unit 2 chapter 4 answers

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Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013

Chapter 4 Integration Try these 4.1

(a) 5 2 5 21d5

x xe x e c− −= +∫

(b) 2 7 2 71d7

x xe x e c− −= − +∫

(c) π 1cos 3 d sin 32 3 2

x x x cπ − = − + ∫

(d) 1tan 5 d ln sec 5

2 5 2x x x c π π + = + +

∫

(e) 7 7

7 1 2 22 d7 ln 2 7 ln 2

x xx x c c

= + = +

∫

(f) 55 dln 5

xx x c= +∫

Try these 4.2

(a) 3

44

1d ln( 5)5 4

x x x cx

= + ++∫

(b) 22

1d ln( 1)1 2

x x x cx

= − +−∫

(c) cos d ln(sin )sin

x x x cx

= +∫

(d) 22 2

3 1 1 6 2 1d d ln 3 2 13 2 1 2 3 2 1 2

x xx x x x cx x x x

+ += = + +

+ + + +∫ ∫

Try these 4.3

(a) 12 2

2d (1 ) d

1x x x x x

x−= +

+∫ ∫

12 21 2 (1 ) d

2x x x−= +∫

12 21 (1 )

12 2

x c−+

= +

21 x c= + +

(b) 8

7 cossin cos d8

xx x x c= − +∫

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(c) 2 32 3

2 1 1d (4 2)(2 2 3) d(2 2 3) 2

x x x x x xx x

−+= + + +

+ +∫ ∫

2 21 (2 2 3)

2 2x x c

−+ += +

−

2 2

14(2 2 3)

cx x

= − ++ +

Try these 4.4

(a) 1 1sin sin

2

1 d1

x x

e x e cx

− −

= +−

∫

(b) 1 1tan tan

2

1 d1

x xe x e cx

− −

= ++∫

(c) 3 3 12 1 1d

3x xx e x e c

++ = +∫

(d) cos cossin dx xxe x e c= − +∫

Exercise 4A

1 7 71d7

x xe x e c= +∫

2 4 2 4 21d4

x xe x e c+ += +∫

3 5 2 5 21d2

x xe x e c− −−= +∫

4 1 1d ln 4 54 5 4

x x cx

= + ++∫

5 3 3d ln 7 27 2 7

x x cx

= − +−∫

6 2 2d ln 4 34 3 3

x x cx

−= − +−∫

7 1tan 2 ln sec 24 2 4

x x cπ π + = + + ∫

8 2 1sec 3 d tan 32 3 2

x x x cπ π − = − − + ∫

9 1 1d cos 2 d sin 24 2 4sec 2

4

x x x x cx

π π = − = − + π −

∫ ∫

10 1 d sin ( 2) d cos ( 2)cosec ( 2)

x x x x cx

= + = − + ++∫ ∫

11 22

1 1d sec (3 1) d tan (3 1)cos (3 1) 3

x x x x cx

= + = + ++∫ ∫

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12 3 3 32 26 d 2 3 d 2x x xx e x x e x e c= = +∫ ∫

13 cos cos cossin d sin dx x xx e x xe x e c= − − = − +∫ ∫

14 1 1d 2 d 22

x x xe x e x e cx x

= = +∫ ∫

15 2 2 21 1d 2 d

2 2x x xxe x xe x e c− − −−= − − = +∫ ∫

16 3 2 6 4 2 6 4 21 1 1( ) d 2 d6 2 2

x x x x x x x xe e x e e e x e e e c− = − + = − + +∫ ∫

17 22 2

1 2 1d d ln 99 2 9 2

x xx x x cx x

= = + ++ +∫ ∫

18 cos 1 2cos 1d d ln 2sin 12sin 1 2 2sin 1 2

x xx x x cx x

= = + ++ +∫ ∫

19 24sec d 2 ln 2 tan 5

2 tan 5x x x c

x= − +

−∫

20 2

33

2 2d ln 55 3

x x x cx

−= − +−∫

21 3

33

1d ln 11 3

xx

xe x e c

e= + +

+∫

22 [ ]2

2

arcsin 1d arcsin21

x x x cx

= +−∫

23 2 2 1tan 3 1 d sec 3 d sec (3 ) d ln sec(3 ) tan(3 )3

x x x x x x x x c+ = = = + +∫ ∫ ∫

24 2

1 1 1sin d cosx cx x x

= + ∫

25 5

4 cossin cos d5

xx x x c−= +∫

26 sin 4 sin 41cos4 d4

x xe x x e c= +∫

27 3 3 32 21 1d 3 d

3 3t t tt e t t e t e c= = +∫ ∫

28 1

12

200

1 1 1 1 10d ln 9 ln 10 ln9 ln9 2 2 2 2 9

x x xx

= + = − = + ∫

Try these 4.5

(a) 21

0dxxe x∫

2Let u x= d 2 du x x=

1 d d2

u x x∴ =

When 0, 0x u= =

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When 1, 1x u= =

21 1 1

1 12 2 00 0

d dx u uxe x e u e ∴ = = ∫ ∫

1 1 12 2 2 ( 1)e e= − = −

21

120

d ( 1)xxe x e∴ = −∫

(b) 1

20

4 1 d( 2)

x xx

++∫

Let 2u x= + d du x= Since 2 2u x x u= + ⇒ = − 4 1 4( 2) 1 4 7x u u∴ + = − + = −

2 2( 2)x u+ =

When 0, 2When 1, 3

x ux u= == =

1 3

2 20 2

4 1 4 7d d( 2)

x ux ux u

+ −∴ =

+∫ ∫

3

22

4 7 duu u

= −∫

3

2

7 7 74ln 4ln 3 4ln 23 2

uu

= + = + − +

3 74ln2 6

= −

Exercise 4B

1 3 d(4 2)

x xx +∫

u = 4x + 2 du = 4 dx

1 d d4

u x=

x = 24

u −

3 3

1 1 2d d(4 2) 4 4

x ux ux u

−= ×

+∫ ∫

2 31 2 ) d16

u u u− −= −∫

2

1 1 1 C16 u u

− = + +

2

1 1 116 4 2 (4 2)

