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Solution manual calculus 8th edition varberg, purcell, rigdon ch07

CHAPTER

7

Techniques of Integration

7.1 Concepts Review

5.

1. elementary function
2.

∫u

5

1

–1 ⎛


x⎞
⎜ ⎟+C
2
⎝ ⎠

6. u = 2 + e x , du = e x dx

du

ex

∫ 2 + e x dx = ∫

3. e x
4.

dx

∫ x2 + 4 = 2 tan

du
u

= ln u + C

2 3
u du
1



= ln 2 + e x + C
= ln(2 + e x ) + C

Problem Set 7.1
1.

∫ ( x – 2)

2.





5

1
dx = ( x – 2)6 + C
6

3 x dx =

1
2
3x ⋅ 3dx = (3x)3 / 2 + C

3
9

3. u = x 2 + 1, du = 2 x dx
When x = 0, u = 1 and when x = 2, u = 5 .
2

∫0 x( x

2

+ 1)5 dx =

1 2 2
( x + 1)5 (2 x dx)
2 ∫0

8.

1 5
= ∫ u 5 du
2 1

15624
= 1302
12

4. u = 1 – x 2 , du = –2 x dx
When x = 0, u = 1 and when x = 1, u = 0 .
1

∫0

x 1 – x 2 dx = –

1 0
= − ∫ u1/ 2 du
2 1
1 1
= ∫ u1/ 2 du
2 0

2t 2

∫ 2t 2 + 1

dt = ∫

= ∫ dt – ∫

5

⎡ u6 ⎤
56 − 16
=⎢ ⎥ =
12
⎣⎢ 12 ⎦⎥1
=

7. u = x 2 + 4, du = 2x dx
x
1 du
∫ x2 + 4 dx = 2 ∫ u
1
= ln u + C
2
1
= ln x 2 + 4 + C
2
1
= ln( x 2 + 4) + C
2

1 1
1 − x 2 (−2 x dx)
2 ∫0

2t 2 + 1 − 1
2t 2 + 1

dt

1

dt
2t + 1
u = 2t , du = 2dt
1
1
du
t–∫
dt = t –

2
2 1 + u2
2t + 1
1
=t–
tan –1 ( 2t ) + C
2
2

9. u = 4 + z 2 , du = 2z dz

∫ 6z

4 + z 2 dz = 3∫ u du

= 2u 3 / 2 + C
= 2(4 + z 2 )3 / 2 + C

1

1
⎡1

= ⎢ u3 / 2 ⎥ =
⎣3
⎦0 3

412

Section 7.1

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


10. u = 2t + 1, du = 2dt
5
5 du
∫ 2t + 1 dt = 2 ∫ u

16. u = 1 – x , du = –
3 / 4 sin

∫0

= 5 u +C
= 5 2t + 1 + C

= 2∫

tan z

∫ cos2 z dz = ∫ tan z sec

2

1

2

1 2
u +C
2
1
= tan 2 z + C
2

17.

12. u = cos z, du = –sin z dz

∫e

cos z

sin z dz = – ∫ ecos z (– sin z dz )

18.

= − ∫ eu du = −eu + C

= – ecos z + C

13. u = t , du =



1
2 t

dt

sin t

dt = 2 ∫ sin u du
t
= –2 cos u + C
= –2 cos t + C

14. u = x 2 , du = 2x dx
2 x dx
du
=∫

4
1– x
1 – u2

= sin –1 u + C
= sin ( x ) + C
–1

2

15. u = sin x, du = cos x dx
π / 4 cos x
2 / 2 du
∫0 1 + sin 2 x dx = ∫0 1 + u 2
= [tan −1 u ]0 2 / 2
2
2
≈ 0.6155

1/ 2

1

sin u du

sin u du

1⎞

= −2 ⎜ cos1 − cos ⎟
2⎠

≈ 0.6746

z dz = ∫ u du

=

dx

= [−2 cos u ]11/ 2

z dz

u = tan z, du = sec2 z dz

∫ tan z sec

dx = –2 ∫

1– x
1/ 2

11.

1– x

1
2 1– x

19.

3x2 + 2 x
1
∫ x + 1 dx = ∫ (3x – 1)dx + ∫ x + 1 dx
3
= x 2 – x + ln x + 1 + C
2
1
x3 + 7 x
2
∫ x – 1 dx = ∫ ( x + x + 8)dx + 8∫ x – 1 dx
1
1
= x3 + x 2 + 8 x + 8ln x – 1 + C
3
2

u = ln 4 x 2 , du =

2
dx
x

sin(ln 4 x 2 )
1
∫ x dx = 2 ∫ sin u du
1
= – cos u + C
2
1
= – cos(ln 4 x 2 ) + C
2

20. u = ln x, du =

1
dx
x

sec 2 (ln x )
1
dx = ∫ sec2 u du
2x
2
1
= tan u + C
2
1
= tan(ln x) + C
2



21. u = e x , du = e x dx

= tan −1

6e x



1 − e2 x

dx = 6 ∫

du
1− u2

du

= 6sin −1 u + C
= 6sin −1 (e x ) + C

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Section 7.1

413

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


22. u = x 2 , du = 2x dx
x
1
du
∫ x4 + 4 dx = 2 ∫ 4 + u 2
1
u
= tan −1 + C
4
2

1
x2 ⎞
= tan –1 ⎜ ⎟ + C
⎜ 2 ⎟
4
⎝ ⎠

27.

= x − ln sin x + C

23. u = 1 – e2 x , du = –2e 2 x dx
3e2 x



1– e

2x

dx = –

3 du
2∫ u

= –3 u + C
= –3 1 – e2 x + C

24.

x3

∫ x4 + 4

dx =

1 4 x3
dx
4 ∫ x4 + 4

28. u = cos(4t – 1), du = –4 sin(4t – 1)dt
sin(4t − 1)
sin(4t − 1)
∫ 1 − sin 2 (4t − 1) dt =∫ cos2 (4t − 1) dt
1 1
=− ∫
du
4 u2
1
1
= u −1 + C = sec(4t − 1) + C
4
4
29. u = e x , du = e x dx

∫e

1
= ln x 4 + 4 + C
4
1
= ln( x 4 + 4) + C
4

25.

1

t2

∫0 t 3

dt =

26.

2

1 1 t2
2t 3 dt
2 ∫0

π / 6 cos x
2
(– sin x dx)
0

sin x dx = – ∫

cos x ⎤ π / 6

⎡ 2
= ⎢–

⎣⎢ ln 2 ⎦⎥ 0
1
=–
(2 3 / 2 – 2)
ln 2
2−2 3/2
ln 2
≈ 0.2559
=

sec e x dx = ∫ sec u du

= ln sec e x + tan e x + C

30. u = e x , du = e x dx

∫e

1

π / 6 cos x

∫0

x

= ln sec u + tan u + C

⎡ 3t ⎤
⎥ = 3 – 1
=⎢
⎢ 2 ln 3 ⎥
2 ln 3 2 ln 3

⎦0
1
=
≈ 0.9102
ln 3
2

sin x − cos x
⎛ cos x ⎞
dx = ∫ ⎜1 −
⎟ dx
sin x
⎝ sin x ⎠
u = sin x, du = cos x dx
sin x − cos x
du
∫ sin x dx = x − ∫ u
= x − ln u + C



x

sec 2 (e x )dx = ∫ sec 2 u du = tan u + C

= tan(e x ) + C

31.



sec3 x + esin x
dx = ∫ (sec2 x + esin x cos x) dx
sec x

= tan x + ∫ esin x cos x dx

u = sin x, du = cos x dx
tan x + ∫ esin x cos x dx = tan x + ∫ eu du
= tan x + eu + C = tan x + esin x + C
32. u = 3t 2 − t − 1 ,
1
du = (3t 2 − t − 1)−1/ 2 (6t − 1)dt
2



(6t − 1) sin 3t 2 − t − 1

3t 2 − t − 1
= –2 cos u + C

dt = 2∫ sin u du

= −2 cos 3t 2 − t − 1 + C

414

Section 7.1

Instructor’s Resource Manual

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


33. u = t 3 − 2 , du = 3t 2 dt



t 2 cos(t 3 − 2)

dt =

39. u = 3 y 2 , du = 6 y dy

1 cos u
du
3 ∫ sin 2 u



sin (t − 2)
v = sin u, dv = cos u du
1 cos u
1
1
du = ∫ v −2 dv = − v −1 + C
3 ∫ sin 2 u
3
3
1
=−
+C
3sin u
1
=−
+C .
3sin(t 3 − 2)

34.



