Más contenido relacionado La actualidad más candente (20) Similar a 51548 0131469657 ism-7 (20) 51548 0131469657 ism-71. CHAPTER 7 Techniques of Integration
7.1 Concepts Review
1. elementary function
2. ∫u5du
3. ex
4. 2 3
∫ u du
1
Problem Set 7.1
1. ( – 2)5 1 ( – 2)6
∫ x dx = x +C
6
2. 3 1 3 3 2 (3 )3/ 2
∫ x dx = ∫ x ⋅ dx = x +C
3 9
3. u = x2 +1, du = 2x dx
When x = 0, u = 1 and when x = 2, u = 5 .
∫ 2 x ( x 2 + 1) 5 dx = 1 ∫ 2 ( x 2 + 1) 5
(2 x dx
)
0 0
2
5 5
1
1
2
= ∫ u du
6 5 6 6
⎡u ⎤ −
= ⎢ ⎥ =
⎢⎣ ⎥⎦
= =
12 12
1
15624 1302
12
5 1
4. u = 1– x2 , du = –2x dx
When x = 0, u = 1 and when x = 1, u = 0 .
∫ 1 x 1– x 2 dx = – 1 ∫ 1 1 − x 2
( − 2 x dx
)
0 0
2
1
2
1
2
0 1/ 2
1
1 1/ 2
0
u du
∫
∫
u du
= −
=
1
1 1
3 3
= ⎡⎢ u 3/ 2
⎤⎥ = ⎣ ⎦
0
dx = 1 tan
⎛ x ⎞ ⎜ ⎟
+ C
x
5. –1
∫
2
+ ⎝ ⎠ 4 2 2
6. u = 2 + ex , du = exdx
∫ ∫
2
+ = ln u +C
ln 2
ln(2 )
x
x
e dx =
du
e u
x
e C
e C
= + +
= + x
+
7. u = x2 + 4, du = 2x dx
x dx du
x u
∫ 2
∫
+ 1 ln
2
1
=
4 2
= u + C
1 ln 2 4
2
= x + +C
1 ln( 2 4)
2
= x + + C
8.
2 2
2 2
t dt t dt
t t
2 2 + 1 −
1
2 1 2 1
∫ =
∫
+ + – 1
=
∫ dt ∫
dt
2 t
2
+ 1
u = 2t, du = 2dt
t – 1 dt t – 1
du
∫ =
∫
+ + – 1 tan–1( 2 )
2 2
2 1 2 1
t u
= t t + C
2
9. u = 4 + z2 , du = 2z dz
∫6z 4 + z2 dz = 3∫ u du
= 2u3/ 2 +C
= 2(4 + z2 )3/ 2 +C
412 Section 7.1 Instructor’s Resource Manual
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2. 10. u = 2t +1, du = 2dt
dt du
5 5
2 1 2
∫ =
∫
+ = 5 u +C
= 5 2t +1 + C
t u
z dz z z dz
z
tan tan sec
cos
11. 2
∫ = 2
∫
u = tan z, du = sec2 z dz
∫ tan z sec2 z dz = ∫u du
1 2
2
= u +C
1 tan2
2
= z +C
12. u = cos z, du = –sin z dz
∫ecos z sin z dz = –∫ecos z (– sin z dz)
= −∫eudu = −eu +C
= –ecos z +C
13. , 1
u t du dt
= =
2
t
sin t dt 2 sin u du
∫ = ∫
t
= –2 cos u + C
= –2cos t +C
14. u = x2 , du = 2x dx
x dx du
x u
2
1– 1–
∫ = ∫
= sin–1 u +C
= sin–1(x2 ) +C
4 2
15. u = sin x, du = cos x dx
x dx du
x u
cos
1 sin 1
π
/ 4 2 / 2
0 2 0 2
∫ =
∫
+ + − u
−
1 2/2
[tan ]
tan 2
0
1
=
≈ 0.6155
2
=
16. 1– , – 1
u xdu dx
2 1–
x
= =
x dx u du
x
sin 1– –2 sin
1–
3/ 4 1/ 2
0 1
∫ = ∫
1
1/ 2
= 2∫ sin u du
1
= [ − 2cos u
]1/ 2
= − 2 ⎛ cos1 − cos 1
⎞ ⎜ ⎟
2
⎝ ⎠
≈ 0.6746
17.
3 2 2 1 (3 –1)
x +
xdx x dx dx
x x
∫ = ∫ +
∫
+ + 3 2 – ln 1
2
1 1
= x x + x + +C
18.
3
x xdx x x dx dx
x x
7 ( 2 8) 8 1
–1 –1
+
∫ = ∫ + + + ∫
1 3 1 2 8 8ln –1
3 2
= x + x + x + x +C
19. u ln 4x2 , du 2 dx
= =
x
sin(ln 4 2 ) 1 sin
x dx u du
x
∫ = ∫
2
– 1 cos
= u +C
2
– 1 cos(ln 4 2 )
2
= x +C
20. u = ln x, du 1 dx
x
=
2
sec (ln ) 1 sec2
x dx u du
x
∫ = ∫
2 2
1 tan
2
= u +C
1 tan(ln )
2
= x +C
21. u = ex , du = exdx
x
x
e dx du du
e u
6 =
6
∫
1 1
2 2
− −
1
1
u C
e C
−
−
6sin
6sin ( )
= +
= x
+
∫
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3. 22. u = x2 , du = 2x dx
x dx 1
du
x u
∫ =
∫
+ + 1 tan 1
4 2
4 2
4 2 4
= − u + C
2
⎛ ⎞
x C
1 tan–1
4 2
= ⎜⎜ ⎟⎟ +
⎝ ⎠
23. u = 1– e2x , du = –2e2xdx
x
x
2
2
e dx du
e u
3 – 3
1– 2
∫ = ∫
= –3 u +C
= –3 1– e2x +C
24.
3 3
4 4
x dx 1 4
x dx
x x
∫ =
∫
+ + 1 ln 4 4
4
4 4 4
= x + +C
1 ln( 4 4)
4
= x + + C
25.
3 1 2 3
1 2 1 2
0 0
∫ t t dt = ∫ t t dt
2
2 1
⎡ t ⎤
= ⎢ ⎥ =
⎢ ⎥
⎣ ⎦
1 0.9102
ln 3
3 3 – 1
2ln3 2ln3 2ln3
0
= ≈
26. / 6 cos / 6 cos
2 x sin x dx – 2 x (– sin x dx) π π
∫ = ∫
0 0
cos / 6
⎡ π =⎢ – 2
x ⎤
⎥
⎢⎣ ln 2
⎥⎦
0
– 1 (2 3 / 2 – 2)
=
ln 2
2 −
2 3 / 2
ln 2
0.2559
=
≈
x x dx x dx
27. sin cos 1 cos
− ⎛ ⎞ = ⎜ − ⎟
∫ ∫
sin x ⎝ sin
x
⎠ u = sin x, du = cos x dx
sin cos
x −
x dx x du
∫ = − ∫
sin
= x − ln u +C
= x − ln sin x +C
x u
28. u = cos(4t – 1), du = –4 sin(4t – 1)dt
t dt t dt
t t
sin(4 − 1) sin(4 −
1)
1 sin (4 1) cos (4 1)
∫ =
∫
− 2 − 2
− 1 1
4
= − ∫
du
u
2
1 1 1 sec(4 1)
4 4
= u− +C = t − +C
29. u = ex , du = exdx
∫ex sec exdx = ∫secu du
= ln secu + tan u + C
= ln sec ex + tan ex +C
30. u = ex , du = exdx
∫ex sec2 (ex )dx = ∫sec2 u du = tan u + C
= tan(ex ) +C
31.
x
3 sin
sec (sec2 sin cos )
x +
e dx x e x x dx
x
∫ = ∫ +
sec
= 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 = 3t2 − t −1 ,
1 (3 2 1) 1/ 2 (6 1)
2
du = t − t − − t − dt
2
t t t dt u du
(6 − 1)sin 3 − −
1 2 sin
∫ =
∫
3 t 2
− t
−
1
= –2 cos u + C
= −2cos 3t2 − 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.
