unit 2 chapter 4 answers
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Macmillan Pure Maths Answers for chapter 4TRANSCRIPT
Page 1 of 46
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= − +∫
Page 2 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
(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
= + = + ++∫ ∫
Page 3 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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= =
Page 4 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
= − + + + +
Page 5 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
Page 6 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
=
Page 7 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
= + +
Page 8 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
Page 9 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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−= − +
Page 10 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
= =
Page 11 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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= =
Page 12 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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 = − +
Page 13 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
∴ = − − ++∫ ∫
Page 14 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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=
Page 15 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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∫
Page 16 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
= −
Page 17 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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=
Page 18 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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= + + + +
Page 19 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
(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= + − +
Page 20 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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+ +∫
Page 21 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
+ −+ −∫
Page 22 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
+≡ +− + + − + +
Page 23 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
+
= −− − + +∫ ∫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
=− + − +
Page 24 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
∴ 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+ + +
+∫
Page 25 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
+ + + +
+
+ +
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
Page 26 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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⇒ + + + = + + + + +
Page 27 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
Page 28 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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− −∫
Page 29 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
= 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 +
∫
Page 30 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
= 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 +∫
Page 31 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
= +
Page 32 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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= − + +
Page 33 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
Page 34 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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 = ∫
Page 35 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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=
Page 36 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
= 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
= +∫ ∫
Page 37 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
/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= + − −
Page 38 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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= +∫
Page 39 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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= − | − | + | − |
Page 40 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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 = − ∫ ∫
Page 41 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
= − − 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= − − +
Page 42 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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
= +
Page 43 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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 = − +
Page 44 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
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= −∫
Page 45 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
/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
= + − ∫ ∫
Page 46 of 46
Unit 2 Answers: Chapter 4 © Macmillan Publishers Limited 2013
/ 2
/ 2
02 cos d 1xe x x e
ππ= −∫
∴/ 2
/ 2
0
1cos d ( 1)2
xe x x eπ
π= −∫