2 3

1 + cos 2 x
2

sin 2 x

dx = ∫

1
2

sin 2 x

dx + ∫

cos 2 x
sin 2 2 x

= ∫ csc2 2 x dx + ∫ cot 2 x csc 2 x dx
1
1
= − cot 2 x − csc 2 x + C
2
2

35. u = t 3 − 2 , du = 3t 2 dt



t 2 cos 2 (t 3 − 2)

dt =

1 cos 2 u
du
3 ∫ sin 2 u

sin 2 (t 3 − 2)
1
1
= ∫ cot 2 u du = ∫ (csc2 u –1)du
3
3
1
= [− cot u − u ] + C1
3
1
= [− cot(t 3 − 2) − (t 3 − 2)] + C1
3
1
= − [cot(t 3 − 2) + t 3 ] + C
3

36. u = 1 + cot 2t, du = −2 csc2 2t



csc2 2t
1 + cot 2t

dt = −

1 1
du
2∫ u

=− u +C
= − 1 + cot 2t + C

37. u = tan −1 2t , du =
e tan

−1

1 + 4t 2

dt

2t

1 u
∫ 1 + 4t 2 dt = 2 ∫ e du
−1
1
1
= eu + C = e tan 2t + C
2
2

38. u = −t 2 − 2t − 5 ,
du = (–2t – 2)dt = –2(t + 1)dt
1
− t 2 − 2t − 5
= − ∫ eu du
∫ (t + 1)e
2
1 u
1 − t 2 − 2t − 5
= − e +C = − e
+C
2
2
Instructor’s Resource Manual

16 − 9 y 4

dy =

1
1
du

6
42 − u 2

1
⎛u⎞
= sin −1 ⎜ ⎟ + C
6
⎝4⎠
⎛ 3y2
1
= sin −1 ⎜
⎜ 4
6



⎟+C



40. u = 3x, du = 3 dx

∫ cosh 3x dx
dx

1
1
(cosh u )du = sinh u + C
3∫
3
1
= sinh 3 x + C
3
=

41. u = x3 , du = 3 x 2 dx
1
2
3
∫ x sinh x dx = 3 ∫ sinh u du
1
= cosh u + C
3
1
= cosh x3 + C
3
42. u = 2x, du = 2 dx
5
5
1
dx = ∫
du

2
9 − 4 x2
32 − u 2
5
⎛u⎞
= sin −1 ⎜ ⎟ + C
2
⎝3⎠
5 −1 ⎛ 2 x ⎞
= sin ⎜ ⎟ + C
2
⎝ 3 ⎠
43. u = e3t , du = 3e3t dt


2

y

e3t
4−e

6t

dt =

1
1
du
3 ∫ 22 − u 2

1
⎛u⎞
= sin −1 ⎜ ⎟ + C
3
⎝2⎠
⎛ e3t
1
= sin −1 ⎜
⎜ 2
3



⎟+C



44. u = 2t, du = 2dt
dt
1
1
= ∫
du

2
2t 4t − 1 2 u u 2 − 1
1
= ⎡sec −1 u ⎤ + C

2⎣
1
= sec −1 2t + C
2

Section 7.1

415

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


45. u = cos x, du = –sin x dx
π/2
0
sin x
1
∫0 16 + cos2 x dx = − ∫1 16 + u 2 du
1
1
=∫
du
0 16 + u 2

50.

51.

1
1
1
= ln(e4 + 1) − ln(e 2 ) − ln 2
2
2
2
4




e +1
1
= ⎜ ln ⎜
⎟ − 2 ⎟ ≈ 0.6625
2 ⎜⎝ ⎜⎝ 2 ⎟⎠ ⎟⎠

47.

1

=

48.

( x + 1) 2 + 22

d ( x + 1)

1

1

1
( x – 2)2 + ( 5)2

d ( x – 2)

⎛ x–2⎞
tan –1 ⎜
⎟+C
5
⎝ 5 ⎠

–( x – 3)2 + 52

x +1

52 – ( x – 3)2

18 x + 18

1

(

52.

)

3– x



16 + 6 x – x

2

dx =

1
6 – 2x
dx
2 ∫ 16 + 6 x – x 2

= 16 + 6 x – x 2 + C

53. u = 2t , du = 2dt
dt
du
∫ 2 =∫ 2 2
t 2t – 9
u u –3
⎛ 2t ⎞
1
⎟+C
= sec –1 ⎜
⎜ 3 ⎟
3



dx

dx

Section 7.1

tan x



2

sec x – 4
sin x

dx = ∫

cos x
tan x
dx
cos x sec2 x – 4

dx
1 – 4 cos 2 x
u = 2 cos x, du = –2 sin x dx
sin x
1
1
dx = − ∫
du

2
2
1 − 4 cos x
1− u2
1
1
= − sin −1 u + C = – sin –1 (2 cos x) + C
2
2

2

⎛ dy ⎞
1 + ⎜ ⎟ dx
a
⎝ dx ⎠

dx

∫ 9 x2 + 18 x + 10 = ∫ 9 x2 + 18 x + 9 + 1

1
= tan –1 (3x + 3) + C
3

54.

L=∫

(3x + 3)2 + 12
u = 3x + 3, du = 3 dx
dx
1
du
∫ (3x + 3)2 + 12 = 3 ∫ u 2 + 12

416

dx

=∫

55. The length is given by

1

=∫

dx

∫ 9 x2 + 18 x + 10 dx = 18 ∫ 9 x2 + 18 x + 10 dx

=∫

∫ x2 – 4 x + 9 dx = ∫ x2 – 4 x + 4 + 5 dx

=

49.

1

1
⎛ x +1⎞
tan –1 ⎜
⎟+C
2
⎝ 2 ⎠

=∫

–( x – 6 x + 9 – 25)

1

∫ x2 + 2 x + 5 dx = ∫ x2 + 2 x + 1 + 4 dx
=∫

2

1
ln 9 x 2 + 18 x + 10 + C
18
1
= ln 9 x 2 + 18 x + 10 + C
18

− e−2 x

1 e4 + 1 1
= ln
− ln 2
2
2
e2

dx

=∫

=

= 2(e2 x − e −2 x )dx

1 e2 + e−2 1
du
∫0 e2 x + e−2 x dx = 2 ∫2
u
1
1
1
e2 + e −2
= ⎡⎣ ln u ⎤⎦
= ln e2 + e −2 − ln 2
2
2
2
2

2

⎛ x –3⎞
= sin –1 ⎜
⎟+C
⎝ 5 ⎠

46. u = e2 x + e−2 x , du = (2e2 x − 2e −2 x )dx
1 e2 x

16 + 6 x – x

=∫

1

⎡1
⎡1

⎛ u ⎞⎤
⎛1⎞ 1
= ⎢ tan −1 ⎜ ⎟ ⎥ = ⎢ tan −1 ⎜ ⎟ − tan −1 0 ⎥
⎝ 4 ⎠⎦0 ⎣ 4
⎝4⎠ 4
⎣4

1
⎛1⎞
= tan −1 ⎜ ⎟ ≈ 0.0612
4
⎝4⎠

dx



b

=∫

π/ 4

=∫

π/ 4

0

0
π/ 4

2

⎡ 1

1+ ⎢
(− sin x) ⎥ dx
⎣ cos x

1 + tan 2 x dx = ∫

π/ 4

0

sec2 x dx

=∫

sec x dx = ⎡⎣ln sec x + tan x ⎤⎦ 0π / 4

= ln

2 + 1 − ln 1 = ln

0

2 + 1 ≈ 0.881

Instructor’s Resource Manual

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


56. sec x =
=

1
1 + sin x
=
cos x cos x(1 + sin x)

sin x + sin 2 x + cos 2 x sin x(1 + sin x) + cos 2 x
=
cos x(1 + sin x)
cos x(1 + sin x)

sin x
cos x
+
cos x 1 + sin x
cos x ⎞
⎛ sin x
∫ sec x = ∫ ⎜⎝ cos x + 1 + sin x ⎟⎠dx
sin x
cos x
dx + ∫
dx
=∫
cos x
1 + sin x
For the first integral use u = cos x, du = –sin x dx,
and for the second integral use v = 1 + sin x,
dv = cos x dx.
sin x
cos x
du
dv
∫ cos x dx + ∫ 1 + sin x dx = – ∫ u + ∫ v
= – ln u + ln v + C
=

= – ln cos x + ln 1 + sin x + C

= ln

57. u = x – π , du = dx
2 π x sin x
π (u + π) sin(u + π)
∫0 1 + cos2 x dx = ∫– π 1 + cos2 (u + π) du
π (u + π) sin u
=∫
du
– π 1 + cos 2 u
π u sin u
π π sin u
=∫
du + ∫
du
2
– π 1 + cos u
– π 1 + cos 2 u
π u sin u
∫– π 1 + cos2 u du = 0 by symmetry.
π π sin u
π π sin u
∫– π 1 + cos2 u du = 2∫0 1 + cos2 u du
v = cos u, dv = –sin u du
–1 π
1
1
−2∫
dv = 2π ∫
dv
2
1 1+ v
–1 1 + v 2
⎡ π ⎛ π ⎞⎤
= 2π[tan −1 v]1−1 = 2π ⎢ − ⎜ − ⎟ ⎥
⎣ 4 ⎝ 4 ⎠⎦
⎛π⎞
= 2π ⎜ ⎟ = π2
⎝2⎠

1 + sin x
+C
cos x

= ln sec x + tan x + C

58.


4 ⎛x+
– π ⎜⎝
4

V = 2π ∫

π⎞
⎟ sin x – cos x dx
4⎠

π
, du = dx
4
π
π⎞
π⎞
π⎞



V = 2π∫ 2π ⎜ u + ⎟ sin ⎜ u + ⎟ – cos ⎜ u + ⎟ du
– ⎝
2⎠
4⎠
4⎠


2
u= x–

π
π⎞ 2
2
2
2

= 2π∫ 2π ⎜ u + ⎟
sin u +
cos u –
cos u +
sin u du
– ⎝
2⎠ 2
2
2
2
2
π
π
π
π⎞

= 2π∫ 2π ⎜ u + ⎟ 2 sin u du = 2 2π ∫ 2π u sin u du + 2π 2 ∫ 2π sin u du
− ⎝


2⎠
2
2
2
π

2 2π∫ 2π u sin u du = 0 by symmetry. Therefore,


2

π

π

V = 2π2 2 ∫ 2 sin u du = 2 2π2 [− cos u ]02 = 2 2π2
0

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Section 7.1

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


6. u = x

dv = sin 2x dx
1
du = dx
v = – cos 2 x
2
1
1
∫ x sin 2 x dx = – 2 x cos 2 x – ∫ – 2 cos 2 x dx
1
1
= – x cos 2 x + sin 2 x + C
2
4

7.2 Concepts Review
1. uv – ∫ v du
2. x; sin x dx
3. 1
4. reduction

7. u = t – 3
dv = cos (t – 3)dt
du = dt
v = sin (t – 3)
∫ (t – 3) cos(t – 3)dt = (t – 3) sin(t – 3) – ∫ sin(t – 3)dt