cx x

= − + + + +

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2 2

d6 8

x xx +∫

u = 6x2 + 8 du = 12x dx

=1 d d12

u x x

2

1 1d d126 8

x x uux

=+∫ ∫

1/ 21 d12

u u−= ∫

1/211122

u c= +

2 1/21 (6 8)6

x c= + +

3 3

31d

(2 1)x x

x −∫

u = 2x – 1 du = 2 dx

1 d d2

u x=

x = 1 ( 1)2

u +

x = 3, u = 6 – 1 = 5 x = 1, u = 2 – 1 = 1

3 5

3 31 1

1 1d d(2 1) 4

x ux ux u

+=

−∫ ∫

5

2 31

1 1 1 d4

uu u

= +∫

5

21

1 1 14 2u u = − −

1 1 1 114 5 50 2 = − − − − −

1 1 1 34 5 50 2 = − − +

1 10 1 75 1 64 84 50 4 50 25

− − + = = × =

4 22

60

3 d1

x xx+∫

y = x3 dy = 3x2 dx 1 + x6 = 1 + y2

x = 2, y = 23 = 8 x = 0, y = 03 = 0

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22

60

3 d1

x xx+∫

8

20

1 d1

yy

=+∫

81

0tan ( )y− =

1tan (8) 1.446−= =

5 2

1

3 1 d3 2x xx+−∫

u = 3x – 2 du = 3 dx

1 d d3

u x=

u + 3 = 3x + 1 x = 2, u = 6 – 2 = 4 x = 1, u = 3 – 2 = 1

2 4

1/ 21 1

3 1 1 3d d33 2

x ux uux

+ +=

−∫ ∫

4

1/ 2 1/ 2

1

1 3 d3

u u u− = + ∫

4

3 / 2 1/ 2

1

1 2 63 3

u u = +

3/2 1/21 2 2(4) 6(4) 63 3 3 = + − +

1 16 212 63 3 3 = + − −

329

=

6 4

2

09 dx x x+∫

u = x2 + 9 du = 2x dx

1 d d2

u x x=

x = 0, u = 9 x = 4, u = 42 + 9 = 25

4

2

09 dx x x+∫

251/ 2

9

1 d2

u u= ∫

25

3/2

9

13

u =

3 / 2 3 / 21 25 93 = −

[ ]1 125 273

= −

983

=

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7 3

4 2

8 12 2 d3

x x xx x x+ +

+ +∫

u = x4 + 3x2 + x du = (4x3 + 6x + 1) dx 2 du = (8x3 + 12x +2) dx

3

1/ 2

4 2

8 12 2 2dd 2 d3

x x ux u uux x x

−+ += =

+ +∫ ∫ ∫

= 4u1/2 + c = 4 (x4 + 3x2 + x)1/2 + c

8 4 d1

x xx+∫

u = x2

du = 2x dx

1 d 2 d2

u x x=

14 2

1 1 1d d tan ( )1 2 1 2

x x u u cx u

−= = ++ +∫ ∫

1 21 tan ( )2

x c−= +

9 sin cos 1 dx x x+∫

u = cos x + 1 du = –sin x dx

1/2 3/22d3

u u u c− = − +∫

3/22 (cos 1)3

x c= − + +

10 + −∫ 5(2 1) (4 1) dx x x

u = 4x – 1 du = 4 dx

1 d d4

u x=

14

u x+=

2x + 1 = 1 12 14 4

u + +

1 32 2

u= +

( ) + − = + ∫ ∫5 51 1 3(2 1) (4 1) d d

4 2 2x x x u u u

6 51 3 ) d8

u u u= +∫

7 61 3

8 7 6u u c

= + +

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7

61 (4 1) 1 (4 1)8 7 2

x x c −= + − +

11 1 22

1 (tan ) d1

x xx

−

+∫

u = tan–1 x

du = 2

1 d1

xx +

2du u∫ 3

3u c= +

1 3(tan )

3x c

−

= +

12 43 5 dxx e x+∫

u = x4 + 5 du = 4x3 dx

31 d d4

u x x=

3 4 5 1d d4

x ux e x e u+ =∫ ∫

14

ue c= +

4 514

xe c+= +

13 3

4 d1

x xx+∫

u = 1 + x4

du = 4x3 dx

31 d d4

u x x=

3

44

1 1 1 1d d ln ln (1 )1 4 4 4

x x u u c x cx u

= = + = + ++∫ ∫

14 24 39 dxx e x−∫

u = 4 – 3x2

du = –6x dx

1 d d6

u x x− =

24 3 99 d d

6x uxe x e u− −=∫ ∫

32

ue c= − +

24 33

2xe c−= − +

15 8

2( 4) dx x x+∫

u = x2 + 4 du = 2x dx

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1 d d2

u x x=

2 8 81( 4) d d2

x x x u u+ =∫ ∫

9118

u c= +

2 91 ( 4)18

x c= + +

16 2/3

1/3 d1

x xx+∫

u = 1 + x1/3

du = 2 / 31 d3

x x−

3x2/3 du = dx x1/3 = u – 1 x2/3 = (u – 1)2

3(u – 1)2 du = dx x1/3 = u – 1

2/3 2 2

1/3

( 1) ( 1)d 3 d1

x u ux ux u

− −=+∫ ∫

=2 2( 2 1) ( 2 1)3 du u u u u

u− + − +

∫

=4 3 2 3 2 2( 2 2 4 2 2 1)3 du u u u u u u u u

u− + − + − + − +

∫

=4 3 2( 4 6 4 1)3 du u u u u

u− + − +

∫

= 3 2 13 4 6 4 du u u uu

− + − + ∫

4 3 21 43 3 4 ln4 3

u u u u u c = − + − + +

( ) ( ) ( )4 3 21/ 3 1/ 3 1/ 33 1 4 1 9 14

x x x= + − + + + ( ) ( )1/3 1/312 1 ln 1x x c− + + + +

17 2

5

sec 4 d(1 3tan 4 )

x xx−∫

u = 1 – 3 tan 4x du = (–12 sec2 4x) dx

2

5 5

sec 4 1 1d d(1 3tan 4 ) 12

x x ux u

= −−∫ ∫

41

12 4u c−

= − + −

4148

u c−= +

( ) 41 1 3tan 448

x c−= − +

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18 1sin ( )

2

3 d1

xe xx

−

−∫

u = sin–1(x)