4. 33. u = t3 − 2 , du = 3t2dt
2 3
t t dt u du
cos( −
2) 1 cos
sin ( 2) 3 sin
∫ =
∫
− v = sin u, dv = cos u du
2 3 2
t u
u du v dv v C
u
1 cos 1 1
3 sin 3 3
2 1
∫ = ∫ − = − − +
2
1
3sin
C
= − +
u
1
= − +
3
t
3sin( 2)
C
−
.
x dx dx x dx
x x x
1 cos2 1 cos2
sin 2 sin 2 sin 2
+
∫ = ∫ + ∫
= ∫csc2 2x dx + ∫cot 2x csc 2x dx
34. 2 2 2
1 cot 2 1 csc 2
2 2
= − x − x +C
35. u = t3 − 2 , du = 3t2dt
2 2 3 2
t t dt u du
cos ( −
2) 1 cos
sin ( 2) 3 sin
∫ =
∫
− 1 cot2
3
2 3 2
t u
= ∫ u du 1 (csc2 –1)
= ∫ u du
3
= − u − u +C
1
1[ cot ]
3
1[ cot( 3 2) ( 3
2)]
3
= − t − − t − + C
1[cot( 3 2) 3]
1
= − t − + t +C
3
36. u = 1 + cot 2t, du = −2csc2 2t
csc2 2 1 1
1 cot 2 2
t dt du
t u
∫ = −
∫
+ = − u + C
= − 1+ cot 2t + C
2
1 4
37. u = tan−1 2t , 2
du dt
t
=
+
tan −
1 2
e dt eudu
∫ + 2
∫
1 1 tan 1 2
2 2
1
t
=
1 4 t
2
−
eu C e t C
= + = +
38. u = −t2 − 2t − 5 ,
du = (–2t – 2)dt = –2(t + 1)dt
2 2 5 1 ( 1)
∫ t + e−t − t− = − ∫eudu
2
1
2
= − eu +C
1 2 2 5
2
= − e−t − t− +C
39. u = 3y2 , du = 6y dy
y dy 1 1
du
y u
∫ =
∫
1 sin 1
6 4
16 9 6 4
4 2 2
− −
= − ⎛ u ⎞ +C ⎜ ⎟
⎝ ⎠
2
− ⎛ y ⎞ C
1 sin 1 3
6 4
= ⎜⎜ ⎟⎟ +
⎝ ⎠
40. u = 3x, du = 3dx
x dx
u du u C
∫
∫
cosh 3
1 (cosh ) 1 sinh
3 3
1 sinh 3
3
= = +
x C
= +
41. u = x3 , du = 3x2dx
2 sinh 3 1 sinh
∫ x x dx = ∫ u du
3
1 cosh
3
= u +C
1 cosh 3
3
= x +C
42. u = 2x, du = 2 dx
5 5 1
9 4 2 3
∫ ∫
5 sin 1
2 3
dx =
du
x 2 2 u
2
− −
= − ⎛ u ⎞ +C ⎜ ⎟
⎝ ⎠
= − ⎛ x ⎞ +C ⎜ ⎟
5 sin 1 2
2 3
⎝ ⎠
43. u = e3t , du = 3e3tdt
t
t
3
6 2 2
e dt 1 1
du
e u
∫ =
∫
1 sin 1
3 2
4 3 2
− −
= − ⎛ u ⎞ +C ⎜ ⎟
⎝ ⎠
3
e t − C ⎛ ⎞
1 sin 1
3 2
= ⎜⎜ ⎟⎟ +
⎝ ⎠
44. u = 2t, du = 2dt
dt 1 1
du
t t u u
∫ =
∫
1 sec 1
2
2 4 2 − 1 2 2
−
1
= ⎣⎡ − u ⎦⎤ +C
= 1 sec − 1 2
t +C
2
Instructor’s Resource Manual Section 7.1 415
© 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.
5. 45. u = cos x, du = –sin x dx
x dx du
x u
sin 1
/ 2 0
0 2 1 2
∫ = −
∫
+ + 16 cos 16
π
1
16
1
0 2
du
u
=
+ ∫
1
⎡ = 1 tan
− 1
⎛ u ⎞⎤ ⎢ 4 ⎜ ⎣ ⎝ 4
⎠⎦
⎟⎥ 0
1 tan 1 1 1 tan 1 0
4 4 4
= ⎡ − ⎛ ⎞ − − ⎤ ⎢ ⎜ ⎟ ⎥ ⎣ ⎝ ⎠ ⎦
1 tan 1 1 0.0612
4 4
= − ⎛ ⎞ ≈ ⎜ ⎟
⎝ ⎠
46. 2 2 x x u e e−
= + , du = (2e2x − 2e−2x )dx
= 2(e2x − e−2x )dx
1 2 x − 2 x e 2 + e
−
2
0 2 x −
2 x
2
e −
e dx 1 1
du
e e 2
u
∫ =
∫
+ 2 2
2
+ − = ⎡⎣ ⎤⎦
1 ln
2
e e u
1 ln 2 2 1 ln 2
2 2
= e + e− −
4
2
e
e
1 +
ln 1 1 ln 2
2 2
= −
1 ln( 4 1) 1 ln( 2 ) 1 ln 2
2 2 2
= e + − e −
1 4 1 ln 2 0.6625
2 2
= ⎛⎜ ⎛⎜ e + ⎞⎟ − ⎞⎟ ≈ ⎜ ⎜ ⎟ ⎟ ⎝ ⎝ ⎠ ⎠
1 1
2 5 2 1 4
+ + + + + ∫ ∫
47. 2 2
dx =
dx
x x x x
1 ( 1)
+ + ∫
1 tan–1 1
2 2
= +
2 2
x
( 1) 2
d x
x C
⎛ + ⎞ = ⎜ ⎟ +
⎝ ⎠
1 1
–4 9 –4 4 5
+ + + ∫ ∫
48. 2 2
dx =
dx
x x x x
1 ( –2)
+ ∫
1 tan–1 – 2
5 5
2 2
x
( –2) ( 5)
d x
=
x C
⎛ ⎞
= ⎜ ⎟ +
⎝ ⎠
dx dx
+ + + + + ∫ ∫
49. =
9 x 2 18 x 10 9 x 2 18 x
9 1
dx
x
=
∫
(3 + 3)2 + 12
u = 3x + 3, du = 3 dx
dx 1
du
x u
∫ =
∫
+ + + 1 tan–1(3 3)
3
2 2 2 2
(3 3) 1 3 1
= x + +C
50.
dx dx
x x x x
∫ =
∫
16 + 6 – 2 –( 2 – 6 +
9 – 25)
dx
x
dx
x
∫ –( – 3)2 52
52 – ( – 3)2
=
+
= ∫
= sin–1 ⎛ x – 3
⎞ ⎜ ⎟
+C 5
⎝ ⎠
x + 1 dx 1 18 x +
18
dx
+ + + + ∫ ∫
51. =
2 2
9 x 18 x 10 18 9 x 18 x
10
1 ln 9 2
18 10
18
1 ln 9 18 10
18
x x C
= + + +
( 2
)
x x C
= + + +
52.
x dx x dx
x x x x
3– 1 6 – 2
16 6 – 2 16 6 –
∫ =
∫
2 2
+ +
= 16 + 6x – x2 +C
53. u = 2t, du = 2dt
dt du
∫ = ∫
2 2 – 9 2 – 32
t t u u
⎛ ⎞
t
–1 1 2 sec
3 3
C
= ⎜ ⎟ +
⎜ ⎟
⎝ ⎠
54.
x dx x x dx
x x x
tan cos tan
sec – 4 cos sec – 4
∫ = ∫
2 2
x dx
= ∫
2
u = 2 cos x, du = –2 sin x dx
sin
1– 4cos
x
x dx
1 1
2 1
∫ 2
2
sin
1 4cos
− x
du
u
= −
−
∫
= − − u +C – 1 sin–1(2cos )
1 sin 1
2
= x +C
2
55. The length is given by
2
L dy dx
= + ⎛ ⎞ ⎜ ⎟
1 b
a
∫
⎝ dx
⎠ / 4 2
1 1 ( sin )
0
π = ∫
+ ⎡ ⎢ − ⎤ ⎣ cos
⎥ ⎦ x dx
x
= ∫ + / 4 2
/ 4 2
0
1 tan x dx π
sec x dx π
= ∫
0
/ 4
0
sec x dx π
= ∫ 0= ⎡ ⎣ ln sec x + tan x ⎤ π / 4
⎦
= ln 2 +1 − ln 1 = ln 2 +1 ≈ 0.881
416 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.