Problem Set 7.2

v = ex

du = dx

∫ xe

x

= (t – 3) sin (t – 3) + cos (t – 3) + C

dv = e x dx

1. u = x

dx = xe − ∫ e dx = xe − e + C
x

x

x

x

dv = e3 x dx
1
du = dx
v = e3 x
3
1
1 3x
3x
3x
∫ xe dx = 3 xe − ∫ 3 e dx
1
1
= xe3 x − e3 x + C
3
9

2. u = x

dv = e5t +π dt
1
du = dt
v = e5t +π
5
1 5t +π
1 5t +π
5t +π
∫ te dt = 5 te – ∫ 5 e dt
1
1
= te5t +π – e5t +π + C
5
25

3. u = t

4. u = t + 7 dv = e2t +3 dt
1
du = dt
v = e 2t + 3
2
1
1 2t + 3
2t + 3
2t + 3
∫ (t + 7)e dt = 2 (t + 7)e – ∫ 2 e dt
1
1
= (t + 7)e2t +3 – e2t +3 + C
2
4
t
13
= e 2 t + 3 + e 2t + 3 + C
2
4
5. u = x
dv = cos x dx
du = dx
v = sin x
∫ x cos x dx = x sin x – ∫ sin x dx

= x sin x + cos x + C

418

Section 7.2

8. u = x – π
dv = sin(x)dx
du = dx
v = –cos x
(
x

π
)
sin(
x
)
dx = –( x – π) cos x + ∫ cos x dx


= ( π – x) cos x + sin x + C
dv = t + 1 dt

9. u = t

v=

du = dt

2
(t + 1)3 / 2
3

2
2
t + 1 dt = t (t + 1)3 / 2 – ∫ (t + 1)3 / 2 dt
3
3
2
4
= t (t + 1)3 / 2 – (t + 1)5 / 2 + C
3
15

∫t

dv = 3 2t + 7dt
3
v = (2t + 7)4 / 3
du = dt
8
3
3
4/3
4/3
3
∫ t 2t + 7dt = 8 t (2t + 7) – ∫ 8 (2t + 7) dt
3
9
(2t + 7)7 / 3 + C
= t (2t + 7)4 / 3 –
8
112

10. u = t

11. u = ln 3x
1
du = dx
x

dv = dx
v=x
1

∫ ln 3x dx = x ln 3x −∫ x x dx
12. u = ln(7 x5 )

du =

5
dx
x

∫ ln(7 x

5

= x ln 3 x − x + C

dv = dx
v=x

5
)dx = x ln(7 x5 ) – ∫ x dx
x

= x ln(7 x5 ) – 5 x + C

Instructor Solutions Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


13. u = arctan x
1
du =
dx
1 + x2

dv = dx

1
du = dt
t

v=x
x

∫ arctan x = x arctanx − ∫ 1 + x2 dx
1 2x
= x arctan x − ∫
dx
2 1 + x2
1
= x arctan x − ln(1 + x 2 ) + C
2

14. u = arctan 5x
5
du =
dx
1 + 25 x 2

5x

∫ arctan 5x dx = x arctan 5x – ∫ 1 + 25 x2 dx
1 50 x dx
= x arctan 5 x – ∫
10 1 + 25 x 2
1
= x arctan 5 x – ln(1 + 25 x 2 ) + C
10
dx

x2
1
1
du = dx
v=–
x
x
ln x
ln x
1⎛1⎞
∫ x2 dx = – x – ∫ – x ⎜⎝ x ⎟⎠ dx
ln x 1
=–
– +C
x
x

16. u = ln 2 x5

du =

x2

1

x2
1
v=−
x

5
dx
x

3 ln 2 x5

∫2

dv =

dx

3

3 1
⎡ 1

dx = ⎢ − ln 2 x5 ⎥ + 5∫
dx
2 x2
⎣ x
⎦2
3

5⎤
⎡ 1
= ⎢ − ln 2 x5 − ⎥
x ⎦2
⎣ x
5⎞ ⎛ 1
5⎞
⎛ 1
= ⎜ − ln(2 ⋅ 35 ) − ⎟ − ⎜ − ln(2 ⋅ 25 ) − ⎟
3⎠ ⎝ 2
2⎠
⎝ 3
1
5
5
5
= − ln 2 − ln 3 − + 3ln 2 +
3
3
3
2
8
5
5
= ln 2 − ln 3 + ≈ 0.8507
3
3
6

Instructor's Resource Manual

e

∫1

e

v=x

dv =

2
v = t3 / 2
3

e2
⎡2

t ln t dt = ⎢ t 3 / 2 ln t ⎥ – ∫ t1/ 2 dt
⎣3
⎦1 1 3

e

=

2 3/ 2
2
⎡4

e ln e – ⋅1ln1 − ⎢ t 3 / 2 ⎥
3
3
9

⎦1

=

2 3/ 2
4
4 2
4
e
− 0 − e3 / 2 + = e3 / 2 + ≈ 1.4404
3
9
9 9
9

dv = dx

15. u = ln x

dv = t dt

17. u = ln t

dv = 2 xdx
1
v = (2 x)3 / 2
3

18. u = ln x3
3
du = dx
x
5

∫1

5

5
⎡1

2 x ln x3 dx = ⎢ (2 x)3 2 ln x3 ⎥ − ∫ 23 2 x dx
1
⎣3
⎦1
5

1
25 / 2 3 / 2 ⎤
= ⎡ (2 x)3 / 2 ln x3 −
x
3
⎢⎣ 3
⎥⎦1

1
25 2 3 / 2 ⎛ 1 3 2 3 25 2 ⎞
= (10)3 2 ln 53 −
5
− ⎜ (2) ln1 −

⎜3
3
3
3 ⎟⎠

=−

4 2 32 4 2
5 +
+ 103 2 ln 5 ≈ 31.699
3
3

dv = z 3dz
1
v = z4
4
1
1 4 1
3
4
∫ z ln z dz = 4 z ln z − ∫ 4 z ⋅ z dz
1
1
= z 4 ln z − ∫ z 3 dz
4
4
1 4
1
= z ln z − z 4 + C
4
16

19. u = ln z
1
du = dz
z

20. u = arctan t
1
du =
dt
1+ t2

dv = t dt
1
v = t2
2
1

∫ t arctan t dt = 2 t

2

arctan t –

1 t2
dt
2 ∫ 1+ t2

1
1 1+ t2 −1
= t 2 arctan t − ∫
dt
2
2 1+ t2
1
1
1
1
= t 2 arctan t − ∫ dt + ∫
dt
2
2
2 1+ t2
1
1 1
= t 2 arctan t − t + arctan t + C
2
2 2

Section 7.2

419

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


⎛1⎞
21. u = arctan ⎜ ⎟
dv = dt
⎝t ⎠
1
v=t
du = –
dt
1+ t2
t
⎛1⎞
⎛1⎞
∫ arctan ⎜⎝ t ⎟⎠ dt = t arctan ⎜⎝ t ⎟⎠ + ∫ 1 + t 2 dt
⎛1⎞ 1
= t arctan ⎜ ⎟ + ln(1 + t 2 ) + C
⎝t ⎠ 2

dv = t 5 dt

7
dt
t

1
v = t6
6

du =

1
7
ln(t 7 )dt = t 6 ln(t 7 ) – ∫ t 5 dt
6
6
1
7
= t 6 ln(t 7 ) – t 6 + C
6
36

∫t

5

dv = csc2 x dx
v = − cot x

23. u = x
du = dx
π/2

22. u = ln(t 7 )

∫π / 6 x csc

π/ 2

x dx = [ − x cot x ]π / 6 + ∫

2

π/ 2

π/6

cot x dx = ⎣⎡− x cot x + ln sin x ⎤⎦

π2
π6

π
π
1
π
= − ⋅ 0 + ln1 +
3 − ln =
+ ln 2 ≈ 1.60
2
6
2 2 3
dv = sec2 x dx
v = tan x

24. u = x
du = dx
π4

∫π 6 x sec
=

2

π4

x dx = [ x tan x ]π 6 − ∫

π4

π6

π
2 ⎛ π
3⎞
π4
= + ln
− ⎜⎜
+ ln
tan x dx = ⎡⎣ x tan x + ln cos x ⎤⎦

π6
4
2 ⎝6 3
2 ⎟⎠

π
π
1 2

+ ln ≈ 0.28
4 6 3 2 3

25. u = x3

dv = x 2 x3 + 4dx
2
v = ( x3 + 4)3 / 2
du = 3 x 2 dx
9
2
2 2 3
2 3 3
4 3
5
3
3 3
3/ 2
3/ 2
3/ 2
5/ 2
∫ x x + 4dx = 9 x ( x + 4) – ∫ 3 x ( x + 4) dx = 9 x ( x + 4) – 45 ( x + 4) + C
dv = x 6 x 7 + 1 dx

26. u = x7

v=

du = 7 x 6 dx

∫x

13

x7 + 1dx =

2 7 7
2
2
4 7
( x + 1)5 / 2 + C
x ( x + 1)3 / 2 – ∫ x6 ( x7 + 1)3 / 2 dx = x 7 ( x 7 + 1)3 / 2 –
21
3
21
105
dv =

4
27. u = t

v=

du = 4t 3 dt
t7

∫ (7 – 3t 4 )3 / 2

420

2 7
( x + 1)3 / 2
21

dt =

Section 7.2

t3
(7 – 3t 4 )3 / 2
1

dt

6(7 – 3t 4 )1/ 2
t4

6(7 – 3t 4 )1/ 2



t3
t4
2
1
dt
=
+ (7 – 3t 4 )1/ 2 + C
4 1/ 2 9
3 ∫ (7 – 3t 4 )1/ 2
6(7 – 3t )