du = 2

1 d1

xx−

3 d 3u ue u e c= +∫

1sin x3 e c−

= +

19 2cos d2

x x xπ + ∫

u = x2 + 2π

du = 2x dx

1 d d2

u x x=

2 1 1cos d cos d sin2 2 2

x x x u u u cπ + = = + ∫ ∫

21 sin2 2

x cπ = + +

20 2

2 1 d1

x xx x

++ −∫

u = x2 + x – 1 du = (2x + 1) dx

2

2 1 1d d ln1

x x u u cx x u

+ = = ++ −∫ ∫

2ln 1x x c= + − +

21 cos 3 d4 sin3

x xx+∫

u = 4 + sin 3x du = 3 cos 3x dx

1 d cos3 d3

u x x=

cos 3 d4 sin3

x xx+∫

1 1 d3

uu

= ∫

1 ln3

u c= +

1 ln 4 sin33

x c= + +

Try these 4.6 (a) cos dx x x∫

d, cosd

vu x xx

= =

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d 1, sindu v xx= =

cos d sin sin dx x x x x x x= −∫ ∫

sin cosx x x c= + +

(b) 2 2

0sin dx x x

π

∫

2 d, sind

vu x xx

= =

d 2 , cosdu x v xx= = −

π/2 π/2

π/22 2

00 0

sin d cos 2 cos dx x x x x x x x = − + ∫ ∫

[ ] π/22

0cos 2 sin cosx x x x x = − + +

[ ]2π π π π πcos 2 sin 2cos 2

4 2 2 2 2

= − + + −

/22

π 2

Hence sinπ 2o

x x dxπ

= −

= −∫

(c) 1sin dx x−∫

1 dsin , 1dvu xx

−= =

2

d 1 ,d 1u v xx x= =

−

1 1

2sin d sin d

1xx x x x

x− −= −

−∫ ∫

11 2 2sin (1 ) dx x x x x−−= − −∫

11 2 21sin 2 (1 ) d

2x x x x x−−= − −∫

1 2sin ( ) 1x x x c−= + − +

1 1 2Hence sin sin 1x dx x x x c− −= + − +∫

Exercise 4C

1 2

1ln dx x x∫

u = ln x, dd

v xx=

2d 1 1,d 2

u v xx x= =

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2

2 22

1 11

1 1ln d ln d2 2

x x x x x x x = − ∫ ∫

2

2 2

1

1 1ln2 4

x x x = −

1 1(2 ln 2 1) ln 12 4

= − − −

32ln 24

= −

2 2 cos dx x x∫

u = x2, d cosd

v xx=

d 2 , sind

u x v xx= =

2 2cos d sin 2 sin dx x x x x x x x= −∫ ∫

sin dx x x∫

u = x, d sind

v xx=

d 1, cosd

u v xx= = −

sind cos cos dx x x x x x= − +∫ ∫

= –x cos x + sin x 2 2cos d sin cos sinx x x x x x x x c= + − +∫

3 1/ 2 ln dx x x∫

u = ln x, 1/ 2dd

v xx=

3 / 2d 1 2,d 3

u v xx x= =

1/ 2 3 / 2 1/ 22 2ln d ln d3 3

x x x x x x x= −∫ ∫

3/2 3/22 4ln3 9

x x x c= − +

4 / 2

0sin2 dx x x

π

∫

u = x, d sin2d

v xx=

d 11, cos2d 2u v xx= = −

/2

/2 /2

0 00

1 1sin 2 d cos2 cos2 d2 2

x x x x x x xπ

π π = − + ∫ ∫

π/2

0

1 1cos2 sin 22 4

x x x = − +

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1cos sin4 4π= − π + π

4π=

5 2 dxxe x∫

u = x, 2dd

xv ex=

2d 11,d 2

xu v ex= =

2 2 2 2 21 1 1 1d d2 2 2 4

x x x x xxe x xe e x xe e c= − = − +∫ ∫

6 2 ln dx x x∫

u = ln x, 2dd

v xx=

3d 1 1,d 3

u v xx x= =

2 3 21 1ln d ln d3 3

x x x x x x x= −∫ ∫

3 31 1ln3 9

x x x c= − +

7 3 ln dx x x∫

u = ln x, 3dd

v xx=

4d 1 1,d 4

u v xx x= =

3 4 3 4 41 1 1 1ln d ln d ln4 4 4 16

x x x x x x x x x x c= − = − +∫ ∫

8 3 arctan dx x x∫

u = arctan x, 3dd

v xx=

4

2

d 1 ,d 1 4

u xvx x= =

+

4

3 42

1 1arctan d arctan d4 4 1

xx x x x x xx

= −+∫ ∫

2

2 4

4 2

2

2

11

1

1

xx x

x x

xx

−+

+

−

− −

3 4 22

1 1 1arctan d arctan 1 d4 4 1

x x x x x x xx

∴ = − − ++∫ ∫

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4 31 1 1arctan arctan4 4 3

x x x x x c = − − + +

4

31 1arctan4 12 4

x xx x c−= − + +

9 1/21 1

0 01 d (1 ) dx x x x x x+ = +∫ ∫

u = x, 1/ 2d (1 )d

v xx= +

3 / 2d 21, (1 )d 3u v xx= = +

1

1 11/2 3/2 3/2

0 00

2 2(1 ) d (1 ) (1 ) d3 3

x x x x x x x + = + − + ∫ ∫

1

3/2 5/2

0

2 4(1 ) (1 )3 15

x x x = + − +

3 / 2 5 / 22 4 4(2) (2)3 15 15

= − +

4 16 42 23 15 15

= − +

( )4 4 42 2 115 15 15

= + = +

10 2 3 dxx e x−∫

u = x2, 3dd

xv ex

−=

3d 12 ,d 3

xu x v ex

−= = −

2 3 2 3 31 2d d3 3

x x xx e x x e xe x− − −= − +∫ ∫

3 dxxe x−∫

u = x, 3dd

xv ex

−=

3d 11,d 3

xu v ex

−= = −

3 3 31 1d d3 3

x x xxe x x e e x− − −= − +∫ ∫

3 31 1 C3 9

x xxe e− −−= − +

∴ 2 3 2 3 3 31 2 2d3 9 27

x x x xx e x x e xe e c− − − −= − − − +∫

3 21 2 23 3 9

xe x x c− = − + + +

11 2

2

1ln dx x x∫

u = ln x, 2dd

v xx=

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3d 1 1,d 3

u v xx x= =

2

2 22 3 2

1 11

1 1ln d ln d3 3

x x x x x x x = − ∫ ∫

2

3 3

1

1 1ln3 9

x x x = −

8 8 1ln 23 9 9

= − +

8 7ln 23 9

= −

12 arccos (2 ) dx x∫

u = arccos(2x), d 1d

vx=

2

d 2 ,d 1 4

u v xx x

−= =

−

2

2arccos(2 ) d arccos(2 ) d1 4

xx x x x xx

= +−∫ ∫

( )1/22arccos(2 ) 2 1 4 dx x x x x= + −∫

2 1/21arccos(2 ) 8 (1 4 ) d4

x x x x x−= + − − −∫

( )1/221arccos(2 ) 1 42

x x x c= − − +

21arccos(2 ) 1 42

x x x c= − − +

13 2 ln(5 ) dx x x∫

u = ln (5x), 2dd

v xx=

3d 1 1,d 3

u v xx x= =

2 3 21 1ln5 d ln(5 ) d3 3

x x x x x x x= −∫ ∫

3 31 1ln(5 )3 9

x x x c= − +

14 2

1(ln ) d

ex x∫

( )2 dln , 1dvu xx

= =

d 2 ln ,d

u x v xx x= =

( ) ( )2 2

1 11ln d ln 2 ln d

ee ex x x x x x = − ∫ ∫

1

ln de

x x∫

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u = ln x, d 1dvx=

d 1 ,du v xx x= =

[ ]11 1ln d ln l d

e eex x x x x= −∫ ∫

[ ]1ln ex x x= −

∴ 2

1(ln ) d

ex x =∫ ( ) [ ]2

11ln 2 ln

e ex x x x x − −

( )2ln 2 ln 2 2e e e e e= − + + 2e= +

15 / 2

0sin dxe x x

π

∫

u = ex, d sindv xx=

d , cosd

xu e v xx= = −

/2 /2/2

00 0sin d cos cos dx x xe x x e x e x x

π ππ = − + ∫ ∫

/ 2

0cos dxe x x

π

∫

u = ex, d cosdv xx=

d , sind

xu e v xx= =

/2 /2/2

00 0cos d sin sin dx x xe x x e x e x x

π ππ = − ∫ ∫

∴ /2 /2/2

00 0sin d cos sin sin dx x x xe x x e x e x e x x

π ππ = − + − ∫ ∫

/2

/2 /2 /2 0 0

02 sin d cos sin cos0 + sin 0

2xe x x e e e e

ππ π ππ ⇒ = − + − − ∫

/ 2 1eπ= +

/ 2

/ 2

0

1sin d 12

xe x x eπ

π ∴ = + ∫

16 1

0arctan dx x∫

u = arctan x, d 1d

vx=

2

d 1 ,d 1

u v xx x= =

+

[ ]1 11

200 0arctan d arctan d

1xx x x x xx

= −+∫ ∫

1

2

0

1arctan ln 12

x x x = − +

1arctan (1) ln 22

= −

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1 ln 24 2π= −

17 5 3 5 3 1/21 d (1 ) dx x x x x x+ = +∫ ∫

u = x3, 2 3 1/ 2d (1 )d

v x xx= +

( )3 / 22 3d 1 23 , 1d 3 3

u x v xx= = + ×

( )3 / 25 3 1/ 2 3 3 2 3 3 / 22 2(1 ) d 1 (1 ) d9 3

x x x x x x x x+ = + − +∫ ∫

( ) ( )3/2 3/23 3 2 32 2 2 51 3 1 d9 9 5 2

x x x x x= + − × × +∫

( ) ( )3/2 5/23 3 32 41 19 45

x x x c= + − + +

18 / 2

03 cos2 dx x x

π

∫

u = 3x, d cos2d

v xx=

d 13, sin2d 2

u v xx= =

/ 2

/ 2 / 2

0 00

3 33 cos2 d sin 2 sin 2 d2 2

x x x x x x xπ

π π= −∫ ∫

/ 2

0

3 3sin 2 cos22 4

x x xπ= +

3 3 3sin cos cos04 4 4π = π + π −

3 3 34 4 2

= − − = −

19 4 4

1/2

1 1ln d ln dθ θ θ θ θ θ=∫ ∫

1/ 2dln ,dvu θ θθ

= =

3 / 2d 1 2,d 3

u v θθ θ= =

4

4 41/2 3/2 1/2

1 11

2 2ln d ln d3 3

θ θ θ θ θ θ θ = − ∫ ∫

4

3/2 3/2

1

2 4ln3 9θ θ θ = −

( )2 4 4(8) ln 4 83 9 9

= − +

16 28ln 43 9

= −

20 2

2

1ln dx x∫

u = ln x2, d 1d

vx=

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d 2 ,d

u v xx x= =

22 2

2 2

1 11ln d ln 2dx x x x x = − ∫ ∫

22

1ln 2x x x = −

(2 ln 4 4) ( 2)= − − − = 2 ln 4 – 2 = ln 16 – 2 Try these 4.7

(a) 2 d d7 12 ( 3)( 4)x xx x

x x x x=

+ + + +∫ ∫

( 3)( 4) 3 4

x A Bx x x x

≡ ++ + + +

( 4) ( 3)x A x B x∴ ≡ + + + When 4, 4 , 4x B B= − − = − = When 3, 3x A= − − =

3 4

( 3)( 4) 3 4x

x x x x∴ ≡ − +

+ + + +

2

3 4 d7 12 3 4x x

x x x x = − + + + + + ∫ ∫

3ln 3 4ln 4x x c= − + + + +

2 3ln 3 4ln 47 12x dx x x c

x x∴ =− + + + +

+ +∫

(b) 2

2

3 1 d( 1)(2 1)

x x xx x

+ ++ +∫

2

2 2

3 1( 1)(2 1) 1 2 1

x x A Bx Cx x x x

+ + +≡ +

+ + + +

2 23 1 (2 1) ( )( 1)x x A x Bx C x∴ + + ≡ + + + + When 1, 3 3 , 1x A A= − = = When 0, 1 0x A C C= = + ⇒ =

Equating coefficients of 2 : 3 2 1x A B B= + ⇒ =

2

2 2

3 1 1( 1)(2 1) 1 2 1

x x xx x x x

+ +∴ ≡ +

+ + + +

2

2 2

3 1 1d d( 1)(2 1) 1 2 1

x x xx xx x x x

+ + = + + + + + ∫ ∫

2

1 1 4d d1 4 2 1

xx xx x

= ++ +∫ ∫

21ln 1 ln 2 14

x x c= + + + +

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(c) 3

2 d3 2x x

x x+ +∫

2 3

3 2

2

2

33 2

3 23 23 9 6

7 6

xx x x

x x xx xx x

x

−+ +

+ +− −

− − −+

3

2

7 633 2 ( 1)( 2)x xx

x x x x+

∴ ≡ − ++ + + +

7 6

( 1)( 2) 1 2x A B

x x x x+

≡ ++ + + +

7 6 ( 2) ( 1)x A x B x∴ + ≡ + + + When 1, 1x A= − − = When 2, 8 8x B B= − − = − ⇒ =