6. x x
1 +
56. sec = =
1 sin
x x x
cos cos (1 +
sin )
sin sin2 cos2 sin (1 sin ) cos2
x + x + x x + x +
x
= =
x x x x
cos (1 + sin ) cos (1 +
sin )
x x
x x
sin cos
cos 1 sin
= +
+
x x x dx
sec sin cos
∫ = ∫
⎛ ⎞ ⎜ + ⎝ cos x 1 + sin
x
⎟ ⎠ x dx x dx
x x
sin cos
cos 1 sin
= ∫ +
∫
+ 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 cos –
cos 1 sin
x dx x dx du dv
x x u v
∫ + ∫ = ∫ +
∫
+ = – ln u + ln v +C
= – ln cos x + ln 1+ sin x +C
1 +
ln sin
= +
cos
= ln sec x + tan x + C
x C
x
57. u = x – π , du = dx
x x u u
sin ( + π ) sin( + π
)
2
π π
0 2 – 2
+ + + π ∫ ∫
dx =
du
x π
u
1 cos 1 cos ( )
u u
( )sin
1 cos
π
π
+ π
+ ∫
– 2
du
u
=
u sin u sin
u
π π
π π
π
+ + ∫ ∫
du du
u u
= +
– 2 – 2
1 cos 1 cos
u sin
u
π
π
∫ 0
by symmetry.
– + 2
1 cos
du
u
=
u u du du
u u
sin sin 2
π π
π π
π
∫ =
∫
+ + v = cos u, dv = –sin u du
– 2 0 2
1 cos 1 cos
π
2 2 1
–1 1
1 2 –1 2
+ + ∫ ∫
dv dv
v v
− = π
1 1
= 2 π [tan − 1 v
] 1
⎡π ⎛ π ⎞⎤ −
1 = 2
π ⎢ − ⎜− ⎣ 4 ⎝ 4
⎟⎥ ⎠⎦
2 2
⎛ π ⎞ = π⎜ ⎟ = π
2
⎝ ⎠
58.
3
4
4 –
π
π
⎛ π ⎞ = π ⎜ + ⎟
⎝ ⎠ ∫
– ,
4
V 2 x sin x – cos
x dx
4
π
u = x du =
dx
π
π
⎛ π ⎞ ⎛ π ⎞ ⎛ π ⎞ = π ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ∫
V u u u du
2
2 –
2 sin – cos
2 4 4
π
π
2 2 sin 2 cos – 2 cos 2 sin
⎛ π ⎞ = π ⎜ + ⎟ + +
⎝ ⎠ ∫
2
2 –
u u u u udu
2 2 2 2 2
π π π
−π −π −π
⎛ π ⎞ = π ⎜ + ⎟ = +
2 2 sin 2 2 sin 2 2 sin
⎝ ⎠ ∫ ∫ ∫
u u du π u u du π u du
2 2 2
2 2
2 2
π
−π
π∫ = by symmetry. Therefore,
2 2 u sin u du 0
2
2
π π
2 2 2 2 2
= π ∫ = π − = π
V 2 2 sin u du 2 2 [ cosu] 2 2
0 0
Instructor’s Resource Manual Section 7.1 417
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7. 7.2 Concepts Review
1. uv – ∫v du
2. x; sin x dx
3. 1
4. reduction
Problem Set 7.2
1. u = x dv = exdx
du = dx v = ex
∫ xexdx = xex − ∫exdx = xex − ex +C
2. u = x dv = e3xdx
du = dx 1 3
v = e x
3
3 1 3 1 3
∫ xe xdx = xe x − ∫ e xdx
3 3
1 3 1 3
3 9
= xe x − e x +C
3. u = t dv = e5t+πdt
du = dt 1 5
v = e t+π
5
5 1 5 – 1 5
∫te t+πdt = te t+π ∫ e t+πdt
5 5
1 5 – 1 5
5 25
= te t+π e t+π +C
4. u = t + 7 dv = e2t+3dt
du = dt v = 1 e 2 t+
3
2
( 7) 2 3 1 ( 7) 2 3 – 1 2 3
∫ t + e t+ dt = t + e t+ ∫ e t+ dt
2 2
1 ( 7) 2 3 – 1 2 3
2 4
= t + e t+ e t+ +C
= t e t+ + e t+ + C
2 3 13 2 3
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
6. u = x dv = sin 2x dx
du = dx – 1 cos 2
v = x
2
sin 2 – 1 cos 2 – – 1 cos 2
∫ x x dx = x x ∫ x dx
2 2
– 1 cos 2 1 sin 2
= x x + x +C
2 4
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
= (t – 3) sin (t – 3) + cos (t – 3) + C
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
9. u = t dv = t +1 dt
du = dt v = 2 ( t +
1)3/ 2
3
1 2 ( 1)3/ 2 – 2 ( 1)3/ 2
∫t t + dt = t t + ∫ t + dt
3 3
2 ( 1)3/ 2 – 4 ( 1)5/ 2
3 15
= t t + t + +C
10. u = t dv = 3 2t + 7dt
du = dt v = 3 (2 t +
7)4 / 3
8
3 2 7 3 (2 7)4 / 3 – 3 (2 7)4 / 3
∫t t + dt = t t + ∫ t + dt
8 8
3 (2 7)4 / 3 – 9 (2 7)7 / 3
8 112
= t t + t + +C
11. u = ln 3x dv = dx
du 1 dx
= v = x
x
ln 3x dx x ln 3x x 1 dx
∫ = −∫ = x ln 3x − x +C
x
12. u = ln(7x5 ) dv = dx
du 5 dx
= v = x
x
ln(7x5 )dx x ln(7x5 ) – x 5 dx
x
∫ = ∫
= x ln(7x5 ) – 5x +C
418 Section 7.2 Instructor Solutions Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8. 13. u = arctan x dv = dx
du dx
2
1
1
x
=
+
v = x
x x x x dx
+ ∫ ∫
2 arctan arctan
1
x
= −
x x x dx
2
arctan 1 2
+ ∫
2 1
x
= −
arctan 1 ln(1 2 )
= x x − + x +C
2
14. u = arctan 5x dv = dx
du dx
2
5
1 25
x
=
+
v = x
x dx x x x dx
2
arctan 5 arctan 5 – 5
+ ∫ ∫
1 25
x
=
x x x dx
2
arctan 5 – 1 50
+ ∫
10 1 25
x
=
arctan 5 – 1 ln(1 25 2 )
= x x + x +C
10
dv dx
15. u = ln x 2
x
=
du 1 dx
= v – 1
x
x
=
ln x dx – ln x – – 1 1 dx
x x x x
∫ 2
∫
⎝ ⎠ – ln x – 1 C
= ⎛ ⎞ ⎜ ⎟
= +
x x
16. u = ln 2x5
dv 1 dx
2
x
=
du 5 dx
= v 1
x
x
= −
3 5 3 3 5
2 2 2 2 2
ln 2x dx 1 ln 2x 5 1 dx
x x x
= ⎡− ⎤ + ⎢⎣ ⎥⎦ ∫ ∫
3
1 ln 2 5 5
1 ln(2 3 ) 5 1 ln(2 2 ) 5
3 3 2 2
1 ln 2 5 ln 3 5 3ln 2 5
3 3 3 2
= ⎡− ⎢⎣ x
− ⎤ x x
⎥⎦
2
= ⎛ − ⋅ 5 − ⎞ ⎜ ⎟ − ⎛− ⎜ ⋅ 5
− ⎞ ⎟
⎝ ⎠ ⎝ ⎠
= − − − + +
8 ln 2 5 ln 3 5 0.8507
3 3 6
= − + ≈
17. u = ln t dv = t dt
du 1dt
= 2 3/ 2
t
v = t
3
e e ∫ t ln t dt = ⎡⎢ 2 t 3/ 2 ln t⎤⎥ – ∫
e 2
t 1/ 2
dt ⎣ 3 ⎦ 3
1 1 1
2 ln – 2 1ln1 4
3 3 9
= e 3/ 2 e ⋅ − ⎡⎢ t 3/ 2
⎤⎥ ⎣ ⎦
2 3/ 2 0 4 3/ 2 4 2 3/ 2 4 1.4404
3 9 9 9 9
e
1
= e − − e + = e + ≈
18. u = ln x3 dv = 2xdx
du 3 dx
= 1 (2 )3/ 2
x
v = x
3
5 3
1
∫ 2x ln x dx
5 5 3 2 3 3 2
1 (2 ) ln 2
3
= ⎡⎢ x x ⎤⎥ − x dx ⎣ ⎦ ∫
1 1
5/ 2 5
= ⎡ 1 ⎢⎣ (2 x ) 3/ 2 ln x 3 − 2
x 3/ 2
⎤
3 3
⎥⎦
1
5 2 5 2
⎛ ⎞
1 (10)3 2 ln 53 2 53/ 2 1 (2)3 2 ln13 2
3 3 3 3
= − − ⎜⎜ − ⎟⎟
⎝ ⎠
4 2 53 2 4 2 103 2 ln 5 31.699
3 3
= − + + ≈
19. u = ln z dv = z3dz
du 1 dz
= 1 4
z
v = z
4
3 ln 1 4 ln 1 4 1
∫ = − ∫ ⋅
1 4 ln 1 3
4 4
z zdz z z z dz
4 4
z
= z z − ∫ z dz
1 4 ln 1 4
4 16
= z z − z +C
20. u = arctan t dv = t dt
du dt
2
1
1
t
=
+
1 2
2
v = t
2
t arctan t dt 1 t 2
arctan t – 1
t dt
2
+ ∫ ∫
2 21
t
=
2
t t t dt
1 2
1 1 + −
arctan 1
2 2 1
t
2
= −
+ ∫
1 2
arctan 1 ∫ 1 ∫
1
2 2 2 1
+ t t dt dt
2
t
= − +
1 2 arctan 1 1 arctan
2 2 2
= t t − t + t +C
Instructor's Resource Manual 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.