Instructor Solutions Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


28. u = x 2

dv = x 4 – x 2 dx
1
du = 2x dx
v = – (4 – x 2 )3 / 2
3
1
1 2
2
3
2
2
2 3/ 2 2
2 3/ 2
2 3/ 2
2 5/ 2
∫ x 4 – x dx = – 3 x (4 – x ) + 3 ∫ x(4 – x ) dx = – 3 x (4 – x ) – 15 (4 – x ) + C

29. u = z 4

dv =

v=

du = 4 z 3 dz
z7

∫ (4 – z 4 )2

dz =

z3
(4 – z 4 ) 2

dz

1
4(4 – z 4 )

z4
4

4(4 – z )

−∫

z3
4– z

4

dz =

z4

1
+ ln 4 – z 4 + C
4(4 – z ) 4
4

30. u = x
dv = cosh x dx
du = dx
v = sinh x
x
cosh
x
dx
=
x
sinh
x – ∫ sinh x dx = x sinh x – cosh x + C

31. u = x
dv = sinh x dx
du = dx
v = cosh x
∫ x sinh x dx = x cosh x – ∫ cosh x dx = x cosh x – sinh x + C
32. u = ln x
dv = x –1/ 2 dx
1
du = dx
v = 2 x1/ 2
x
ln x
1
∫ x dx = 2 x ln x – 2∫ x1/ 2 dx = 2 x ln x – 4 x + C
33. u = x

dv = (3 x + 10)49 dx

1
(3x + 10)50
150
x
1
x
1
49
50
50
50
51
∫ x(3x + 10) dx = 150 (3x + 10) – 150 ∫ (3x + 10) dx = 150 (3x + 10) – 22,950 (3x + 10) + C

du = dx

34. u = t
du = dt

v=

dv = ( t − 1) dt
12

v=

1
( t − 1)13
13
1

1 1
⎡t
13 ⎤
13
12
∫0 t (t − 1) dt = ⎢⎣13 ( t − 1) ⎥⎦0 − 13 ∫0 ( t − 1) dt
1

1

1
1
⎡t
13
= ⎢ ( t − 1) −
( t − 1)14 ⎤⎥ =
13
182
182

⎦0
dv = 2 dx
1 x
du = dx
v=
2
ln 2
x x
1
x
x
∫ x2 dx = ln 2 2 – ln 2 ∫ 2 dx
x x
1
=
2 –
2x + C
2
ln 2
(ln 2)

35. u = x

x

dv = a z dz
1 z
du = dz
v=
a
ln a
z z
1
z
z
∫ za dz = ln a a – ln a ∫ a dz
z z
1
a –
az + C
=
2
ln a
(ln a)

36. u = z

37. u = x 2

du = 2x dx

∫x

dv = e x dx

v = ex

e dx = x 2 e x − ∫ 2 xe x dx

2 x

u=x

dv = e x dx

du = dx

v = ex

∫x

(

e dx = x 2 e x − 2 xe x − ∫ e x dx

2 x

)

= x 2 e x − 2 xe x + 2e x + C
Instructor's Resource Manual

Section 7.2

421

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


41. u = et

2

dv = xe x dx
1 2
v = ex
2

38. u = x 4
du = 4 x3 dx

du = et dt

∫e

5 x2

1 4 x2
3 x2
∫ x e dx = 2 x e – ∫ 2 x e dx

∫ ln

2

v = –cos t
t
t
∫ e cos t dt = e sin t − ⎡⎣−e cos t + ∫ e cos t dt ⎤⎦
t

2

∫ ln

2

t

∫e

∫e

at

dv = sin t dt
v = –cos t

sin t dt = – e cos t + a ∫ e at cos t dt
at

dv = dz

u = e at

v=z

du = ae dt

dv = cos t dt
at

(

z dz = z ln z – 2 z ln z – ∫ dz

t

1 t
e (sin t + cos t ) + C
2

du = ae dt

v = sin t

∫ e sin t dt = – e cos t + a ( e sin t – a ∫ e sin t dt )
at
at
at
2 at
∫ e sin t dt = – e cos t + ae sin t – a ∫ e sin t dt
at

)

= z ln 2 z – 2 z ln z + 2 z + C

at

at

at

dv = dx

(1 + a 2 ) ∫ e at sin t dt = – e at cos t + ae at sin t + C

v=x

∫e

du =

40 ln x 20
dx
x

∫ ln

x dx = x ln 2 x 20 – 40 ∫ ln x 20 dx

at

sin t dt =

– e at cos t
a2 + 1

+

aeat sin t
a2 + 1

+C

2 20

u = ln x 20
20
du =
dx
x

∫ ln

cos t dt =

t

t

at

z dz = z ln 2 z – 2 ∫ ln z dz

40. u = ln 2 x 20

t

42. u = e at

v=z

2

t

∫ e cos t dt = e sin t + e cos t − ∫ e cos t dt
2∫ et cos t dt = et sin t + et cos t + C

dv = dz

u = ln z
1
du = dz
z

dv = sin t dt

du = e dt

v = ex
1 4 x2 ⎛ 2 x2

x2
5 x2
∫ x e dx = 2 x e – ⎜⎝ x e − ∫ 2 xe dx ⎟⎠
2
2
2
1
= x4e x – x2e x + e x + C
2

39. u = ln 2 z
2 ln z
du =
dz
z

t

t

dv = 2 xe x dx

du = 2x dx

v = sin t

cos t dt = e sin t − ∫ et sin t dt

t

u = et

2

u = x2

dv = cos t dt

43. u = x 2
du = 2 x dx

dv = dx

v=x

(

x dx = x ln 2 x 20 – 40 x ln x 20 – 20∫ dx

2 20

)

= x ln 2 x 20 – 40 x ln x 20 + 800 x + C

∫x

2

2

v = sin x

cos x dx = x 2 sin x − ∫ 2 x sin x dx

u = 2x
du = 2dx

∫x

dv = cos x dx

dv = sin x dx
v = − cos x

(

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

)

= x sin x + 2 x cos x − 2sin x + C
2

44. u = r 2
du = 2r dr

∫r

2

sin r dr = – r 2 cos r + 2∫ r cos r dr

u=r
du = dr

∫r

422

2

dv = sin r dr
v = –cos r
dv = cos r dr
v = sin r

(

)

sin r dr = – r 2 cos r + 2 r sin r – ∫ sin r dr = – r 2 cos r + 2r sin r + 2 cos r + C

Section 7.2

Instructor Solutions Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


45. u = sin(ln x)

dv = dx

1
du = cos(ln x) ⋅ dx
x

v=x

∫ sin(ln x)dx = x sin(ln x) − ∫ cos(ln x) dx
u = cos (ln x)

dv = dx

1
du = − sin(ln x) ⋅ dx
x

v=x

∫ sin(ln x)dx = x sin(ln x) − ⎡⎣ x cos(ln x) − ∫ − sin(ln x)dx ⎤⎦
∫ sin(ln x)dx = x sin(ln x) − x cos(ln x) − ∫ sin(ln x)dx
2∫ sin(ln x)dx = x sin(ln x) − x cos(ln x) + C
x

∫ sin(ln x)dx = 2 [sin(ln x) − cos(ln x)] + C
46. u = cos(ln x)

dv = dx

1
du = – sin(ln x) dx
x

v=x

∫ cos(ln x)dx = x cos(ln x) + ∫ sin(ln x)dx
u = sin(ln x)

dv = dx

1
du = cos(ln x) dx
x

v=x

∫ cos(ln x)dx = x cos(ln x) + ⎡⎣ x sin(ln x) – ∫ cos(ln x)dx ⎤⎦
2∫ cos(ln x)dx = x[cos(ln x ) + sin(ln x )] + C
x

∫ cos(ln x)dx = 2 [cos(ln x) + sin(ln x)] + C
47. u = (ln x)3
du =

3ln 2 x
dx
x

∫ (ln x)

3

dv = dx
v=x

dx = x(ln x)3 – 3∫ ln 2 x dx

= x ln 3 x – 3( x ln 2 x – 2 x ln x + 2 x + C )

= x ln 3 x – 3x ln 2 x + 6 x ln x − 6 x + C
48. u = (ln x)4
du =

4 ln 3 x
dx
x

∫ (ln x)

4

dv = dx
v=x

dx = x (ln x )4 – 4∫ ln 3 x dx = x ln 4 x – 4( x ln 3 x – 3 x ln 2 x + 6 x ln x − 6 x + C )

= x ln 4 x – 4 x ln 3 x + 12 x ln 2 x − 24 x ln x + 24 x + C
49. u = sin x

dv = sin(3x)dx
1
du = cos x dx
v = – cos(3 x)
3
1
1
∫ sin x sin(3 x)dx = – 3 sin x cos(3 x) + 3 ∫ cos x cos(3x)dx
u = cos x
dv = cos(3x)dx
1
v = sin(3 x)
du = –sin x dx
3

Instructor's Resource Manual

Section 7.2

423

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1 ⎡1

1



1

∫ sin x sin(3x)dx = – 3 sin x cos(3x) + 3 ⎢⎣ 3 cos x sin(3x) + 3 ∫ sin x sin(3x)dx ⎥⎦
1
1
1
= – sin x cos(3 x ) + cos x sin(3 x) + ∫ sin x sin(3 x)dx
3
9
9
8
1
1
sin x sin(3 x) dx = – sin x cos(3 x) + cos x sin(3 x) + C
9∫
3
9
3
1
∫ sin x sin(3x)dx = – 8 sin x cos(3x) + 8 cos x sin(3x) + C