3

2

1 8d 3 d3 2 1 2x x x x

x x x x∴ = − − +

+ + + +∫ ∫

21 3 ln 1 8ln 22

x x x x c= − − + + + +

Exercise 4D

1 2 2 2d d 1 d 2 lnx xx x x x x cx x x x+ = + = + = + +∫ ∫ ∫

2 2 3 2 4 7 2( 2) 7d d d2 2 2 2 2

x x xx x xx x x x+ − + −= = +− − − −∫ ∫ ∫

72 d 2 7ln 22

x x x cx

= + = + − +−∫

3 5 7 d2 1x xx+−∫

52

2 1 5 75529 2

x x

x

− +

−

5 7 d2 1x xx+−∫

5 19 / 2 d2 2 1

xx

= +−∫

5 19 ln 2 12 4

x x c= + − +

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4 2

d2

x xx −∫

+

−

−

−

2

2

22

2

22 4

4

xx x

x x

xx

2

d2

x xx −∫

42 d2

x xx

= + +−∫

21 2 4ln 22

x x x c= + + − +

5 3 2 d

1x x x

x+ ++∫

− ++ + +

+

− + +

− −

++

2

3

3 2

2

2

21 2

2

2 22 2

0

x xx x x

x x

x xx x

xx

3 2 d

1x x x

x+ ++∫ 2 2 dx x x= − +∫

3 21 1 23 2

x x x c= − + +

6 1 d d( 2) ( 3) 2 3

A Bx xx x x x

= ++ − + −∫ ∫

1 / 5 1 / 5 d2 3

xx x−= ++ −∫

1 1ln 2 ln 35 5

x x c= − + + − +

7 4 d( 3) ( 7)

xx x− −∫ d

3 7A B x

x x= +

− −∫

1 1 d3 7

xx x−

= +− −∫

ln 3 ln 7x x c= − − + − +

8 3 d(2 3)( 1)

x xx x+ +∫

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d2 3 1

A B xx x

= ++ +∫

9 3 9d ln 2 3 3 ln 12 3 1 2

x x x cx x

= − = + − + ++ +∫

9 2 d d5 6 ( 2)( 3)x xx x

x x x x=

+ + + +∫ ∫

d2 3

A B xx x

= ++ +∫

2 3 d2 3

xx x−

= ++ +∫

2ln 2 3 ln 3x x c= − + + + +

10 2

2 d3 8 4

x xx x

+− +∫

2 d(3 2)( 2)

x xx x

+=

− −∫

d3 2 2

A B xx x

= +− −∫

2 1 d3 2 2

xx x−

= +− −∫

2 ln 3 2 ln 23

x x c−= − + − +

11 2

5 2 d6 2

x xx x

−+ −∫

5 2 d(3 2) (2 1)

x xx x

−=

+ −∫

d3 2 2 1

A B xx x

= ++ −∫

16 / 7 1/ 7 d3 2 2 1

xx x

= ++ −∫

16 1ln 3 2 ln 2 121 14

x x c= + + − +

12 2

4 4d d d( 4)( 3) ( 2)( 2)( 3) 2 2 3

x x A B Cx x xx x x x x x x x

= = + +− − − + − − + −∫ ∫ ∫

2 2 / 5 12 / 5 2 12d 2ln 2 ln 2 ln 32 2 3 5 5

x x x x cx x x−= − + = − − − + + − +− + −∫

13 4 2 d d( 1)( 2)( 3) 1 2 3

x A B Cx xx x x x x x

+ = + +− + + − + +∫ ∫

1/ 2 2 5 / 2 d1 2 3

xx x x

= + −− + +∫

1 5ln 1 2ln 2 ln 32 2

x x x c= − + + − + +

14 2

2

8 2 24 d( 4 )( 2)

x x xx x x

+ −+ −∫

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28 2 24 d

( 4) ( 2)x x x

x x x+ −

=+ −∫

d4 2

A B C xx x x

= + ++ −∫

3 4 1 d4 2

xx x x

= + ++ −∫

3ln 4ln 4 ln 2x x x c= + + + − +

15 2

2

3 8 8 d( 2)(2 3 2)

x x xx x x

+ −+ − −∫

23 8 8 d

( 2)(2 1)( 2)x x x

x x x+ −

=+ + −∫

d2 2 1 2

A B C xx x x

= + ++ + −∫

1 3 1 d2 2 1 2

xx x x−= + ++ + −∫

3ln 2 ln 2 1 ln 22

x x x c= − + + + + − +

16 2

2

43 22 3 d(2 – 7 + 3) ( 2)

x x xx x x

− −+∫

243 22 3 d

(2 1)( 3) ( 2)x x x

x x x− −

=− − +∫

d2 1 3 2

A B C xx x x

= + +− − +∫

5 2 3 d2 1 3 2

xx x x−

= − +− − +∫

5 ln 2 1 2ln 3 3 ln 22

x x x c−= − − − + + +

17 2

1 d( 1)

xx x +∫

2 2

1( 1) 1

A Bx Cx x x x

+≡ ++ +

2 2

1 1d d( 1) 1

xx xx x x x

∴ = −+ +∫ ∫

= ln x – 21 ln 12

x c+ +

18 −∫ 3

1 d8

xx

=− + +∫ 2

1 d( 2)( 2 4)

xx x x

2 2

1( 2)( 2 4) 2 2 4

A Bx Cx x x x x x

+≡ +− + + − + +

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+

= −− − + +∫ ∫3 2

1 11 1 / 12 12 3d d

8 2 2 4

xx x

x x x x

1/12 1d ln 22 12

x xx

= −−∫

2 2

1 11 412 3 d d

2 4 12 2 4

x xx xx x x x

+ +=+ + + +∫ ∫

2

1 2 8 d24 2 4

x xx x

+=+ +∫

2 2

1 2 2 1 6d d24 2 4 24 2 4

x x xx x x x

+= ++ + + +∫ ∫

2 2

1 2 2 1 1d d24 2 4 4 2 4

x x xx x x x

+= ++ + + +∫ ∫

22

1 1 1ln 2 4 d24 4 ( 1) 3

x x xx

= + + ++ +∫

( )

22

2

1 1 1ln 2 4 d24 4 ( 1) 3

x x xx

= + + ++ +

∫

2 11 1 1 1ln 2 4 tan24 4 3 3

xx x c− += + + + × +

2 13

1 1 1 3 1d ln 2 ln 2 4 tan8 12 24 12 3

xx x x x cx

− +∴ = − + + + + + − ∫

19 2

2

5 4 4 d( 2) ( 2 2)

x x xx x x

− ++ − +∫

Using the methods from the chapter on partial fractions,

16 9 6, ,5 5 5

A B C −= = =

2

2 2

9 65 4 4 16 / 5 5 5

( 2) ( 2 2) 2 2 2

xx xx x x x x x

−− +∴ ≡ ++ − + + − +

2

2 2

9 65 4 4 16 / 5 5 5d d

( 2)( 2 2) 2 2 2

xx x x xx x x x x x

−− + = ++ − + + − +∫ ∫

= 2

16 1 1 9 6d d5 2 5 2 2

xx xx x x

−+

+ − +∫ ∫

= 2

9 (2 2) 316 1 1 2d d5 2 5 2 2

xx x

x x x

− ++

+ − +∫ ∫

= 2 2

16 1 9 (2 2) 3 1d d d5 2 10 2 2 5 2 2

xx x xx x x x x

−+ +

+ − + − +∫ ∫ ∫

Now 2 2

1 12 2 ( 1) 1x x x

=− + − +

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∴ 2

2 2 2

5 4 4 16 1 9 2 2 3 1d d d d( 2)( 2 2) 5 2 10 2 2 5 ( 1) 1

x x xx x x xx x x x x x x

− + −= + ++ − + + − + − +∫ ∫ ∫ ∫

2 116 9 3ln 2 ln 2 2 tan ( 1)5 10 5

x x x x c−= + + − + + − +

20 24 4 1 d( 1)