9. 21. u arctan 1
= ⎛ ⎞ ⎜ ⎟
t
⎝ ⎠
dv = dt
du dt
2
– 1
1
t
=
+
v = t
dt t t dt
t t t
2
arctan 1 arctan 1
∫ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ = ⎜ ⎟
+ ∫
⎝ ⎠ ⎝ ⎠ 1
+ arctan 1 1 ln(1 2 )
= ⎛ ⎞ + + + ⎜ ⎟
t t C
2
t
⎝ ⎠
22. u = ln(t7 ) dv = t5dt
du 7 dt
= 1 6
t
v = t
6
5 ln( 7 ) 1 6 ln( 7 ) – 7 5
∫t t dt = t t ∫t dt
6 6
= 1 t 6 ln( t 7 ) – 7 t 6
+C
6 36
23. u = x dv = csc2 x dx
du = dx v = −cot x
[ ] / 2 2 / 2 / 2
/ 6 / 6 / 6
x csc x dx x cot x cot x dx π π π
π π π
∫ = − + ∫ 2
6 cot ln sin x x x π
π = ⎡⎣− + ⎤⎦
0 ln1 3 ln 1 ln 2 1.60
2 6 2 2 3
π π π
= − ⋅ + + − = + ≈
24. u = x dv = sec2 x dx
du = dx v = tan x
[ ] 4 2 4 4
6 6 6
x sec x dx x tan x tan x dx π π π
π π π
π ⎛ π ⎞
tan ln cos ln 2 ln 3
∫ = − ∫ x x π
4
x = ⎡⎣ + ⎤⎦ = + − ⎜⎜ + ⎟⎟
6
4 2 6 3 2
π
⎝ ⎠
1 ln 2 0.28
π π
= − + ≈
4 6 3 2 3
25. u = x3 dv = x2 x3 + 4dx
du = 3x2dx v = 2 ( x 3 +
4)3/ 2
9
∫ x 5 x 3 + 4 dx = 2 x 3( x 3 + 4)3/ 2 – ∫ 2 x 2 ( x 3 + 4)3/ 2
dx 2 3( 3 4)3/ 2 – 4 ( 3 4)5 / 2
9 3
= x x + x + +C
9 45
26. u = x7 dv = x6 x7 +1 dx
du = 7x6dx v = 2 ( x 7 +
1)3/ 2
21
∫ x 13 x 7 + 1 dx = 2 x 7 ( x 7 + 1)3/ 2 – ∫ 2 x 6 ( x 7 + 1)3/ 2
dx 2 7 ( 7 1)3/ 2 – 4 ( 7 1)5/ 2
21 3
= x x + x + +C
21 105
27. u = t4
3
dv t dt
(7 – 3 t
4 )3/ 2
=
du = 4t3 dt
4 1/2
1
t
6(7 – 3 )
v
=
7 4 3
4 3/ 2 4 1/ 2 4 1/ 2
t dt t – 2
t dt
t t t
∫ = ∫
(7 – 3 ) 6(7 – 3 ) 3 (7 – 3 )
4
t 1 (7 – 3 t 4 )
1/2
C
t
= + +
4 1/2
6(7 – 3 ) 9
420 Section 7.2 Instructor Solutions Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10. 28. u = x2 dv = x 4 – x2 dx
du = 2x dx v = – 1 (4 – x
2 )3/ 2
3
∫ x 3 4 – x 2 dx = – 1 x 2 (4 – x 2 )3/ 2 + 2 ∫ x (4 – x 2 )3/ 2
dx – 1 2 (4 – 2 )3/ 2 – 2 (4 – 2 )5 / 2
3 3
= x x x +C
3 15
29. u = z4
3
dv z dz
(4 – z
4 )2
=
du = 4z3dz
4
1
z
4(4 – )
v
=
7 4 3
z dz z z dz
z z z
∫ = − ∫
(4 – 4 )2 4(4 – 4 ) 4 – 4
4
z z 4
C
z
= + +
4
1 ln 4 –
4(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/ 2dx
du 1 dx
= v = 2x1/ 2
x
ln x dx 2 x ln x – 2 1 dx
x x
∫ = ∫ = 2 x ln x – 4 x +C
1/ 2
33. u = x dv = (3x +10)49 dx
du = dx v = 1 (3 x +
10)50
150
∫ x (3 x + 10)49 dx = x (3 x + 10)50 – 1 ∫ (3 x + 10)50
dx (3 10)50 – 1 (3 10)51
150 150
= x x + x + + C
150 22,950
34. u = t dv = (t −1)12 dt
du = dt 1 ( 1)13
v = t −
13
( ) ( )
1 1 1 12 13 13
0 0 0
t ( t 1) dt t t 1 1 t 1
dt
− = ⎡ − ⎤ − − ⎢⎣ ⎥⎦
∫ ∫
13 13
t t t
1 1 1 1
= ⎡ 13 14
⎤ ⎢⎣ ( − ) − ( − )
⎥⎦
= 1
13 182 182
0
35. u = x dv = 2x dx
du = dx 1 2
v = x
ln 2
∫ x 2 x dx = x 2 x – 1 ∫ 2
x dx
ln 2 ln 2
= x 2 x – 1 2
x +C
2
ln 2 (ln 2)
36. u = z dv = azdz
du = dz 1
v az
ln
a
=
zazdz z az – 1
azdz
∫ = ∫
a a
ln ln
z az az C
a a
= +
2
– 1
ln (ln )
37. u = x2 dv = exdx
du = 2x dx v = ex
∫ x2exdx = x2ex − ∫ 2xexdx
u = x dv = exdx
du = dx v = ex
∫ x2ex dx = x2ex − 2(xex − ∫ex dx)
= x2ex − 2xex + 2ex +C
Instructor's Resource Manual Section 7.2 421
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
11. 38. u = x4
x2 dv = xe dx
du = 4x3dx
1 2
2
v = ex
5 2 1 4 2 3 2 – 2
∫ x ex dx = x ex ∫ x ex dx
2
2
u = x2
dv = 2xex dx
du = 2x dx
x2 v = e
5 2 1 4 2 2 2 2 – 2
x ex dx = x ex ⎛ x ex − xex dx ⎞ ⎜ ⎟
∫ ∫
2
⎝ ⎠ 1 4 2 2 2 –
2 2
= x ex x ex + ex +C
39. u = ln2 z dv = dz
du 2ln z dz
= v = z
z
∫ln2 z dz = z ln2 z – 2∫ln z dz
u = ln z dv = dz
du 1 dz
= v = z
z
∫ln2 z dz = z ln2 z – 2(z ln z – ∫ dz)
= z ln2 z – 2z ln z + 2z +C
40. u = ln2 x20 dv = dx
40ln x20 du dx
= v = x
x
∫ln2 x20dx = x ln2 x20 – 40∫ln x20dx
u = ln x20 dv = dx
du 20 dx
= v = x
x
∫ln2 x20dx = x ln2 x20 – 40(x ln x20 – 20∫ dx)
= x ln2 x20 – 40x ln x20 + 800x +C
41. u = et dv = cos t dt
du = etdt v = sin t
∫et cos t dt = et sin t − ∫et sin t dt
u = et dv = sin t dt
du = etdt v = –cos t
∫et cost dt = et sin t − ⎡⎣−et cost + ∫et cos t dt⎤⎦
∫et cos t dt = et sin t + et cos t − ∫et cos t dt
2∫et cos t dt = et sin t + et cos t +C
cos 1 (sin cos )
∫et t dt = et t + t +C
2
42. u = eat dv = sin t dt
du = aeatdt v = –cos t
∫eat sin t dt = –eat cos t + a∫eat cos t dt
u = eat dv = cos t dt
du = aeatdt v = sin t
∫eat sin t dt = –eat cos t + a (eat sin t – a∫eat sin t dt )
∫eat sin t dt = –eat cos t + aeat sin t – a2 ∫eat sin t dt
(1+ a2 )∫eat sin t dt = –eat cos t + aeat sin t + C
at at
eat sin t dt – e cos t ae sin
t C
+ + ∫
= + +
2 2
a a
1 1
43. u = x2 dv = cos x dx
du = 2x dx v = sin x
∫ x2 cos x dx = x2 sin x − ∫ 2x sin x dx
u = 2x dv = sin x dx
du = 2dx v = −cos x
∫ x2 cos x dx = x2 sin x − (−2x cos x + ∫ 2cos x dx)
