 63 2

0 0

/ 6 2

0

/ 6313 0

31 1 12 3 2

1 1 112 24 24

cos d cos 1 sin d

cos cos sin d

sin sin

0

x x x x x

x x x

x x

x

(iii) − −

=− −∫ ∫

1.5 1.5

2 9 21.5 1.52

1 1 d d

9 2 2x x

x x

π π

π

−

⎡ ⎤= ⎢ ⎥

⎣ ⎦

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

1.5

1.5

1 2arcsin

2 3

1 2arcsin arcsin

2 2 2

1

2 4 4

2 2

x

−2

(iv) =+ +∫ ∫

7.5 7.5

2 2 752.5 2.54

1 1 1 d d

4 75 4x x

x x

π π

π

⎡ ⎤⎛ ⎞= ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞⎛ ⎞= − ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎛ ⎞= −⎜ ⎟⎝ ⎠

=

7.5

2.5

1 2 2arctan

4 5 3 5 3

1 3arctan arctan

10 3 3 3

1

10 3 3 6

60 3

x

1

© MEI, 07/06/05 2/5

Further Pure 2

2. (a) =− −∫ ∫

1 1

40 03

1 1d d

4 3 ² 3 ²x x

x x.

π

π

⎡ ⎤= ⎢ ⎥⎣ ⎦

⎛ ⎞= −⎜ ⎟

⎝ ⎠⎛ ⎞= −⎜ ⎟⎝ ⎠

=

1

0

1 3arcsin

3 2

1 3arcsin arcsin 0

3 2

10

3 3

3 3

x

(b) = +d

1 4 ²d

yy

x

( )( )

=+

= ++

× =

= +

∫ ∫

∫

2

214

12

1d d

1 4

1 1

4

12 arctan 2

4

arctan 2

y xy

dy x cy

+y x c

y x c

When x = 4, y = o ⇒ =

⇒ = −

12 arctan 0 4

4

c

c

+

( )( )

( )

( )

= −

= −

= −

= −

12

12

arctan 2 4

arctan 2 2 8

2 tan 2 8

tan 2 8

y x

y x

y x

y x

3. (a) − −

=+ +∫ ∫

0.4 0.4

2 2 40.4 0.425

1 1 1d d

25 4 25x x

x x

( )( )

π π

π

−

⎡ ⎤⎛ ⎞= × ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

= −

⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

0.4

0.4

1 5 5arctan

25 2 2

1arctan 1 arctan 1

10

1

10 4 4

20

x

−

© MEI, 07/06/05 3/5

Further Pure 2

(b) (i) ( )

⎛ ⎞⎛ ⎞ = ×⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ −

=−

=−

2

22

4

2

d 1arcsin

d 2 2 1

1

2 1

1

4

x

x

xx

1

x

(ii) ⎡ ⎤⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ −∫ ∫3

3 3

20 00

arcsin d arcsin d2 2 4

x x xx x x

x

( )

π

π

⎡ ⎤⎛ ⎞= + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞= + −⎜ ⎟

⎝ ⎠

= × + −

= −

12

32

0

arcsin 42

33 arcsin 1 0 4

2

3 1 23

13

xx x

−

4. θ=3 2 tanx θ

θ θ

⇒ =

⇒ + = + =

2 2

2 2

3 4 tan

3 4 4 tan 4 4sec

x2x

θ=3 2 tanx θ θ θθ

⇒ = ⇒ =2 2d 23 2 sec d sec d

d 3

xx

When θ= =0, 0x

When πθ θ= = ⇒2, tan 33

x =

( ) ( )

[ ]

π

π

π

π

π

θ θθ

θ θθ

θθ

θ θ

θ

= ×+

= ×

=

=

=

= ×

=

∫ ∫

∫

∫

∫

3 32 2

2 / 3 2

2 20 0

/ 3 230

/ 3

0

/ 3

0

/ 3

0

1 1 2d s

33 4 4sec

1 2sec d

8 sec 31

d4 3 seccos

d4 3

1sin

4 3

1 3

4 3 21

8

xx

ec d

© MEI, 07/06/05 4/5

Further Pure 2 5. (i)

( )= arccos 2y x

(ii) ( )( )( )

−= ×

−−

=−

2

2

d 1arccos 2 2

d 1 2

2

1 4

xx x

x

(iii) ( ) ( )

( )

( ) ( )( )

−= −

−

= +−

= − −

= − −

+

+

∫12

2

2

212

212

2arccos 2 d arccos 2 d

1 42

arccos 2 d1 4

arccos 2 1 4

arccos 2 1 4

∫ ∫x

x x x x xx

xx x x

x

x x x c

x x c

x

© MEI, 07/06/05 5/5