x x xx x+ ++∫

+ + +

+

2 2

2

44 4 14 4

1

x x x xx x

∴ + + = ++ +

24 4 1 14( 1) ( 1)

x xx x x x

41

A Bx x

= + ++

= + −+

1 141x x

24 4 1 1 1d 4 d( 1) 1

x x x xx x x x+ + = + −+ +∫ ∫

4 ln ln 1x x x c= + − + +

21 4

2

3 2 d1

x x xx+ −−∫

2

2 4

4 2

2

2

11 3 2

3 2 1

3 1

xx x x

x x

x xx

x

+− + −

−

+ −

−

−

4

22

3 2 3 111 ( 1)( 1)

x x xxx x x+ − −≡ + +− − +

2 11 1

A Bxx x

≡ + + +− +

2 1 211 1

xx x

≡ + + +− +

4

22

3 2 1 2d 1 d1 1 1

x x x x xx x x+ − = + + +− − +∫ ∫

31 ln 1 3 ln 13

x x x x c= + + − + + +

22 3 2

2

2 1 d( 1)

x x x xx x+ + +

+∫

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+ + + +

+

+ +

3 3 2

3

2

12 1

1

x x x x xx x

x x

3 2 2

3 2

2 1 11( 1)

x x x x xx x x x

+ + + + +≡ ++ +

2

2 2

1( 1) 1

x x A Bx Cx x x x+ + +≡ ++ +

2

1 111x x

≡ + ++

+ + + = + ++ +∫ ∫

3 2

3 2

2 1 1 1d 1 d1

x x x x xx x x x

1ln tan ( )x x x c−= + + +

23 2

2

3 3 d2 1

x x xx x+ ++ +∫

+ + + ++ +

+

2 2

2

12 1 3 3

2 12

x x x xx x

x

3

2 2

3 3 2d 1 d2 1 2 1

x x xx xx x x x+ + +

= ++ + + +∫ ∫

2 2 2

2 22 1 ( 1) 1 ( 1)

x x A Bx x x x x

+ += ≡ ++ + + + +

2

1 11 ( 1)x x

≡ ++ +

∴ 3

2 2

3 3 1 1d 1 d2 1 1 ( 1)

x x x xx x x x+ +

= + ++ + + +∫ ∫

= + + − ++1ln 1 C

1x x

x

24 2

2 2

3 3 2(2 1) ( 1) 2 1 1

x x A Bx Cx x x x

+ + +≡ ++ + + +

⇒ 3x2 + 3x + 2 ≡ A (x2 + 1) + (Bx + C) (2x + 1)

When 1 3 3 5, 22 4 2 4

x A= − − + =

5 54 4

A=

A = 1 When x = 0, 2 = A + C ⇒ C = 1 Equating coefficients of x2, 3 = A + 2B ⇒ B = 1

2

2 2

3 3 2 1 1(2 1)( 1) 2 1 1

x x xx x x x

+ + +≡ ++ + + +

+ + = + ++ + + + +∫ ∫21 1

2 2 20 0

3 3 2 1 1d d(2 1)( 1) 2 1 1 1

x x xx xx x x x x

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1

2 1

0

1 1ln 2 1 ln 1 tan ( )2 2

x x x− = + + + +

1 1ln (3) ln(2)2 2 4

π= + +

1 ln 62 4

π= +

25 2

2 2

4 5 6( 2)( 9) 2 9

x x A Bx Cx x x x

+ + +≡ ++ + + +

2 24 5 6 ( 9) ( ) ( 2)x x A x Bx C x⇒ + + = + + + + When x = –2, 16 – 10 + 6 = 13A

1213

A =

When x = 0, 6 = 9A + 2C ⇒ 6 – 10813

= 2C

30 213

C− =

1513

C = −

Equating coefficients of x2, 4 = A + B ⇒ B = 4 – 12 4013 13

=

2

2 2

40 154 5 6 12 / 13 13 13

( 2)( 9) 2 9

xx xx x x x

−+ + = ++ + + +

∴ + ++ +∫

2

2

4 5 6 d( 2)( 9)

x x xx x 2

40 1512 / 13 13 13 d

2 9

xx

x x

−= +

+ +∫

2 2 2

4012 / 13 15 / 1313 d

2 9 3

xx

x x x= + −

+ + +∫

2 112 20 15 1ln 2 ln 9 tan13 13 13 3 3

xx x c− = + + + − +

2 112 20 5ln 2 ln 9 tan13 13 13 3

xx x c− = + + + − +

26 + + ++ +∫

3 2

4 2

2 d3 2

x x x xx x

++ + +

+

+

+

2

2 4 2

4 2

2

2

21 3 2

2 22 2

0

xx x x

x x

xx

3 2

2 2 2 2

2( 1)( 2) 1 2x x x Ax B Cx Dx x x x+ + + + +≡ ++ + + +

3 2 2 22 ( ) ( 2) ( ) ( 1)x x x Ax B x Cx D x⇒ + + + = + + + + +

of 46


Equating coefficients of x3, 1 = A + C [1] Equating coefficients of x2, 1 = B + D [2] Equating coefficients of x, 1 = 2A + C [3] Equating constants, 2 = 2B + D [4] Using [1] and [2], C = 1 ∴ A = 0 Using [3] and [4], D = 0 ∴ B = 1

∴ + + + = ++ + + +∫ ∫

3 2

4 2 2 2

2 1d d3 2 1 2

x x x xx xx x x x

1 21tan ( ) ln 22

x x c−= + + +

27 −+ + −∫

21

20

3 5 d( 1)( 2)

x xx x x

2

2 2

3 5( 1)( 2) 1 2

x Ax B Cx x x x x x

− +≡ ++ + − + + −

2 23 5 ( ) ( 2) ( 1)x Ax B x C x x⇒ − = + − + + + When x = 2, 7 = 7C ⇒ C = 1 Equating coefficients of x2, 3 = A + C ⇒ A = 2 Equating constants –5 = –2B + C ⇒ B = 3