= x2 sin x + 2x cos x − 2sin x +C
44. u = r2 dv = sin r dr
du = 2r dr v = –cos r
∫ r2 sin r dr = –r2 cos r + 2∫ r cos r dr
u = r dv = cos r dr
du = dr v = sin r
∫ r2 sin r dr = –r2 cos r + 2(r sin r – ∫sin r dr ) = –r2 cos r + 2r sin r + 2cos r +C
422 Section 7.2 Instructor Solutions Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12. 45. u = sin(ln x) dv = dx
du cos(ln x) 1 dx
= ⋅ v = x
x
∫sin(ln x)dx = x sin(ln x) − ∫cos(ln x) dx
u = cos (ln x) dv = dx
du sin(ln x) 1 dx
= − ⋅ v = x
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
sin(ln ) [sin(ln ) cos(ln )]
∫ x dx = x x − x +C
2
46. u = cos(ln x) dv = dx
du – sin(ln x) 1 dx
= v = x
x
∫cos(ln x)dx = x cos(ln x) + ∫sin(ln x)dx
u = sin(ln x) dv = dx
du cos(ln x) 1 dx
= v = x
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
cos(ln ) [cos(ln ) sin(ln )]
∫ x dx = x x + x + C
2
47. u = (ln x)3 dv = dx
3ln2 x du dx
= v = x
x
∫(ln x)3dx = x(ln x)3 – 3∫ln2 x dx
= x ln3 x – 3(x ln2 x – 2x ln x + 2x +C)
= x ln3 x – 3x ln2 x + 6x ln x − 6x +C
48. u = (ln x)4 dv = dx
4ln3 x du dx
= v = x
x
∫(ln x)4 dx = x(ln x)4 – 4∫ln3 x dx = x ln4 x – 4(x ln3 x – 3x ln2 x + 6x ln x − 6x +C)
= x ln4 x – 4x ln3 x +12x ln2 x − 24x ln x + 24x +C
49. u = sin x dv = sin(3x)dx
du = cos x dx v = – 1 cos(3 x
)
3
sin sin(3 ) – 1 sin cos(3 ) 1 cos cos(3 )
∫ x x dx = x x + ∫ x x dx
3 3
u = cos x dv = cos(3x)dx
du = –sin x dx v = 1 sin(3 x
)
3
Instructor's Resource Manual Section 7.2 423
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13. sin sin(3 ) – 1 sin cos(3 ) 1 1 cos sin(3 ) 1 sin sin(3 )
∫ x x dx = x x + ⎡⎢ x x + ∫
x x dx⎤⎥ 3 33 ⎣ 3
⎦ – 1 sin cos(3 ) 1 cos sin(3 ) 1 sin sin(3 )
= x x + x x + ∫ x x dx
3 9 9
8 sin sin(3 ) – 1 sin cos(3 ) 1 cos sin(3 )
9 3 9
∫ x x dx = x x + x x + C
sin sin(3 ) – 3 sin cos(3 ) 1 cos sin(3 )
∫ x x dx = x x + x x +C
8 8
50. u = cos (5x) dv = sin(7x)dx
du = –5 sin(5x)dx v = – 1 cos(7 x
)
7
cos(5 )sin(7 ) – 1 cos(5 ) cos(7 ) – 5 sin(5 ) cos(7 )
∫ x x dx = x x ∫ x x dx
7 7
u = sin(5x) dv = cos(7x)dx
du = 5 cos(5x)dx v = 1 sin(7 x
)
7
cos(5 )sin(7 ) – 1 cos(5 ) cos(7 ) – 5 1 sin(5 )sin(7 ) – 5 cos(5 )sin(7 )
∫ x x dx = x x ⎡⎢ x x ∫
x x dx⎤⎥ 7 77 ⎣ 7
⎦ – 1 cos(5 ) cos(7 ) – 5 sin(5 )sin(7 ) 25 cos(5 )sin(7 )
= x x x x + ∫ x x dx
7 49 49
24 cos(5 )sin(7 ) – 1 cos(5 ) cos(7 ) – 5 sin(5 )sin(7 )
49 7 49
∫ x x dx = x x x x +C
cos(5 )sin(7 ) – 7 cos(5 ) cos(7 ) – 5 sin(5 )sin(7 )
∫ x x dx = x x x x + C
24 24
51. u = eα z dv = sin βz dz
du =αeα zdz v – 1 cosβ z
β
=
eα z sin z dz – 1 eα z cos z α eα z cos z dz
∫ = + ∫
u = eα z dv = cos βz dz
du =αeα zdz v 1 sinβ z
β β β
β β
β
=
⎡ ⎤
eα z sin z dz 1 eα z cos z α 1 eα z sin z α eα z sin z dz
⎣ ⎦ ∫ ∫
β β β β
= − + ⎢ − ⎥
β β β β
2
– 1 eα z cos z α eα z sin z –α eα z sin z dz
= + ∫
2 2
β β β
2 2
β β β
β α eα z sin z dz – 1 eα z cos z α eα z sin z C
∫ = + +
β β β
+
2 2
β β β
e α z ( α sin β z – β cos β
z) C
eα z sin z dz –β eα z cos z α eα z sin z C
∫ β = β + β
+
2 + 2 2 + 2
2 2
α β α β
= +
α +
β
424 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.
14. 52. u = eα z dv = cosβ z dz
du =αeα zdz v 1 sinβ z
β
=
eα z cos z dz 1 eα z sin z – α eα z sin z dz
∫ = ∫
u = eα z dv = sin βz dz
du =αeα zdz v – 1 cosβ z
β β β
β β
β
=
⎡ ⎤
eα z cos z dz 1 eα z sin z α 1 eα z cos z α eα z cos z dz
⎣ ⎦ ∫ ∫
β β β β
= − ⎢− + ⎥
β β β β
2
1 eα z sin z α eα z cos z –α eα z cos z dz
= + ∫
2 2
β β β
2 2
β β β
α β eα z cos z dz α eα z cos z 1 eα z sin z C
∫ = + +
β β β
+
2 2
β β β
z
α
α z e ( α cos β z +
β β
e cos z dz sin z )
C
+ ∫
= +
2 2
β
α β
53. u = ln x dv = xα dx
du 1 dx
x
=
1
1
α
α
v x
+
=
+
, α ≠ –1
α
+
1 ln ln –
1 x x dx x x x dx
α α
1 1
+ +
x x x C
α α
+ + ∫ ∫
α α
1 1
=
α
2 ln – , –1
1 ( 1)
= + ≠
α α
+ +
54. u = (ln x)2 dv = xα dx
du 2ln x dx
x
=
1
1
α
α
v x
+
=
+
, α ≠ –1
1
α
+
x (ln x )2 dx x (ln x )2 – 2 x ln
x dx
1 1 1
+ ⎡ + + ⎤
x α α α
x x x x C
α α α α
(ln ) 2 ln
1 1 1 ( 1)
α α
+ + ∫ ∫
α α
1 1
=
2
= − ⎢ − ⎥ +
2
+ + ⎢⎣ + + ⎥⎦
1 1 1
+ + +
α α α
x x 2
x x x C
α
2 3 (ln ) – 2 ln 2 , –1
1 ( 1) ( 1)
= + + ≠
α + α + α
+
Problem 53 was used for xα ln x dx. ∫
55. u = xα dv = eβ x dx
du =α xα –1dx v 1 eβ x
β
=
α β
x α e β xdx x e α – x α –1
e β
xdx
x
∫ = ∫
β β
56. u = xα dv = sin βx dx
du =α xα –1dx v – 1 cosβ x
β
=
α
x sin x dx – x cos x x –1 cos x dx
α β α α
∫ = + ∫
β β
β β
Instructor's Resource Manual Section 7.2 425
© 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.