50. u = cos (5x)

dv = sin(7x)dx
1
du = –5 sin(5x)dx v = – cos(7 x)
7
1
5
∫ cos(5 x) sin(7 x)dx = – 7 cos(5 x) cos(7 x) – 7 ∫ sin(5 x) cos(7 x)dx
u = sin(5x)
dv = cos(7x)dx
1
v = sin(7 x)
du = 5 cos(5x)dx
7
1
5 ⎡1
5

∫ cos(5 x) sin(7 x)dx = – 7 cos(5 x) cos(7 x) – 7 ⎢⎣ 7 sin(5 x) sin(7 x) – 7 ∫ cos(5 x) sin(7 x)dx ⎥⎦
1
5
25
= – cos(5 x) cos(7 x) – sin(5 x) sin(7 x) + ∫ cos(5 x) sin(7 x)dx
7
49
49
24
1
5
cos(5 x ) sin(7 x)dx = – cos(5 x ) cos(7 x ) – sin(5 x ) sin(7 x ) + C

49
7
49
7
5
∫ cos(5 x) sin(7 x)dx = – 24 cos(5 x) cos(7 x) – 24 sin(5 x) sin(7 x) + C

51. u = eα z

dv = sin βz dz
1
v = – cos β z

du = α eα z dz
αz

∫e

β

sin β z dz = –
αz

du = α eα z dz
αz

=–


1 αz
α ⎡1
α
e cos β z + ⎢ eα z sin β z − ∫ eα z sin β z dz ⎥
β
β ⎣β
β


1 αz
α αz
α2 αz
e cos β z +
e sin β z –
∫ e sin β z dz

β

β

2

αz

424

β

β

sin β z dz = −

β 2 +α2

∫e

β

dv = cos βz dz
1
v = sin β z

u=e

∫e

1 αz
α
e cos β z + ∫ eα z cos β z dz

β2

αz

∫e

β2

sin β z dz = –

sin β z dz =

Section 7.2

–β

α2 + β2

1 αz
α αz
e cos β z +
e sin β z + C

β

eα z cos β z +

β2

α
α2 + β 2

eα z sin β z + C =

eα z (α sin β z – β cos β z )

α2 + β2

+C

Instructor Solutions Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


52. u = eα z

dv = cos β z dz

du = α eα z dz
αz

∫e

v=

cos β z dz =

=

β

dv = sin βz dz
1
v = – cos β z

β

cos β z dz =


1 αz
α⎡ 1
α
e sin β z − ⎢ − eα z cos β z + ∫ eα z cos β z dz ⎥
β⎣ β
β


β

1 αz
α αz
α2 αz
e sin β z +
e cos β z –
∫ e cos β z dz

β

β2

α2 + β2
β

2

αz

∫e

α2 + β2

+C

dv = xα dx

1
dx
x

v=

xα +1
, α ≠ –1
α +1

xα +1
xα +1
xα +1
1
α
x
ln x –
x
dx
=
ln

+ C , α ≠ –1
α +1
α +1 ∫
α +1
(α + 1)2

α
∫ x ln x dx =

dv = xα dx

54. u = (ln x)2

2 ln x
dx
x

du =

α αz
1
e cos β z + eα z sin β z + C
2
β
β

eα z (α cos β z + β sin β z )

53. u = ln x
du =

β2

cos β z dz =

αz
∫ e cos β z dz =

α
2
∫ x (ln x) dx =

=

sin β z

1 αz
α
e sin β z – ∫ eα z sin β z dz

du = α eα z dz
αz

β

β

u = eα z

∫e

1

v=

xα +1
, α ≠ –1
α +1

xα +1
2
xα +1
xα +1 ⎤
2 ⎡ xα +1
α
2
(ln x)2 –
ln
x
x
dx
(ln
)
ln
=
x

x


⎥+C
α +1
α +1 ∫
α +1
α + 1 ⎣⎢ α + 1
(α + 1) 2 ⎦⎥

xα +1
xα +1
xα +1
(ln x) 2 – 2
ln x + 2
+ C , α ≠ –1
α +1
(α + 1)2
(α + 1)3

Problem 53 was used for
55. u = xα

α βx

e

dx =

v=

xα e β x

β

56. u = xα
du = α xα –1dx
α

∫x

ln x dx.

dv = e β x dx

du = α xα –1dx

∫x

α

∫x

sin β x dx = –



1 βx
e

β

α α –1 β x
x e dx
β∫

dv = sin βx dx
1
v = – cos β x

β

xα cos β x

β

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+

α α –1
cos β x dx
x
β∫

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425

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57. u = xα

dv = cos βx dx
1
v = sin β x

du = α xα –1dx
α

∫x

β

α

x sin β x

cos β x dx =

β



α α –1
sin β x dx
x
β∫

58. u = (ln x)α

dv = dx
α –1

α (ln x)

du =

α

∫ (ln x)

v=x

dx

x

dx = x(ln x)α – α ∫ (ln x)α –1 dx

59. u = (a 2 – x 2 )α

dv = dx
2 α –1

du = –2α x(a – x )
2

∫ (a

2

v=x

dx

– x 2 )α dx = x (a 2 – x 2 )α + 2α ∫ x 2 (a 2 – x 2 )α –1 dx

60. u = cosα –1 x

dv = cos x dx
α –2

du = –(α – 1) cos
α

∫ cos

v = sin x

x sin x dx

x dx = cosα –1 x sin x + (α – 1) ∫ cosα –2 x sin 2 x dx

= cosα −1 x sin x + (α − 1) ∫ cosα − 2 x(1 − cos 2 x) dx = cosα –1 x sin x + (α – 1) ∫ cosα –2 x dx – (α – 1) ∫ cosα x dx

α ∫ cosα x dx = cosα −1 x sin x + (α − 1) ∫ cosα − 2 x dx
α
∫ cos x dx =

cosα –1 x sin x α – 1
α –2
+
∫ cos x dx

α

α

61. u = cosα –1 β x

dv = cos βx dx

du = – β (α – 1) cosα –2 β x sin β x dx
α
∫ cos β x dx =

=
=

β
cosα –1 β x sin β x

β

∫x

sin β x

+ (α – 1) ∫ cosα –2 β x sin 2 β x dx

+ (α – 1) ∫ cosα –2 β x dx – (α – 1) ∫ cosα β x dx

β

+ (α − 1) ∫ cosα − 2 β x dx

cosα –1 β x sin β x α – 1
α –2
+
∫ cos β x dx

αβ

α

1 4 3x 4 3 3x
1
4 ⎡1

x e – ∫ x e dx = x 4 e3 x – ⎢ x3e3 x – ∫ x 2 e3 x dx ⎥
3
3 ⎣3
3
3

4
4 ⎡1
2
1
4
4
8 ⎡1
1


– x3e3 x + ⎢ x 2 e3 x – ∫ xe3 x dx ⎥ = x 4 e3 x – x3e3 x + x 2 e3 x – ⎢ xe3 x – ∫ e3 x dx ⎥
9
3 ⎣3
3
3
9
9
9
3
3



4 3 3x 4 2 3x 8 3x 8 3x
– x e + x e –
xe + e + C
9
9
27
81

e dx =

4 3x

1 4 3x
x e
3
1
= x 4 e3 x
3

426

1

β

+ (α − 1) ∫ cosα − 2 β x(1 − cos 2 β x) dx

cosα −1 β x sin β x

α
∫ cos β x dx =

=

β

cosα −1 β x sin β x

α ∫ cosα β x =

62.

cosα –1 β x sin β x

v=

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


63.

64.

1 4
4
1
4⎡ 1

x sin 3 x – ∫ x3 sin 3 x dx = x 4 sin 3 x – ⎢ – x3 cos 3x + ∫ x 2 cos 3 x dx ⎥
3
3
3
3⎣ 3

1
4
4 ⎡1
2

= x 4 sin 3 x + x3 cos 3 x – ⎢ x 2 sin 3x − ∫ x sin 3 x dx ⎥
3
9
3 ⎣3
3

1 4
4 3
4 2
8⎡ 1
1

= x sin 3 x + x cos 3 x – x sin 3 x + ⎢ – x cos 3 x + ∫ cos 3 x dx ⎥
3
9
9
9⎣ 3
3

1 4
4 3
4 2
8
8
= x sin 3 x + x cos 3 x – x sin 3 x –
x cos 3 x + sin 3 x + C
3
9
9
27
81

∫x

4

cos 3 x dx =

1
5
1
5⎡ 1
3

cos5 3 x sin 3 x + ∫ cos 4 3x dx = cos5 3x sin 3x + ⎢ cos3 3x sin 3 x + ∫ cos 2 3 x dx ⎥
18
6 ⎣12
4
18
6

1
5
5
1
1


= cos5 3 x sin 3 x + cos3 3 x sin 3 x + ⎢ cos 3 x sin 3 x + ∫ dx ⎥
18
72
8 ⎣6
2

x
1
5
5
5
= cos5 3x sin 3x + cos3 3 x sin 3 x + cos 3 x sin 3 x + + C
18
72
48
16

∫ cos

6

3 x dx =

65. First make a sketch.

From the sketch, the area is given by
e

∫1 ln x dx
u = ln x
1
du = dx
x

dv = dx
v=x

∫1 ln x dx = [ x ln x ]1 − ∫1 dx = [ x ln x − x]1 = (e – e) – (1 · 0 – 1) = 1
e

e

e

e

e

66. V = ∫ π(ln x) 2 dx
1

u = (ln x)2

du =

2 ln x
dx
x

dv = dx
v=x

e
e
e
e


π∫ (ln x) 2 dx = π ⎜ ⎡ x(ln x) 2 ⎤ − 2∫ ln x dx ⎟ = π ⎡ x(ln x)2 − 2( x ln x − x) ⎤ = π[ x (ln x) 2 − 2 x ln x + 2 x]1e
⎦1
1
1

⎦1
⎝⎣

= π[(e − 2e + 2e) − (0 − 0 + 2)] = π(e − 2) ≈ 2.26

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67.