− += ++ + − + + −

2

2 2

3 5 2 3 1( 1)( 2) 1 2

x xx x x x x x

21 1

2 20 0

3 5 2 3 1d d( 1)( 2) 1 2

x xx xx x x x x x

− += ++ + − + + −∫ ∫

+ += ++ + −∫

1

20

2 1 2 1 d1 2

x xx x x

+= + ++ + + + −∫

1

2 20

2 1 2 1 d1 1 2

x xx x x x x

+= + ++ + − + +

∫1

220

2 1 2 1 d1 21 3

2 4

x xx x x

x

1

2 1

0

11 2ln 1 2 tan ln ( 2)3 3

2 2

xx x x−

+

= + + + × + −

1 1

3 14 3 4 32 2ln 3 tan ln 1 ln 1 tan ln 2

3 33 32 2

− −

= + + − − + + −

4 3 4 3ln3 ln 23 3 3 6

π π = + − −

3 4 3ln2 18

= + π

3 2ln 32 9

= + π

of 46


Try these 4.8 (a) 5sin dx x∫ = 4sin sin dx x x∫

= 2 2sin (1 cos ) dx x x−∫

= 2 4sin (1 2cos cos ) dx x x x− +∫

= 2 4sin 2sin cos sin cos dx x x x x x− +∫

= 3 52 1cos cos cos3 5

x x x c− + − +

(b) 5cos dx x∫ = 4cos cos dx x x∫

= 2 2cos (1 sin ) dx x x−∫

= 2 4cos (1 2sin sin ) dx x x x− +∫

= 2 4cos 2cos sin cos sin dx x x x x x− +∫

= 3 52 1sin sin sin3 5

x x x c− + +

(c) 4cos dx x∫ = 21 cos 2 d

2x x+

∫

= 21 1 2cos 2 cos 2 d4

x x x+ +∫

= 1 1 cos 41 2cos 2 d4 2

xx x++ +∫

= 1 3 12cos 2 cos 4 d4 2 2

x x x+ +∫

= 1 3 1sin 2 sin 44 2 8

x x x c + + +

Try these 4.9 (a) 4tan dx x∫ = 2 2tan (sec 1)dx x x−∫

= )2 2 2tan sec tan dx x x x−∫

= )2 2 2tan sec sec 1 dx x x x− +∫

= 31 tan tan3

x x x c− + +

(b) 5tan dx x = 3 2tan tan dx x x∫

= 3 2tan (sec 1) dx x x−∫

= 3 2 3tan sec tan dx x x x−∫

= 3 2 2tan sec tan tan dx x x x x−∫

= 3 2 2tan sec tan (sec 1) dx x x x x− −∫

of 46


= 3 2 2tan sec tan sec tan dx x x x x x− +∫

= 4 21 1tan tan ln sec4 2

x x x c− + +

Try these 4.10 (a) cos 6 sin 3 dx x x∫

2cos 6sin 3x = sin 9 sin 3x x−

cos 6 sin 3x x = 1 1sin 9 sin 32 2

x x−

∴ cos 6 sin 3 dx x x∫ = 1 1sin 9 sin 3 d2 2

x x x−∫

= 1 1cos9 cos3

18 6x x c− + +

(b) cos8 cos 2 dx x x∫

2cos8 cos 2x x = cos10 cos 6x x+

cos8 cos 2 dx x x∫ = 1 cos10 cos 6 d2

x x x+∫

=1 1 1sin10 sin 62 10 6

x x c + +

(c) sin10 sin dx x x∫

2sin10 sinx x− = cos11 cos9x x−

sin10 sin dx x x∴∫ = 1 cos11 cos9 d2

x x x− −∫

= 1 1 1sin11 sin 92 11 9

x x c − − +

Try these 4.11

(a) 12 2 2

1 1 1d d tan9 3 3 3

xx x cx x

− = = + + + ∫ ∫

(b) 12 2 2

1 1 1 5d d tan4 25 2 (5 ) 10 2

xx x cx x

− = = + + + ∫ ∫

(c) 2

4 d9 6 16

xx x+ +∫ =

2

4 d1 1593 9

xx

+ +

∫

= 22

4 1 d9 1 15

3 9

x

x +

∫

of 46


= 1

14 9 3tan9 15 15

9

xc−

+

× +

= 1

14 3tan

3 15 159

xc−

+

+

(d) 2

1 d2

xx x− −

∫

2 22 ( 2 )x x x x− − = − +

21 ( 2 1)x x= − + +

2 2

1 1d d2 1 ( 1)

x xx x x

∴ =− − − +

∫ ∫

1sin ( 1) cx− + + Exercise 4E

1 4

3 2 3 2 (tan )tan sec d (tan ) sec d4xx x x x x x c= = +∫ ∫

2 1 1 1sin 7 sin3 d (cos10 cos4 ) d sin10 sin 42 20 8

x x x x x x x x c= − − = − + +∫ ∫

1 1sin 4 sin108 20

x x c= − +

3 (a) 2

1 d25 4

xx +∫

2

1 1 d42525

xx

=+

∫

22

1 1 d25 2

5

xx

= +

∫

11 5 tan 225 2 5

x c− = × +

11 5tan10 2

x c− = +

(b) 2

1 d16 9

xx +∫

of 46


2

1 1 d91616

xx

=+

∫

22

1 1 d16 3

4

xx

= +

∫

11 4 tan16 3 3 / 4

x c− = × +

11 4tan12 3

x c− = +

(c) 2 2

1 1 1d d2 6 2 3

x xx x

=+ +∫ ∫

( )2

2

1 1 d2 3

xx

=+

∫

11 tan2 3 3

x c− = +

4 2 4cos sin dx x x∫

21 cos2 1 cos2 d

2 2x x x+ − = ∫

21 (1 cos2 ) (1 2cos2 cos 2 ) d8

x x x x= + − +∫

2 2 31 1 2cos 2 cos 2 cos 2 2cos 2 cos 2 d8

x x x x x x= − + + − +∫

2 31 1 cos2 cos 2 cos 2 d8

x x x x= − − +∫

21 1 1 cos4 11 cos2 d d cos2 (1 sin 2 ) d8 8 2 8

xx x x x x x+= − − + −∫ ∫ ∫

31 1 1 1 1 1 sin 2 1 sin 2sin 2 sin 4

8 2 8 2 8 8 2 8 2 3x xx x x x c = − − + + − + ×

31 1 1sin 4 sin 216 64 48

x x x c= − − +

5 / 4

2 4

0tan sec dx x x

π

∫

/ 4

2 2 2

0tan (1 tan ) sec dx x x x

π= +∫

/4

2 2 4 2

0tan sec tan sec dx x x x x

π= +∫

/ 4

3 5

0

1 1tan tan3 5

x xπ= +

( ) ( )3 51 1tan tan4 43 5π π= +

1 13 5

= +

of 46


815

=

6 2

1 d6 13

xx x+ +∫

2

1 d( 3) 4

xx

=+ +∫

2 2

1 d( 3) 2

xx

=+ +∫

11 3tan2 2

x c− + = +

7 24 dx x−∫

x = 2 sinθ so θ = sin-1

2x

dx = 2 cosθ dθ 2 24 4 4sinx θ− = −

24(1 sin )θ= −

24cos θ= = 2 cosθ 24 d 2cos (2cos ) dx x θ θ θ− =∫ ∫

24 cos dθ θ= ∫

1 cos24 d2

θ θ+= ∫

2 1 cos2 dθ θ= +∫

12 sin 22

cθ θ = + +

[ ]2 sin cos cθ θ θ= + +

2

1 42 sin2 2 2x x x c−

− = + +

8 (a) 1cos8 cos6 d (cos14 cos2 ) d2

x x x x x x= +∫ ∫

1 1sin14 sin 228 4

x x c= + +

(b) 1sin7 cos3 d (sin10 sin 4 ) d2

x x x x x x= +∫ ∫

1 1cos10 cos420 4

x x c= − − +

(c) 1cos6 sin2 d (sin8 sin 4 ) d2

x x x x x x= −∫ ∫

1 1cos8 cos416 8

x x c= − + +

of 46


9 1

20

1 d4 4 10

xx x+ +∫

4x2 + 4x + 10 = 4(x2 + x) + 10

214 10 1

2x = + + −

214 9

2x = + +

1 1

220 02

1 1d d4 4 10 14 3

2

x xx x

x=

+ + + +

∫ ∫

1

2 20

1 1 d4 1 3

2 2

xx

= + +

∫

1

1

0

11 2 2tan 34 3

2

x−

+ = ×

1 11 1 1tan (1) tan6 6 3

− − = −

= 0.077 10 (a) –x2 + 2x + 3 = – (x2 – 2x) + 3 = – (x – 1)2 + 3 + 1 = 4 – (x – 1)2

2 2

1 1d d3 2 4 ( 1)

x xx x x

=+ − − −∫ ∫

1 1sin2

x c− − = +

(b) –x2 – 4x + 5 = – (x2 + 4x) + 5 = – (x + 2)2 + 5 + 4 = 9 – (x + 2)2

2 2

1 1d d5 4 9 ( 2)

x xx x x

=− − − +∫ ∫

1 2sin3

x c− + = +

(c) –x2 – 6x + 7 = – (x2 + 6x) + 7 = – (x + 3)2 + 7 + 9 = 16 – (x + 3)2

2 2

1 1d d7 6 16 ( 3)

x xx x x

=− − − +∫ ∫

= sin–1 34

x c+ +

11 2 2

1 d4

xx x+∫

x = 2 tanθ

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dx = 2 sec2θ dθ x2 = 4 tan2θ 2 24 4 4 tanx θ+ = +

24 (1 tan )θ= +

24sec θ= = 2 secθ

2 2

1 d4

xx x+∫

22

1 2sec d4 tan (2sec )

θ θθ θ

= ∫

2

1 sec d4 tan

θ θθ

= ∫

2

2

1 1 cos d4 cos sin

θ θθ θ

= ×∫

2

1 cos d4 sin

θ θθ

= ∫

2

2 2

1 1d cos (sin ) d44

xx x

θ θ θ−=+∫ ∫

11 (sin )