15. 57. u = xα dv = cos βx dx
du =α xα –1dx v 1 sinβ x
β
=
α
x cos x dx x sin x – x –1 sin x dx
α β α α
∫ = ∫
β β
β β
58. u = (ln x)α dv = dx
(ln x) –1 du dx
α α
= v = x
x
∫(ln x)α dx = x(ln x)α –α ∫(ln x)α –1dx
59. u = (a2 – x2 )α dv = dx
du = –2α x(a2 – x2 )α –1dx v = x
∫(a2 – x2 )α dx = x(a2 – x2 )α + 2α ∫ x2 (a2 – x2 )α –1dx
60. u = cosα –1 x dv = cos x dx
du = –(α –1) cosα –2 x sin x dx v = sin x
∫cosα x dx = cosα –1 x sin x + (α –1)∫cosα –2 x sin2 x dx
= cosα −1 x sin x + (α −1)∫cosα −2 x(1− cos2 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
–1
α
cos x dx cos x sin x –1 cos –2 x dx
α α α
∫ = + ∫
α α
61. u = cosα –1 β x dv = cos βx dx
du = –β (α –1) cosα –2 β x sinβ x dx v 1 sinβ x
β
=
–1
α
cos x dx cos x sin x ( –1) cos –2 x sin2 x dx
α β β α
β α β β
∫ = + ∫
β
1
−
α
cos β x sin β x ( 1) cos α
2 x(1 cos2 x) dx
= + − ∫ − −
α β β
β
–1
α
cos x sin x ( –1) cos –2 x dx – ( –1) cos x dx
β β α α
= + ∫ ∫
α β α β
β
1
−
α
cos x cos x sin x ( 1) cos 2 x dx
α β β α
∫ = + − ∫ −
α β α β
β
–1
α
cos x dx cos x sin x –1 cos –2 x dx
α β β α α
β β
∫ = + ∫
αβ α
62. 4 3 1 4 3 – 4 3 3
∫ x e xdx = x e x ∫ x e xdx 1 4 3 – 4 1 3 3 – 2 3
3 3
= x e x ⎡ ⎢⎣ x e x ∫
x e xdx⎤ 3 33
⎥⎦ = 1 x 4 e 3 x – 4 x 3 e 3 x + 4 ⎡ 1 x 2 e 3 x – 2 ∫ xe 3
xdx⎤ 1 4 3 – 4 3 3 4 2 3 – 8 1 3 – 1 3
3 9 33 ⎢⎣ 3
⎥⎦ = x e x x e x + x e x ⎡ ⎢⎣ xe x ∫
e xdx⎤ 3 9 9 93 3
⎥⎦ 1 4 3 – 4 3 3 4 2 3 – 8 3 8 3
3 9 9 27 81
= x e x x e x + x e x xe x + e x +C
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16. 63. 4 cos3 1 4 sin 3 – 4 3 sin 3
∫ x x dx = x x ∫ x x dx 1 4 sin 3 – 4 – 1 3 cos3 2 cos3
3 3
= x x ⎡⎢ x x + ∫
x x dx⎤⎥ 3 3 ⎣ 3
⎦ 1 4 sin 3 4 3 cos3 – 4 1 2 sin 3 2 sin 3
3 9 3 3 3
= x x + x x ⎡⎢ x x − x x dx⎤⎥ ⎣ ⎦ ∫
1 4 sin 3 4 3 cos3 – 4 2 sin 3 8 – 1 cos3 1 cos3
3 9 9 9 3 3
= x x + x x x x + ⎡⎢ x x + ∫
x dx⎤⎥ ⎣ ⎦ = 1 x 4 sin 3 x + 4 x 3 cos3 x – 4 x 2 sin 3 x – 8 x cos3 x + 8 sin 3
x + C
3 9 9 27 81
64. cos6 3 1 cos5 3 sin 3 5 cos4 3
∫ x dx = x x + ∫ x dx 1 cos5 3 sin 3 5 1 cos3 3 sin 3 3 cos2 3
18 6
= x x + ⎡⎢ x x + ∫
x dx⎤⎥ 18 6 ⎣ 12 4
⎦ 1 cos5 3 sin 3 5 cos3 3 sin 3 5 1 cos3 sin 3 1
18 72 8 6 2
= x x + x x + ⎡⎢ x x + ∫
dx⎤⎥ ⎣ ⎦ = 1 cos5 3 x sin 3 x + 5 cos3 3 x sin 3 x + 5 cos3 x sin 3 x + 5
x +C
18 72 48 16
65. First make a sketch.
From the sketch, the area is given by
∫ e ln x dx
1
u = ln x dv = dx
du 1 dx
= v = x
x
ln ln ∫e x dx = x x e − ∫e dx 1 [ ln ]e
[ ]1 1 1
= x x − x = (e – e) – (1 · 0 – 1) = 1
(ln ) e V = ∫ π x dx
u = (ln x)2 dv = dx
du 2ln x dx
66. 2
1
= v = x
x
π ∫ e 2 e (ln x ) dx = π⎛ 2
e ⎞ 2
1 ⎜ ⎡ ⎤ ⎝ ⎣ x (ln x ) ⎦ − 2 ∫ ln
x dx 1 1
⎟ ⎠ e
⎡x x x x x ⎤
⎣ ⎦ = π − − 2
1
(ln ) 2( ln )
1 [ (ln ) 2 ln 2 ]e
= π x x − x x + x
= π[(e − 2e + 2e) − (0 − 0 + 2)] = π(e − 2) ≈ 2.26
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17. 67.
3 –9 – 1
e x dx = e x ⎛ dx ⎞ ⎜ ⎟
9 – / 3 9 – / 3
0 0
–9[e x ] – 9 9
∫ 0
∫ – /3 ⎝ 3
⎠ 9= = + ≈ 8.55
3
e
68. 9 – / 3 2 9 –2 / 3
V = ∫ π(3e x ) dx = 9π∫ e x dx
0 0
9 – 3 – 2
= π⎛ ⎞ e x ⎛ dx ⎞ ⎜ ⎟ ⎜ ⎟
9 –2 / 3
0
– 27 π [ e x
] – 27 π 27 π
42.31
2 2 2
∫ –2 / 3 9
⎝ 2 ⎠ ⎝ 3
⎠ = = + ≈
0 6
e
69. / 4 / 4 / 4
(x cos x – x sin x)dx x cos x dx – x sin x dx π π π
∫ = ∫ ∫
0 0 0
4 4
0 0 xsin x sin x dx ⎛ π π ⎞ = ⎜ ⎡⎣ ⎤⎦ ⎟ ⎝ ⎠
− ∫ [ ] 4 4
xπ
−⎜ ⎛ − x cos x π ∫
π + cos x dx ⎞ ⎟
⎝ 0 0
⎠ [ x sin x cos x x cos x – sin ] / 4
0 = + + 2 –1
= ≈ 0.11
4
π
Use Problems 60 and 61 for ∫ x sin x dx and ∫ x cos x dx.