9

∫0 3e

–x/3

9
9
⎛ 1 ⎞
dx = –9∫ e – x / 3 ⎜ – dx ⎟ = –9[e – x / 3 ]90 = – + 9 ≈ 8.55
0
3


e3

9

9

0

0

68. V = ∫ π(3e – x / 3 ) 2 dx = 9π ∫ e –2 x / 3 dx
27π –2 x / 3 9
27π 27 π
⎛ 3⎞ 9
⎛ 2 ⎞
= 9π ⎜ – ⎟ ∫ e –2 x / 3 ⎜ – dx ⎟ = –
[e
]0 = –
+
≈ 42.31
0
2
3
2
2




2e 6

69.

π/4

∫0

( x cos x – x sin x)dx = ∫

π/ 4

0

x cos x dx – ∫

π/4

0

x sin x dx

π4
π4
π4


π4
= ⎜ ⎣⎡ x sin x ⎦⎤ 0 − ∫ sin x dx ⎟ − ⎛⎜ [ − x cos x ]0 + ∫ cos x dx ⎞⎟
0
0


⎠ ⎝


–1 ≈ 0.11
4
Use Problems 60 and 61 for ∫ x sin x dx and ∫ x cos x dx.
= [ x sin x + cos x + x cos x – sin x]0π / 4 =

⎛ x⎞
x sin ⎜ ⎟ dx
⎝2⎠
x
u=x
dv = sin dx
2
x
du = dx v = –2 cos
2


⎛⎡


x⎤
x ⎞
x⎤ ⎞

V = 2π ⎜ ⎢ –2 x cos ⎥ + ∫ 2 cos dx ⎟ = 2π ⎜ 4π + ⎢ 4sin ⎥ ⎟ = 8π2
0
⎜⎣

2 ⎦0
2 ⎟
2 ⎦0 ⎟






70. V = 2π∫



0

71.

e

∫1 ln x

2

e

dx = 2 ∫ ln x dx
1

u = ln x
1
du = dx
x

dv = dx
v=x

(

)

e
e ⎞

2∫ ln x dx = 2 ⎜ [ x ln x]1e − ∫ dx ⎟ = 2 e − [ x]1e = 2
1
1


e

∫1 x ln x

2

e

dx = 2 ∫ x ln x dx
1

u = ln x
1
du = dx
x

dv = x dx
1
v = x2
2
e
e


⎛1
e
e1
⎡1

⎡1 ⎤ ⎞ 1
2∫ x ln x dx = 2 ⎜ ⎢ x 2 ln x ⎥ – ∫ x dx ⎟ = 2 ⎜ e2 – ⎢ x 2 ⎥ ⎟ = (e2 + 1)
1
⎜⎣2

⎜2
⎦1 1 2
⎣ 4 ⎦1 ⎟⎠ 2




428

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1 e
(ln x)2 dx
2 ∫1
u = (ln x)2
du =

dv = dx

2 ln x
dx
x

v=x

e
1 e
1
1
(ln x)2 dx = ⎛⎜ [ x(ln x) 2 ]1e – 2∫ ln x dx ⎞⎟ = (e – 2)
1
2 ∫1
2⎝
⎠ 2

x=
y=

72. a.

1 (e 2
2

+ 1)

=

2
1 (e – 2)
2

2

=

e2 + 1
4

e–2
4

u = cot x

dv = csc2 x dx

du = – csc2 x dx

v = –cot x

∫ cot x csc

2

x dx = − cot 2 x − ∫ cot x csc2 x dx

2∫ cot x csc2 x dx = − cot 2 x + C



1
cot x csc2 x dx = − cot 2 x + C
2

b. u = csc x
du = –cot x csc x dx

∫ cot x csc

2

dv = cot x csc x dx
v = –csc x

x dx = − csc2 x − ∫ cot x csc2 x dx

2∫ cot x csc2 x dx = − csc2 x + C


c.
73. a.

1
cot x csc2 x dx = − csc 2 x + C
2

1
1
1
1
– cot 2 x = – (csc 2 x – 1) = – csc2 x +
2
2
2
2
p ( x ) = x3 − 2 x
g ( x) = e x

All antiderivatives of g ( x) = e x

∫ (x
b.

3

− 2 x)e x dx = ( x3 − 2 x)e x − (3x 2 − 2)e x + 6 xe x − 6e x + C

p( x) = x 2 − 3x + 1
g(x) = sin x
G1 ( x) = − cos x
G2 ( x) = − sin x
G3 ( x) = cos x

∫ (x

2

− 3 x + 1) sin x dx = ( x 2 − 3x + 1)(− cos x) − (2 x − 3)(− sin x) + 2 cos x + C

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


74. a.

We note that the nth arch extends from x = 2π (n − 1) to x = π (2n − 1) , so the area of the nth arch is
A(n) =

π (2 n −1)

∫2π (n−1) x sin x dx .

u=x

Using integration by parts:

dv = sin x dx

du = dx

v = − cos x

π (2 n −1)
π (2 n −1)
π (2 n −1)
1)
1)
A(n) = ∫2ππ(2( nn−−1)
x sin x dx = − x cos x 2π ( n −1) − ∫2ππ(2( nn−−1)
− cos x dx = − x cos x 2π ( n −1) + sin x 2π ( n −1)

[

]

[

]

[

]

= [ −π (2n − 1) cos(π (2n − 1)) + 2π (n − 1) cos(2π (n − 1)) ] + [sin(π (2n − 1)) − sin(2π (n − 1)) ]
= −π (2n − 1)(−1) + 2π (n − 1)(1) + 0 − 0 = π [ (2n − 1) + (2n − 2) ] .

b.

V = 2π ∫

So

A(n) = (4n − 3)π

3π 2



x sin x dx

u = x2
dv = sin x dx
du = 2x dx
v = –cos x







V = 2π ⎜ ⎡ – x 2 cos x ⎤ + ∫ 2 x cos x dx ⎟ = 2π ⎜ 9π2 + 4π2 + ∫ 2 x cos x dx ⎟


π
2
π
2





u = 2x
dv = cos x dx
du = 2 dx
v = sin x



V = 2π ⎜13π2 + [2 x sin x]32ππ – ∫ 2sin x ⎟




(

)

= 2π 13π2 + [2 cos x]32ππ = 2π(13π2 – 4) ≈ 781
dv = sin nx dx

75. u = f(x)

1
v = − cos nx
n

du = f ′( x)dx

π

an =


1⎡ ⎡ 1
1 π

cos(nx) f ′( x)dx ⎥
− cos(nx) f ( x) ⎥ + ∫


n −π
π⎣ ⎣ n


−π 



Term 1

Term 1 =

Term 2

1
1
cos(nπ)( f (−π) − f (π)) = ± ( f (−π) − f (π))
n
n

Since f ′( x) is continuous on [– ∞ , ∞ ], it is bounded. Thus,

π

∫– π cos(nx) f ′( x)dx

is bounded so

π
1 ⎡
± ( f (−π) − f (π)) + ∫ cos(nx) f ′( x) dx ⎥⎤ = 0.

−π
n→∞ πn ⎣


lim an = lim

n →∞

1/ n

Gn [(n + 1)(n + 2) ⋅⋅⋅ (n + n)]1 n ⎡⎛ 1 ⎞⎛ 2 ⎞ ⎛ n ⎞ ⎤
=
76.
= ⎢⎜1 + ⎟⎜ 1 + ⎟ …⎜1 + ⎟ ⎥
n
[n n ]1 n
⎣⎝ n ⎠⎝ n ⎠ ⎝ n ⎠ ⎦

⎛ G ⎞ 1 ⎡⎛ 1 ⎞⎛ 2 ⎞ ⎛ n ⎞ ⎤
ln ⎜ n ⎟ = ln ⎢⎜1 + ⎟⎜ 1 + ⎟ …⎜ 1 + ⎟ ⎥
⎝ n ⎠ n ⎣⎝ n ⎠⎝ n ⎠ ⎝ n ⎠ ⎦
=

1⎡ ⎛ 1⎞
⎛ 2⎞
⎛ n ⎞⎤
ln ⎜1 + ⎟ + ln ⎜ 1 + ⎟ + ⋅⋅⋅ + ln ⎜1 + ⎟ ⎥
n ⎢⎣ ⎝ n ⎠
n


⎝ n ⎠⎦

⎛G
lim ln ⎜ n
n →∞ ⎝ n

⎛G
lim ⎜ n
n →∞ ⎝ n

430

2

⎟ = ∫1 ln x dx = 2 ln 2 –1


4
⎞ 2 ln 2–1
= 4e –1 =
⎟=e
e


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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


77. The proof fails to consider the constants when integrating 1t .

The symbol

∫ (1 t ) dt is a

family of functions, all of who whom have derivative 1t . We know that any two of these

functions will differ by a constant, so it is perfectly correct (notationally) to write
78.

∫ (1 t ) dt = ∫ (1 t ) dt + 1

d 5x
[e (C1 cos 7 x + C2 sin 7 x) + C3 ] = 5e5 x (C1 cos 7 x + C2 sin 7 x) + e5 x (–7C1 sin 7 x + 7C2 cos 7 x)
dx
= e5 x [(5C1 + 7C2 ) cos 7 x + (5C2 – 7C1 ) sin 7 x]
Thus, 5C1 + 7C2 = 4 and 5C2 – 7C1 = 6.