4 1cθ −

= +−

14sin

cθ

= − +

tan θ = 2

, so sin2 4x x

xθ =

+

2

2 2

1 1 4d44

xx cxx x+∴ = − +

+∫

12 1 cos dx x−∫

2 sin2θ = 1 – cos 2θ

⇒ 2 sin2 2θ

= 1 – cosθ

21 cos d 2sin d2θθ θ θ − = ∫ ∫

2 sin d2θ θ = ∫

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2 2 cos2

cθ = − +

13 21

20d

1x x

x+∫

x = tanθ dx = sec2θ dθ 1 + x2 = 1 + tan2θ = sec2θ When x = 0, tanθ = 0 ⇒ θ = 0

x = 1, tanθ = 1 ⇒ θ = 4π

2 21

2 20

tand1 sec

x xx

θθ

∴ =+∫

/ 42

0sec θ

π

∫ dθ

/ 4

2

0tan dθ θ

π= ∫

/4

2

0sec 1 dθ θ

π= −∫

[ ]π/40tan θ θ= −

tan 14 4 4π π π= − = −

14 sin d1 cos

x xx+∫

Let u = 1 + cos x du = –sin x dx ∴ –du = sin x dx 1 cos x u+ =

sin 1d d1 cos

x x ux u

= −+∫ ∫

12 du u

−= − ∫

1/2

1 / 2u c−= +

1/22u c= − + Since u = 1 + cos x

⇒ sin d 2 1 cos1 cos

x x x cx

= − + ++∫

15 2 2

1 d9

xx x−∫

x = 3 sinθ dx = 3 cosθ dθ x2 = 9 sin2θ

2 29 9 9sinx θ− = −

29 (1 sin )θ= −

29 cos θ=

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= 3 cosθ

2 2

1 3cosd9

xx x

θ∴ =

−∫ 29sin ( 3cosθ θd

)θ∫

2

1 d9sin

θθ

= ∫

21 cosec d9

θ θ= ∫

21 1cosec d cot9 9

cθ θ= − +∫

Since sinθ = 3x ,

29cot

xx

θ−

=

2

2 2

91 1d99

xx c

xx x−

∴ = − +−∫

Review exercise 4 1 (a) 1/ 2(1 ) dx x x+∫

u = x, 1/ 2d (1 )dv xx= +

3 / 2d 21, = (1 )d 3u v xx= +

1/ 2

3 / 2 3 / 22 2(1 ) d (1 ) (1 ) d3 3

x x x x x x x+ = + − +∫ ∫

5/2

3/22 2 (1 )(1 )3 3 5 / 2

xx x c+= + − +

3/2 5/22 4(1 ) (1 )3 15

x x x c= + − + +

(b) (i) / 4

2

0

2cos 4 dx xπ

∫

Using 2 1 cos2cos2

xx +=

2 1 cos8cos 42

xx +=

/4 /4

2

0 0

2cos 4 d 1 cos8 dx x x xπ π

= +∫ ∫

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/4

0

1 sin 88

x xπ

= +

1 sin 24 8π= + π

4π=

(ii) 2

2 2 3

11

1 1(ln ) d (ln )3

x x xx

= ∫

3 31 1(ln 2) (ln1)3 3

= −

31 (ln2)3

=

= 0.111 2 2cosec ( ) dx x x∫

u = x, 2d cosecd

v xx=

d 1, cotdu v xx= = −

2cosec d cot cot dx x x x x x x= − +∫ ∫

= –x cot x + ln sinx + c

3 (a) 2 2 2

11 1

2 1 3d 2 d 2 3 ln 22 2

x x x x xx x+ = − = − + + +∫ ∫

= (4 – 3 ln 4) – (2 – 3 ln 3) = 4 – 3 ln 4 – 2 + 3 ln 3 = 2 + 3(ln 3 - ln 4)

3= 2 + 3 ln4

(b) 1

0

1 d(1 )(2 )

xx x+ −∫

1(1 )(2 ) 1 2

A Bx x x x

≡ ++ − + −

⇒ 1 = A(2 – x) + B (1 + x)

When x = 2, 1 = 3B ⇒ B = 13

When x = –1, 1 = 3A ⇒ A = 13

∴

1 11 3 3

(1 )(2 ) 1 2x x x x= +

+ − + −

1 1

0 0

1 1 1 1d d(1 )(2 ) 3 1 2

x xx x x x

= ++ − + −∫ ∫

[ ]101 ln (1 ) ln (2 )3

x x= + − −

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1

0

1 1 1 2 1 1ln ln ln3 2 3 1 3 2

xx

+ = = − −

1 ln (2 2)3

= ×

1 ln 43

=

4 (a) 2 3 / 2

1 d( 9)

xx −∫

x = 3 secθ

d 3sec tandx= θ θ

θ

dx = 3 secθ tan θ dθ 2 3 / 2 2 3 / 2( 9) (9sec 9)x − = −θ

3 / 229(sec 1) = − θ

= (9 tan2θ)3/2

= 27 tan3θ

∴ 2 3/2 3

1 1d 3sec tan d( 9) 27 tan

xx

θ θ θθ

=−∫ ∫

2

1 sec d9 tan

= ∫θ θθ

2

2

1 1 cos d9 cos sin

= ×∫θ θ

θ θ

21 (sin ) cos d9

−= ∫ θ θ θ

11 (sin )

9 1cθ −

= +−

19sin

cθ

−= +

Now 3sec , so cos

3x

xθ θ= =

2 9sin xx

θ −∴ =

2 3/2 2

1 1( 9) 9 9

x cx x

= − +− −

∫

(b) 2 1/ 2(1 3 ) dx x x+∫ 2 1/ 21 6 (1 3 ) d

6x x x= +∫

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2 3/21 (1 3 )

6 3 / 2x c += +

2 3/21 (1 3 )9

x c= + +

5 1

0

4 d2

x xx−

∫

x = 4 sin2θ dx = 8 sinθ cosθ dθ

2 2

2 2

4 4 4sin 4(1 sin )2 8sin 8sin

xx− − −

= =θ θ

θ θ

2

2

4cos 1 cos8sin sin2

= =θ θθ θ

Change the limits: When x = 0, sinθ = 0 ⇒ θ = 0

When x = 1, sinθ = 12

⇒ θ = 6π

2 / 6

0

4 1 cosd2 sin2

x xx

θθ

π− = ∫1

08 sinθ×∫ cos dθ θ

/6

2

0

8 cos d2

θ θπ

= ∫

/ 6

0

8 1 cos2 d22

θ θπ += ∫

/ 6

0

4 1 sin 222

θ θπ

= +

4 1 sin6 2 32π π = +

32 26 4 π= +

323 2

π= +

6 5

24

2 d5 6

xx x− +∫

2

2 25 6 ( 2)( 3)x x x x

=− + − −

2( 2)( 3) 2 – 3

A Bx x x x

≡ +− − −

∴ 2 = A (x – 3) + B (x – 2) When x = 3, 2 = B When x = 2, 2 = –A, A = –2

5 5

24 4

2 2 2d d5 6 2 3

x xx x x x

−∴ = +− + − −∫ ∫

]54[ 2 ln 2 2ln 3x x= − | − | + | − |

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5

4

32ln2

xx

−= −

2 12 ln 2 ln3 2

= −

2 12ln3 2

= ÷

42 ln3

=

24 16ln ln

3 9 = =

7 (a) 2 1 cos8sin 4 d d2

xx x x+=∫ ∫

1 1 sin82 8

x x c = + +

(b) ln ( 4) dx x+∫

u = ln (x + 4), d 1dvx=

d 1 ,d 4u v xx x= =

+

ln ( 4) d ln ( 4) d4

xx x x x xx

+ = + −+∫ ∫

4ln ( 4) 1 d4

x x xx

= + − −+∫

ln ( 4) 4 ln ( 4)x x x x c= + − + + +

(c) 3 dxxe x∫

u = x, 3dd

xv ex=

= 3d 11, =d 3

xu v ex

3 3 31 1d d3 3

x x xxe x xe e x= −∫ ∫

3 31 13 9

x xxe e c= − +

8 3

1ln d

ex x x∫

u = lnx, 3dd

v xx=

4d 1 1,d 4

u v xx x= =

3 4 3

1 11

1 1ln d ln d4 4

ee ex x x x x x x = − ∫ ∫

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= − − 4 4

1

1 1 1ln ln14 4 16

e

e e x

4 41 1 14 16 16

e e = − −

41 316 16

e= +

41 (1 3 )16

e= +

9 4

04 dx x∫

Let y = 4x

ln y = x ln 4

1 d ln 4dy

y x=

d 4 ln 4d

xyx=

∴ d 4 4 ln 4d

x x

x =

[ ]44

0 04 4 ln 4 dxx x∴ = ∫

4

4 0

04 4 ln 4 4 dx x⇒ − = ∫

[ ]4

0

1 63 4 dln 4

x x= ∫

4

0

634 dln 4

x x∴ =∫

10 2 2

5( 1)( 4) 1 4

A Bx Cx x x x

+≡ ++ + + +

25 ( 4) ( ) ( 1)A x Bx C x∴ = + + + + When 1, 5 5 1x A A= − = ⇒ = When 0, 5 4 1x A C C= = + ⇒ = 2Equating coefficients of , 0 1x A B B= + ⇒ = −