V x x dx π ⎛ ⎞ = π ⎜ ⎟
70. 2
⎝ ⎠ ∫
2 sin2
0
dv = x dx
u = x sin
2
v = x
du = dx –2cos
2
2 2
0 0
⎛ ⎡ ⎤ π π ⎞ = π⎜ ⎢ ⎥ + ⎟ ⎜ ⎣ ⎦ ⎟ ⎝ ⎠
V x x x dx
∫
2 –2 cos 2cos
2 2
2
2
⎛ π ⎞ = 2 π⎜ 4 π + ⎡ 4sin x ⎤ ⎢ ⎥ ⎟ = 8
π ⎜ ⎝ ⎣ 2
⎦ ⎟ 0
⎠
ln 2 ln e e ∫ x dx = ∫ x dx
u = ln x dv = dx
du 1 dx
71. 2
1 1
= v = x
x
( ) 1 1 1 1
2 ln 2 [ ln ] 2 [ ] 2 e x dx ⎛ x x e e dx⎞ e x e ⎜ − ⎟
∫ = ∫ = − =
⎝ ⎠
∫ e x ln x 2
dx = 2 ∫ e x ln x dx
1 1
u = ln x dv = x dx
du = 1 dx
1 2
x
v = x
2
e e e x xdx x x xdx
⎛ ⎡ ⎤ ⎞ = ⎜ ⎢ ⎥ ⎟ ⎜ ⎣ ⎦ ⎟ ⎝ ⎠
2 ln 2 1 ln – 1
e
⎛ ⎡ ⎤ ⎞ = ⎜ ⎢ ⎥ ⎟ = + ⎜ ⎣ ⎦ ⎟ ⎝ ⎠
2 1 – 1 1 ( 1)
2 4 2
∫ 2
∫ 2 2 2
2 2
1 1 1
e x e
1
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18. ∫e x 2
dx
u = (ln x)2 dv = dx
du 2ln x dx
1 (ln )
2
1
= v = x
x
1 (ln ) 1 [ (ln ) ] – 2 ln
2 2
∫ e x 2 dx = ⎛ ⎜ x x 2
e e 1 ∫ x dx ⎞ ⎟
= 1( e
–2)
1 ⎝ 1
⎠ 2
1 2 2
2( 1) 1
e e x
+ +
= =
2 4
1
( –2) 2– 2
e e y = =
2 4
72. a. u = cot x dv = csc2 x dx
du = – csc2 x dx v = –cot x
2 2 2
∫ ∫
∫
∫
x xdx x x xdx
x xdx x C
x xdx x C
cot csc = − cot −
cot csc
2 cot csc 2 = − cot
2
+
cot csc 2 = − 1 cot
2
+
2
b. u = csc x dv = cot x csc x dx
du = –cot x csc x dx v = –csc x
2 2 2
∫ ∫
∫
∫
x xdx x x xdx
x xdx x C
x xdx x C
cot csc = − csc −
cot csc
2 cot csc 2 = − csc
2
+
cot csc 2 = − 1 csc
2
+
2
c. – 1 cot2 – 1 (csc2 –1)
x = x – 1 csc2 1
2 2
= x +
2 2
73. a. p(x) = x3 − 2x
g(x) = ex
All antiderivatives of g(x) = ex
∫(x3 − 2x)exdx = (x3 − 2x)ex − (3x2 − 2)ex + 6xex − 6ex + C
b. p(x) = x2 − 3x +1
g(x) = sin x
G1(x) = −cos x
G2 (x) = −sin x
G3(x) = cos x
∫(x2 − 3x +1)sin x dx = (x2 − 3x +1)(−cos x) − (2x − 3)(−sin x) + 2cos x + C
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19. 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
( ) sin n
(2 −
1)
2 ( −
1)
A n x xdx π
= ∫ . Using integration by parts:
n
π
u = x dv =
sin
xdx
du = dx v =−
cos
x
A n x xdx x x xdx x x x
n n n n n
n n n n n
(2 − 1) (2 − 1) (2 − 1) (2 − 1) (2 −
1)
2 ( − 1) 2 ( − 1) 2 ( − 1) 2 ( − 1) 2 ( −
1)
π [ ] π π [ ] π [ ]
π
π π π π π
( ) sin cos cos cos sin
= ∫ = − − ∫ − = − +
= − − − + − − +[ − − − ]
[ ]
n n n n n n
))
(2 1) cos( (2 1)) 2 ( 1) cos(2 ( 1)) sin( (2 1)) sin(2 ( 1
π π π π π π
[ ]
= −π (2n −1)(−1) + 2π (n −1)(1) + 0 − 0 =π (2n −1) + (2n − 2) . So A(n) = (4n − 3)π
b. 3 2
V 2 x sin x dx π
= π∫
2
π
u = x2 dv = sin x dx
du = 2x dx v = –cos x
2 3 3
π π
π π
⎛ ⎡ ⎤ ⎞ ⎜ ⎣ ⎦ ⎟ ⎝ ⎠
= π + ∫ 2 2 3
V 2 –x cos x 2x cos x dx
2 2
2 2 9 4 2x cos x dx π
⎛ ⎞
⎜ ⎝ π
⎟
⎠
= π π + π + ∫
u = 2x dv = cos x dx
du = 2 dx v = sin x
2 3 3
2 2 V 2 13 [2xsin x] – 2sin x π π
⎛ ⎞
⎜ ⎟
⎝ ⎠
= π π + π ∫
π
( 2 3 ) 2
2 13 [2cos x]2 2 (13 – 4) π
= π π + π = π π ≈ 781
75. u = f(x) dv = sin nx dx
du = f ′(x)dx v 1 cos nx
n
= −
π π
−π −π
⎡ ⎡ ⎤ ′ ⎤ = ⎢ ⎢− ⎥ + ⎥ π ⎣ ⎣ ⎦ ⎦ ∫
1 1 cos( ) ( ) 1 cos( ) ( ) an nx f x nx f x dx
n n
Term 1 Term 2
Term 1 = 1 cos(n π )( f ( −π ) − f ( π )) =± 1 ( f ( −π ) − f ( π
))
n n
Since f ′(x) is continuous on [–∞ ,∞ ], it is bounded. Thus,
cos(nx) f (x)dx π
π
∫ ′ is bounded so
–
a 1 ⎡± f π
lim n
= lim ⎢⎣ ( ( −π ) − f ( π )) + ∫
cos( nx ) f ′ ( x ) dx
⎤ ⎥⎦ = 0. n n
π n
→∞ →∞ −π
76.
1
G n n n n
n n
[( + 1)( + 2) ⋅⋅⋅ ( +
)]
n 1
n
[ ]
n
n
=
n n
1/ 1 1 1 2 1
= ⎡⎛ ⎢⎜ + ⎞⎛ ⎟⎜ + ⎟…⎜ ⎞ ⎛ + ⎞⎤ ⎣⎝ n ⎠⎝ n ⎠ ⎝ n
⎟⎥ ⎠⎦
ln Gn 1 ln 1 1 1 2 1 n
⎛ ⎞ ⎡⎛ ⎞⎛ ⎜ ⎟ = + + ⎞ ⎟…⎜ ⎛ + ⎞⎤ ⎝ n ⎠ n ⎢⎜ n ⎟⎜ ⎣⎝ ⎠⎝ n ⎠ ⎝ n
⎠⎦
⎟⎥ 1 ln 1 1 ln 1 2 ln 1 n
n n n n
⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ = ⎢ ⎜ + ⎟ + ⎜ + ⎟ + ⋅⋅⋅+ ⎜ + ⎟⎥ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦
G xdx
⎛ ⎞ 2
⎜ ⎟
= ∫
= ⎝ ⎠ 1
lim ln n ln 2ln 2 –1
n
→∞ n
G e e
lim n 2ln 2–1 4 –1 4
n
⎛ ⎞ = = = ⎜ ⎟
⎝ ⎠
→∞ n e
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20. 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 ∫(1 t )dt = ∫(1 t )dt +1
[ ( 1 cos 7 2 sin 7 ) 3] 5 ( 1 cos7 2 sin 7 ) (–7 1 sin 7 7 2 cos 7 ) d e x C x C x C e x C x C x e x C x C x
dx
78. 5 5 5
+ + = + + +
5
[(5 1 7 2 ) cos 7 (5 2 – 7 1)sin 7 ] = e x C + C x + C C x
Thus, 5C1 + 7C2 = 4 and 5C2 – 7C1 = 6.