Solving, C1 = –

11
29
; C2 =
37
37

79. u = f(x)
du = f ′( x)dx
b

∫a

dv = dx
v=x

f ( x)dx = [ xf ( x)]a – ∫ xf ′( x)dx
b

b

a

Starting with the same integral,
u = f(x)
dv = dx
du = f ′( x)dx
v=x–a
b

∫a

f ( x) dx = [ ( x – a) f ( x) ]a – ∫ ( x – a) f ′( x)dx
b

b

a

80. u = f ′( x)
du = f ′′( x)dx

dv = dx
v=x–a

b
f (b) – f (a ) = ∫ f ′( x)dx = [ ( x – a) f ′( x) ]a – ∫ ( x – a ) f ′′( x)dx = f ′(b)(b – a ) – ∫ ( x – a ) f ′′( x)dx
b

b

b

a

a

a

Starting with the same integral,
u = f ′( x)
dv = dx
du = f ′′( x)dx
v=x–b
f (b) − f (a) = ∫ f ′( x)dx = [ ( x – b) f ′( x) ]a – ∫ ( x – b) f ′′( x)dx = f ′(a)(b − a) – ∫ ( x – b) f ′′( x)dx
b

b

a

b

b

a

a

81. Use proof by induction.
t

t

a

a

n = 1: f (a) + f ′(a )(t – a) + ∫ (t – x) f ′′( x)dx = f (a ) + f ′(a )(t – a ) + [ f ′( x)(t – x)]ta + ∫ f ′( x)dx

= f (a) + f ′(a)(t – a) – f ′(a)(t – a) + [ f ( x)]ta = f (t )

Thus, the statement is true for n = 1. Note that integration by parts was used with u = (t – x), dv = f ′′( x)dx.
Suppose the statement is true for n.
n
t (t – x ) n ( n +1)
f (i ) ( a )
(t – a)i + ∫
( x)dx
f (t ) = f (a ) + ∑
f
a
i!
n!
i =1
Integrate

(t – x)n ( n +1)
∫a n ! f ( x)dx by parts.
t

u = f ( n +1) ( x)

dv =

(t – x)n
dx
n!

du = f ( n + 2) ( x)

v=–

(t – x)n +1
(n + 1)!
t

⎡ (t – x)n +1 ( n +1) ⎤
t (t – x) n +1 ( n + 2)
(t – x) n ( n +1)
=
+
(
)

(
)
( x)dx
f
x
dx
f
x


∫a n !
∫a (n + 1)! f
⎣⎢ (n + 1)!
⎦⎥
t

a

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431

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


=

t (t – x ) n +1 ( n + 2)
(t – a )n +1 ( n +1)
(a) + ∫
( x)dx
f
f
a (n + 1)!
(n + 1)!

t (t – x) n +1 ( n + 2)
f (i ) ( a )
(t – a) n +1 ( n +1)
(t – a)i +
(a) + ∫
( x)dx
f
f
a ( n + 1)!
(n + 1)!
i!
i =1
n +1 (i )
t (t – x ) n +1 ( n + 2)
f (a)
(t – a)i + ∫
( x)dx
= f (a) + ∑
f
a ( n + 1)!
i!
i =1
n

Thus f (t ) = f (a ) + ∑

Thus, the statement is true for n + 1.
82. a.

1
B (α , β ) = ∫ xα −1 (1 − x) β −1 dx where α ≥ 1, β ≥ 1
0

x = 1 – u, dx = –du
1 α −1

∫0 x

(1 − x) β −1 dx = ∫ (1 − u )α −1 (u ) β −1 (− du ) = ∫ (1 − u )α −1 u β −1du = B ( β , α )
0

1

1

0

Thus, B(α, β) = B(β, α).
b.

1
B (α , β ) = ∫ xα −1 (1 − x) β −1 dx
0

α −1

dv = (1 − x) β −1 dx

u=x

du = (α − 1) xα − 2 dx

v=−

1

β

(1 − x) β

1

⎡ 1
⎤ α − 1 1 α −2
α − 1 1 α −2
B (α , β ) = ⎢ − xα −1 (1 − x) β ⎥ +
x
(1 − x) β dx =
x
(1 − x) β dx


0
0
β
β
β

⎦0
α −1
=
B (α − 1, β + 1)

(*)

β

Similarly,
1
B (α , β ) = ∫ xα −1 (1 − x) β −1 dx
0

u = (1 − x)

β −1

dv = xα −1dx

du = − ( β − 1) (1 − x) β − 2 dx

v=

1 α
x

α

1

β −1 1 α
β −1
β −1 1 α
⎡1

x (1 − x) β − 2 dx =
B (α + 1, β − 1)
B (α , β ) = ⎢ xα (1 − x) β −1 ⎥ +
x (1 − x) β − 2 dx =


0
0
α
α
α
⎣α
⎦0
c.

Assume that n ≤ m. Using part (b) n times,
( n − 1) (n − 2)
n −1
B (n, m) =
B (n − 1, m + 1) =
B (n − 2, m + 2)
m
m(m + 1)
=…=

( n − 1) (n − 2) ( n − 3)…⋅ 2 ⋅1

m(m + 1) ( m + 2 )… (m + n − 2)
1

B(1, m + n − 1) = ∫ (1 − x)m+ n − 2 dx = −
0

Thus, B (n, m) =

Section 7.2

1
1
[(1 − x)m+ n −1 ]10 =
m + n −1
m + n −1

( n − 1) (n − 2) ( n − 3)…⋅ 2 ⋅1

m(m + 1) ( m + 2 ) … (m + n − 2) ( m + n − 1)

If n > m, then B (n, m) = B (m, n) =

432

B(1, m + n − 1).

( n − 1)!( m − 1)!
(n + m − 1)!

=

( n − 1)!( m − 1)! ( n − 1)!( m − 1)!
(m + n − 1)!

=

(n + m − 1)!

by the above reasoning.

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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


dv = f ′′(t )dt
v = f ′(t )

83. u = f(t)
du = f ′(t )dt
b

∫a

f ′′(t ) f (t )dt = [ f (t ) f ′(t ) ]a – ∫ [ f ′(t )]2 dt
b

b

a

b

b

a

a

= f (b) f ′(b) − f (a) f ′(a) − ∫ [ f ′(t )]2 dt = − ∫ [ f ′(t )]2 dt
b

[ f ′(t )]2 ≥ 0, so − ∫ [ f ′(t )]2 ≤ 0 .
a

84.

x⎛ t



∫0 ⎜⎝ ∫0 f ( z )dz ⎟⎠ dt
t

u = ∫ f ( z )dz dv = dt
0

du = f(t)dt

v=t

x⎛ t





t



x

x

x

x

∫0 ⎜⎝ ∫0 f ( z )dz ⎟⎠ dt = ⎣⎢t ∫0 f ( z )dz ⎦⎥0 – ∫0 t f (t )dt = ∫0 x f ( z)dz – ∫0 t f (t )dt
By letting z = t,
x⎛ t

x

x

∫0 x f ( z )dz = ∫0 x f (t )dt ,



x

so

x

x

∫0 ⎜⎝ ∫0 f ( z )dz ⎟⎠ dt = ∫0 x f (t )dt – ∫0 t f (t )dt = ∫0 ( x – t ) f (t )dt
x t1
t
⋅⋅⋅ n −1
0 0
0

85. Let I = ∫





f (tn ) dtn ...dt2 dt1 be the iterated integral. Note that for i ≥ 2, the limits of integration of the

integral with respect to ti are 0 to ti −1 so that any change of variables in an outer integral affects the limits, and
hence the variables in all interior integrals. We use induction on n, noting that the case n = 2 is solved in the
previous problem.
Assume we know the formula for n − 1 , and we want to show it for n.
x t1 t2
t
⋅⋅⋅ n −1
0 0 0
0

I =∫

∫ ∫



where F ( tn −1 ) = ∫

tn −1

0

t

t

0

0

tn − 2
0

f (tn ) dtn ...dt3 dt2 dt1 = ∫ 1 ∫ 2 ⋅⋅⋅∫

F (tn −1 ) dtn −1...dt3 dt2 dt1

f ( tn ) dn .

By induction,
x
1
n−2
I=
F ( t1 )( x − t1 ) dt1

0

n
2
!
(
)
u = F ( t1 ) = ∫ f ( tn ) dtn ,
t1

0

du = f ( t1 ) dt1 ,

I=

dv = ( x − t1 )
v=−

n−2

1
( x − t1 )n −1
n −1

t1 = x
t
1 ⎧⎪ ⎡ 1
1 x
⎪⎫
n −1
f ( t1 )( x − t1 ) dt1 ⎬ .
( x − t1 )n −1 ∫01 f ( tn ) dtn ⎤⎥ +
⎨⎢−

0
( n − 2 )! ⎩⎪ ⎣ n − 1
⎦ t1 = 0 n − 1
⎭⎪

x
1
f (t1 )( x − t1 ) n −1 dt1

0
(n − 1)!
(note: that the quantity in square brackets equals 0 when evaluated at the given limits)

=

Instructor’s Resource Manual

Section 7.3

433

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


86. Proof by induction.
n = 1:
u = P1 ( x)
du =

∫e

x

dv = e x dx

dP1 ( x)
dx
dx

v = ex

P1 ( x)dx = e x P1 ( x) – ∫ e x

Note that

dP1 ( x)
dP ( x )
dP ( x)
dx = e x P1 ( x) – 1 ∫ e x dx = e x P1 ( x) – e x 1
dx
dx
dx

dP1 ( x)
is a constant.
dx

Suppose the formula is true for n. By using integration by parts with u = Pn +1 ( x) and dv = e x dx,

∫e

x

Pn +1 ( x)dx = e x Pn +1 ( x) – ∫ e x

Note that

dPn +1 ( x)
dx
dx

dPn +1 ( x)
is a polynomial of degree n, so
dx

j +1
j
n
n


Pn +1 ( x )
j d
x
x
x
j d ⎛ dPn +1 ( x ) ⎞ ⎥
x
x

(
)
(
)
(
1)
1
=


=


e
P
x
dx
e
P
x
e
e
P
x
e
(
)
(
)


n +1
n +1
⎜ dx ⎟
∫ n+1
j
⎢⎣ j = 0
⎠ ⎥⎦
dx ⎝
dx j +1
j =0
n +1

d j Pn +1 ( x )

j =1

dx j

= e x Pn +1 ( x) + e x ∑ (−1) j

87.

n +1

d j Pn +1 ( x)

j =0

dx j

= e x ∑ (−1) j

4

d j (3 x 4 + 2 x 2 )

j =0

dx j

x
j
4
2 x
∫ (3x + 2 x )e dx = e ∑ (–1)

= e x [3x 4 + 2 x 2 – 12 x3 – 4 x + 36 x 2 + 4 – 72 x + 72]
= e x (3x 4 – 12 x3 + 38 x 2 – 76 x + 76)

7.3 Concepts Review
1.



1 + cos 2 x
dx
2

2.