∴ 2 2

5 1 1( 1)( 4) 1 4

xx x x x

−= +

+ + + +

2 2

2 20 0

5 1 1d d( 1)( 4) 1 4

xx xx x x x

−= ++ + + +∫ ∫

2

2 20

1 1 d1 4 4

x xx x x

= + −+ + +∫

2

1 2

0

1 1ln 1 tan ln 42 2 2

xx x− = + − − +

− − = − − − − − 1 11 1 1 1ln 3 tan (1) ln 8 ln 1 tan (0) ln 4

2 2 2 2

1 1ln3 ln8 ln 48 2 2π= − − +

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1 1ln3 ln8 2 2π= − +

3ln82

π= −

11 2 2

1( 1) ( 2) 1 ( 1) – 2

A B Cx x x x x

≡ + +− − − −

∴ 1 = A(x – 1) (x – 2) + B (x – 2) + C (x – 1)2

When x = 1, 1 = –B, B = –1 When x = 2, 1 = C Equating coefficients of x2 0 = A + C ⇒ A = – 1

∴ 2 2

1 1 1 1( 1) ( 2) 1 ( 1) 2x x x x x

= − − +− − − − −

2 2

1 1 1 1d d( 1) ( 2) 1 ( 1) 2

x xx x x x x

= − − +− − − − −∫ ∫

1ln 1 ln 21

x x cx

= − − + + | − | +−

1 ln 2 ln 11

x x cx

= + | − | − − +−

12 3

2 20

1 d( 9)

xx +∫

x = 3 tan θ dx = 3 sec2θ dθ (x2 + 9)2 = (9 tan2θ + 9)2

= [9 (1 + tan2θ)]2

= 81 sec4θ

When x = 3 , tan θ = 33

⇒ θ = 6π

When x = 0, tan θ = 0 ⇒ θ = 0

23 / 6

2 2 40 0

1 3sec dd( 9) 81sec

xx

θ θθ

π=

+∫ ∫

/ 6

20

1 1 d27 sec

θθ

π= ∫

/6

2

0

1 cos d27

θ θπ

= ∫

/ 6 / 6

2

0 0

1 1 1 cos2cos d d27 27 2

θθ θ θπ π +=∫ ∫

/ 6

0

1 1 sin 254 2

θ θπ

= +

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1 1 sin54 6 2 3

π π = +

1 1 354 6 2 2

π= +

1 354 6 4

π= +

13 –x2 – 6x + 16 = – (x2 + 6x) + 16 = – (x2 + 6x + (3)2) + 16 – (–1) (3)2 = 25 – (x + 3)2

⇒ –1

2 2

1 1 3d in56 16 25 ( 3)

xx s cx x x

+ = = + − − + − +∫ ∫

14 7

4

1 d( 4) (7 )

xx x− −∫

x = 4 cos2θ + 7 sin2θ dx = (–8 cosθ sinθ + 14 sinθ cosθ) dθ dx = 6 sinθ cosθ dθ (x – 4) (7 – x) = (4 cos2θ + 7 sin2θ – 4) (7 – 4 cos2θ – 7 sin2θ) = (–4 (1 – cos2θ) + 7 sin2θ) (7 (1 – sin2θ) – 4 cos2θ) = (–4 sin2θ + 7 sin2θ) (7 cos2θ – 4 cos2θ) = (3 sin2θ) (3 cos2θ) = 9 sin2θ cos2θ θ θ θ θ− − = =2 2( 4)(7 ) 9sin cos 3sin cosx x When x = 7, 7 = 4 cos2θ + 7 sin2θ 7 = 4(1 – sin2θ) + 7 sin2θ

3 = 3 sin2θ ⇒ sin θ = 1, θ = 2π

When x = 4, 4 = 4 cos2θ + 7 sin2θ 4 = 4 (1 – sin2θ) + 7 sin2θ 4 = 4 + 3 sin2θ sin θ = 0 ⇒ θ = 0

∴ [ ]7 /2 /2 /2

04 0 0

1 6sin cosd d 2d 2 23sin cos 2( 4)(7 )

xx x

θ θ θ θ θθ θ

π π π π = = = = = π − − ∫ ∫ ∫

15 /12

0sin3 dx x x

π

∫

u = x, d sin3d

v xx=

d 11, = cos 3d 3

u v xx= −

/12

/12 /12

0 00

1 1sin3 d cos3 cos3 d3 3

x x x x x x xπ

π π = − + ∫ ∫

/12

0

1 3 1cos sin33 12 12 9

xπ

π π = − +

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2 1 sin36 2 9 4−π π= +

2 272 18π= − +

2 (4 )72

= − π

16 2

2 2

15 13 4(1 ) (4 ) 1 (1 ) 4

x x A B Cx x x x x

− + ≡ + +− − − − −

∴ 15 – 13x + 4x2 = A(1 – x) (4 – x) + B (4 – x) + C (1 – x)2

When x = 1, 6 = 3B, B = 2 When x = 4, 15 – 52 + 64 = 9C, 27 = 9C, C = 3 Equating coefficients of x2, 4 = A + C ⇒ A = 1

∴ 2

2 2

15 13 4 1 2 3(1 ) (4 ) 1 (1 ) 4

x xx x x x x

− + ≡ + +− − − − −

23

22

15 13 4 d(1 ) (4 )

x x xx x

− +− −∫

3

22

1 2 3 d1 (1 ) 4

xx x x

= + +− − −∫

3

2

2ln 1 3ln 41

x xx

= − − + − − −

2ln 2 3ln 1 ln 1 2 3ln 22

= − − + − − − − − − −

= –ln 2 – 1 + 2 + 3 ln 2 = 1 + 2 ln2

17 / 6

2

0sec 2 dx x

π

∫

/6

0

1 tan 22

xπ

=

1 3tan2 3 2

π= =

/6 /6

2 2

0 0tan 2 d sec 2 1 dx x x x

π π= −∫ ∫

[ ] /6

0

32

x π= −

32 6

π= −

18 21

2 20

1 d(1 )

x xx

−+∫

x = tanθ dx = sec2θ dθ 1 – x2 = 1 – tan2θ 1 + x2 = 1 + tan2θ = sec2θ

When x = 1, tan θ = 1 ⇒ θ = π4

When x = 0, tanθ = 0 ⇒ θ = 0

2/ 4

240

1 tan sec dsec

θ θ θθ

π −∫

/ 42 2

0(1 tan ) cos dθ θ θ

π= −∫

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/4

2 2

0cos sin dθ θ θ

π= −∫

/4

0cos2 dθ θ

π= ∫

= /4

0

1 sin 22

θπ

1 1sin2 2 2

π= =

19 2

2 2

7( 2 2) ( 1) 2 2 1

x x Ax B Cx x x x x x

+ − +≡ ++ + − + + −

∴ x2 + x – 7 = (Ax + B) (x – 1) + C (x2 + 2x + 2) When x = 1, –5 = 5C ⇒ C = –1 Equating coefficients of x2, 1 = A + C ⇒ A = 2 When x = 0, –7 = –B + 2C ⇒ B = 5

2

2 2

7 2 5 1( 2 2)( 1) 2 2 1

x x xx x x x x x

+ − +∴ ≡ −+ + − + + −

2 2

2 2 3 12 2 2 2 1

xx x x x x

+= + −

+ + + + −

2

2

7 d( 2 2)( 1)

x x xx x x

+ −+ + −∫

2 2

2 2 3 1d d d2 2 ( 1) 1 1

x x x xx x x x

+= + −+ + + + −∫ ∫ ∫

2 1ln 2 2 3 tan ( 1) ln 1x x x x c−= + + + + − − +

20 / 2

0

cos dxe x xπ

∫

d, cosd

x vu e xx

= =

du , sind

xe v xx= =

/ 2

0

cos dxe x xπ

∫/2/2

0 0sin sin dx xe x e x x

ππ = − ∫

Now look at /2

0

sin dxe x xπ

∫

d, sind

x vu e xx

= =

du , cosd

xe v xx= = −

So /2 /2

π/2

00 0

sin d cos cos dx x xe x x e x e x xπ π

= − + ∫ ∫

∴ /2 /2

π/2

00 0

cos d sin cos cos dx x x xe x x e x e x e x xπ π

= + − ∫ ∫

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/ 2

/ 2

02 cos d 1xe x x e

ππ= −∫

∴/ 2

/ 2

0

1cos d ( 1)2

xe x x eπ

π= −∫

unit 2 chapter 4 answers

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