∫ (1 – sin

3.

∫ sin

2

2

2

x) cos x dx

x(1 – sin 2 x) cos x dx
1
[cos(m + n) x + cos(m − n) x ]
2

Problem Set 7.3

∫ sin

2

x dx = ∫

1 ⎛ 1 – cos 2u ⎞

⎟ du
6 ∫⎝
2

1
=
(1 – 2 cos 2u + cos 2 2u )du
24 ∫
1
1
1
=
du –
2 cos 2u du + ∫ (1 + cos 4u )du


24
24
48
3
1
1
=
du –
2 cos 2u du +
4 cos 4u du


48
24
192 ∫
3
1
1
sin 24 x + C
= (6 x) – sin12 x +
48
24
192
3
1
1
= x – sin12 x +
sin 24 x + C
8
24
192

=

4. cos mx cos nx =

1.

2. u = 6x, du = 6 dx
1
4
4
∫ sin 6 x dx = 6 ∫ sin u du

1 – cos 2 x
dx
2

1
1
dx – ∫ cos 2 x dx

2
2
1
1
= x – sin 2 x + C
2
4
=

434

Section 7.2

Instructor Solutions Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


3.

∫ sin x dx = ∫ sin x(1 − cos x)dx
= ∫ sin x dx − ∫ sin x cos 2 x dx
3

2

5.

cos5 θ dθ = ∫

π/2

0

=∫

π/ 2

0

1
= − cos x + cos3 x + C
3
4.

π/2

∫0

(1 – sin 2 θ )2 cosθ dθ

(1 – 2sin 2 θ + sin 4 θ ) cos θ dθ
π/2

2
1


= ⎢sin θ – sin 3 θ + sin 5 θ ⎥
3
5

⎦0
2
1
8


= ⎜1 – + ⎟ – 0 =
15
⎝ 3 5⎠

∫ cos x dx =
= ∫ cos x (1 − sin 2 x)dx
= ∫ cos x dx − ∫ cos x sin 2 x dx
3

1
= sin x − sin 3 x + C
3

6.

π/2

∫0

sin 6 θ dθ = ∫

π / 2 ⎛ 1 – cos 2θ

0




2

3


⎟ dθ


1 π/ 2
(1 – 3cos 2θ + 3cos 2 2θ – cos3 2θ )dθ

0
8
1 π/ 2
3 π/2
3 π/ 2 2
1 π/2 3
= ∫
dθ – ∫
2 cos 2θ dθ + ∫
cos 2θ – ∫
cos 2θ dθ
0
0
0
8
16
8
8 0
1
3
3 π / 2 ⎛ 1 + cos 4θ ⎞
1 π/2
2
= [θ ]0π / 2 – [sin 2θ ]0π / 2 + ∫

⎟ dθ – ∫0 (1 – sin 2θ ) cos 2θ dθ
0
8
16
8
2
8


1 π 3 π/ 2
3 π/2
1 π/2
1 π/2 2
dθ + ∫
4 cos 4θ dθ – ∫
2 cos 2θ dθ + ∫
sin 2θ ⋅ 2 cos 2θ dθ
= ⋅ + ∫
0
0
0
8 2 16
64
16
16 0
π 3π 3
1
1

= + + [sin 4θ ]0π / 2 – [sin 2θ ]0π / 2 + [sin 3 2θ ]0π / 2 =
16 32 64
16
48
32
=

7.

∫ sin
=–

8.

5

4 x cos 2 4 x dx = ∫ (1 – cos 2 4 x) 2 cos 2 4 x sin 4 x dx = ∫ (1 – 2 cos 2 4 x + cos 4 4 x) cos 2 4 x sin 4 x dx

1
1
1
1
(cos 2 4 x – 2 cos 4 4 x + cos6 4 x)(–4sin 4 x)dx = – cos3 4 x + cos5 4 x – cos7 4 x + C
12
10
28
4∫

∫ (sin

3

2t ) cos 2tdt = ∫ (1 – cos 2 2t )(cos 2t )1/ 2 sin 2t dt = –

1
[(cos 2t )1/ 2 – (cos 2t )5 / 2 ](–2sin 2t )dt
2∫

1
1
= – (cos 2t )3 / 2 + (cos 2t )7 / 2 + C
3
7
9.

∫ cos

3

3θ sin –2 3θ dθ = ∫ (1 – sin 2 3θ ) sin –2 3θ cos 3θ dθ =

1
(sin −2 3θ − 1)3cos 3θ dθ
3∫

1
1
= – csc3θ – sin 3θ + C
3
3
10.

∫ sin

1/ 2

=

2 z cos3 2 z dz = ∫ (1 – sin 2 2 z ) sin1/ 2 2 z cos 2 z dz

1
1
1
(sin1/ 2 2 z – sin 5 / 2 2 z )2 cos 2 z dz = sin 3 / 2 2 z – sin 7 / 2 2 z + C

3
7
2

Instructor’s Resource Manual

Section 7.3

435

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


2

11.

2

1
⎛ 1 – cos 6t ⎞ ⎛ 1 + cos 6t ⎞
4
4
2
4
∫ sin 3t cos 3t dt = ∫ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ dt = 16 ∫ (1 – 2 cos 6t + cos 6t )dt
1 ⎡
1
1
1

= ∫ ⎢1 – (1 + cos12t ) + (1 + cos12t )2 ⎥ dt = – ∫ cos12t dt + ∫ (1 + 2 cos12t + cos 2 12t )dt
16 ⎣
4
16
64

1
1
1
1
=–
12 cos12t dt + ∫ dt +
12 cos12t dt +
(1 + cos 24t )dt
192 ∫
64
384 ∫
128 ∫
1
1
1
1
1
3
1
1
sin12t + t +
sin12t +
sin 24t + C =
sin12t +
sin 24t + C
=–
t+
t–
192
64 384
128 3072
128 384
3072
3

12.

13.

1
⎛ 1 + cos 2θ ⎞ ⎛ 1 – cos 2θ ⎞
6
2
3
4
∫ cos θ sin θ dθ = ∫ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ dθ = 16 ∫ (1 + 2 cos 2θ – 2 cos 2θ – cos 2θ )dθ
1
1
1
1
= ∫ dθ + ∫ 2 cos 2θ dθ – ∫ (1 – sin 2 2θ ) cos 2θ dθ – ∫ (1 + cos 4θ )2 dθ
16
16
8
64
1
1
1
1
1
= ∫ dθ + ∫ 2 cos 2θ dθ – ∫ 2 cos 2θ dθ + ∫ 2sin 2 2θ cos 2θ dθ – ∫ (1 + 2 cos 4θ + cos 2 4θ )dθ
16
16
16
16
64
1
1
1
1
1
4 cos 4θ dθ –
(1 + cos8θ )dθ
= ∫ dθ + ∫ sin 2 2θ ⋅ 2 cos 2θ dθ – ∫ dθ –
16
16
64
128 ∫
128 ∫
1
1
1
1
1
1
θ−
sin 4θ −
sin 8θ + C
= θ + sin 3 2θ − θ −
16
48
64
128
128
1024
5
1
1
1
=
θ + sin 3 2θ –
sin 4θ –
sin 8θ + C
128
48
128
1024
1

=

14.

1

∫ sin 4 y cos 5 y dy = 2 ∫ [sin 9 y + sin(− y)] dy = 2 ∫ (sin 9 y − sin y)dy
1⎛ 1
1
1

⎜ − cos 9 y + cos y ⎟ + C = cos y − cos 9 y + C
2⎝ 9
2
18

1

∫ cos y cos 4 y dy = 2 ∫ [cos 5 y + cos(−3 y)]dy

=

1
1
1
1
sin 5 y − sin(−3 y) + C = sin 5 y + sin 3 y + C
10
6
10
6

2

15.

16.

1
⎛ 1 – cos w ⎞ ⎛ 1 + cos w ⎞
4 ⎛ w⎞
2⎛ w⎞
2
3
∫ sin ⎜⎝ 2 ⎟⎠ cos ⎜⎝ 2 ⎟⎠ dw = ∫ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ dw = 8 ∫ (1 – cos w – cos w + cos w)dw
1 ⎡
1
1 ⎡1 1


= ∫ ⎢1 – cos w – (1 + cos 2w) + (1 – sin 2 w) cos w⎥ dw = ∫ ⎢ – cos 2 w – sin 2 w cos w⎥dw
8 ⎣
2
8
2
2



1
1
1
= w – sin 2 w – sin 3 w + C
16
32
24

1

∫ sin 3t sin t dt = ∫ − 2 [cos 4t − cos 2t ] dt

(

)

1
cos 4tdt − ∫ cos 2tdt
2 ∫
1⎛1
1

= − ⎜ sin 4t − sin 2t ⎟ + C
2⎝4
2

1
1
= − sin 4t + sin 2t + C
8
4
=−

436

Section 7.3

Instructor's Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


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