Solving, 1 2
– 11 ; 29
37 37
C = C =
79. u = f(x) dv = dx
du = f ′(x)dx v = x
( ) [ ( )] – ( ) b b b
a a a
∫ f x dx = xf x ∫ xf ′ x dx
Starting with the same integral,
u = f(x) dv = dx
du = f ′(x)dx v = x – a
( ) [( – ) ( )] – ( – ) ( ) b b b
a a a
∫ f x dx = x a f x ∫ x a f ′ x dx
80. u = f ′(x) dv = dx
du = f ′′(x)dx v = x – a
( )– ( ) ( ) b
f b f a = ∫ f ′ x dx [( – ) ( )] – ( – ) ( ) b b
a
= x a f ′ x ∫ x a f ′′ x dx ( )( – ) – ( – ) ( ) b
a a
= f ′ b b a ∫ x a f ′′ x dx
a
Starting with the same integral,
u = f ′(x) dv = dx
du = f ′′(x)dx v = x – b
( ) ( ) ( ) [( – ) ( )] – ( – ) ( ) b b b
f b − f a = ∫ f ′ x dx = x b f ′ x ∫ x b f ′′ x dx ( )( ) – ( – ) ( ) b
a a a
= f ′ a b − a ∫ x b f ′′ x dx
a
81. Use proof by induction.
n = 1: ( ) ( )( – ) ( – ) ( ) ( ) ( )( – ) [ ( )( – )] ( ) t t t
f a + f ′ a t a + ∫ t x f ′′ x dx = f a + f ′ a t a + f ′ x t x + ∫ f ′ x dx
a a a
( ) ( )( – ) – ( )( – ) [ ( )]t ( )
= f a + f ′ a t a f ′ a t a + f x a = 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 ( i )
n i t n
f ( t ) f ( a ) f ( a ) ( t – a ) ( t – x ) f ( 1)
( x )
dx
= +Σ + ∫
1
i n
! a
!
i
+
=
t n n
a
t x f x dx
n
Integrate ( – ) ( 1) ( )
+ ∫ by parts.
!
t x n dv dx
u = f (n+1) (x) ( – )
!
n
=
du = f (n+2) (x)
t x n v
–
( – ) +
1 ( n
1)!
=
+
1 1
n n + t t t n +
n n n
a a
+ ⎡ + ⎤ +
t x f x dx t x f x t x f x dx
n n n
( – ) ( 1) ( ) – ( – ) ( 1) ( ) ( – ) ( 2) ( )
∫ ∫
= ⎢ ⎥ +
⎢⎣ + ⎥⎦ +
! ( 1)! ( 1)!
a
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21. n + 1 t n +
1
n n
t a f a t x f x dx
n n
( – ) ( 1) ( ) ( – ) ( 2) ( )
( 1)! ( 1)!
= + + +
∫
+ a
+ Thus
n ( i ) n + 1 t n +
1
i n n
f ( t ) f ( a ) f ( a ) ( t – a ) ( t – a ) f ( a ) ( t – x ) f ( x )
dx
+ + Σ ∫
( 1) ( 2)
= + + +
1
+ +
i n n
! ( 1)! a
( 1)!
i
=
n + 1 ( i ) t n +
1
i n
f ( a ) f ( a ) ( t – a ) ( t – x ) f ( 2)
( x )
dx
+ Σ ∫
= + +
1
i n
! a
( 1)!
i
+
=
Thus, the statement is true for n + 1.
82. a. 1 1 1
B(α , β ) = ∫ xα − (1− x)β − dx where α ≥ 1,β ≥ 1
0
x = 1 – u, dx = –du
1 1 1 0 1 1
0 1
∫ xα − (1− x)β − dx = ∫ (1− u)α − (u)β − (−du) 1 1 1
= ∫ (1− u)α − uβ − du = B(β , α )
0
Thus, B(α, β) = B(β, α).
b. 1 1 1
B(α , β ) = ∫ xα − (1− x)β − dx
0
u = xα −1 dv = (1− x)β −1dx
du = (α −1) xα −2dx v = − 1 (1 −
x)β
β
1
⎡ − ⎤ − − − −
= ⎢− − ⎥ + − = −
⎣ ⎦
( , ) 1 (1 ) 1 (1 ) 1 (1 )
α β α α β α α β
1 1 2 1 2
∫ ∫
B x x x x dx x x dx
β β 0 β
0
0
1 B
( 1, 1)
α β
−
α
= − +
α β
β
(*)
Similarly,
1 1 1
0
B(α , β ) = ∫ xα − (1− x)β − dx
u = (1− x)β −1 dv = xα −1dx
du = −(β −1)(1− x)β −2 dx v 1 xα
α
=
1 1 1 2
0 0
B( , ) 1 xα (1 x)β β 1 xα (1 x)β dx
β − 1 α β − β −
x(1 x)dx 1 B( 1, 1)
= ⎡ − − ⎤ + − − − ⎢⎣ ⎥⎦ ∫ 1 2
α β
α α
= ∫ − = + −
α 0
α
α β
c. Assume that n ≤ m. Using part (b) n times,
( )
n n n B n m B n m B n m
− 1 − 1 ( −
2) ( , ) = ( − 1, + 1) = ( − 2, +
2)
m mm
( 1)
( n ) n ( n
)
+
1 ( 2) 3 2 1
− − − …⋅ ⋅
= … = + −
( )
B m n
(1, 1).
m m m m n
( 1) 2 ( 2)
+ + … + −
1 2 11
0 0
(1, 1) (1 ) 1 [(1 ) ] 1
B m n x m n dx x m n
+ − = − + − = − − + − =
∫ + − + −
m n m n
1 1
Thus,
( ) ( )
( ) ( ) (
1 ( 2) 3 2 1 1)!( 1)! ( 1)!( 1)!
B n m
( , )
n n n n m n m
− − − …⋅ ⋅ − − − −
= = =
m m m m n m n m n n m
( + 1) + 2 … ( + − 2) + − 1 ( + − 1)! ( + −
1)!
If n m, then
( 1)!( 1)!
B n m B m n
( , ) ( , )
n m
− −
n m
( 1)!
= =
+ −
by the above reasoning.
432 Section 7.2 Instructor Solutions Manual
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22. 83. u = f(t) dv = f ′′(t)dt
du = f ′(t)dt v = f ′(t)
( ) ( ) [ ( ) ( )] – [ ( )]2 b b b
a a a
∫ f ′′ t f t dt = f t f ′ t ∫ f ′ t dt
= f ( b ) f ′ ( b ) − f ( a ) f ′ ( a ) − ∫ b
[ f ′ ( t )]2 dt b
[ ( )]2 a
= −∫ f ′ t dt
a
[ ( )]2 0, so [ ( )]2 0 b
f ′ t ≥ − ∫ f ′ t ≤ .
a
84.
∫ x ⎛ ⎜ ∫
t f ( z ) dz ⎞ ⎟
dt
0 ⎝ 0
⎠ u = ∫ t f ( z ) dz dv = dt
0
du = f(t)dt v = t
∫ x ⎛⎜ x ∫ t f ( z ) dz ⎞⎟ dt = ⎡⎢t ∫ t f ( z ) dz⎤⎥ – ∫ x t f ( t )
dt ⎝ ⎠ ⎣ ⎦ 0 0
0 0 0 0 0
( ) – ( ) x x = ∫ x f z dz ∫ t f t dt
By letting z = t,
( ) ( ) , x x ∫ x f z dz = ∫ x f t dt so
0 0
∫ x ⎛⎜ ∫ t f ( z ) dz ⎞⎟ dt = ∫ x x f ( t ) dt – ∫ x t f ( t ) dt
0 ⎝ 0 ⎠ 0 0
0
( – ) ( ) x = ∫ x t f t dt
( ) ... x t tn
I f tn dtn dt dt = ∫ ∫ ⋅⋅⋅∫ − be the iterated integral. Note that for i ≥ 2, the limits of integration of the
85. Let 1 1
0 0 0 2 1
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.
I x t 1 t 2 tn − 1 f ( tn ) dtn ... dt dt dt t 1 t 2 tn
−
2
F ( tn ) dtn ... dt dt dt = ∫ ∫ ∫ ⋅⋅⋅∫ = ∫ ∫ ⋅⋅⋅∫ − −
where ( ) 1 ( )
0 0 0 0 3 2 1 0 0 0 1 1 3 2 1
tn
− = ∫ .
F tn f tn dn −
1 0
By induction,
1
2 !
I x F t x t n dt
( ) ( )( ) 2
− = −
− ∫
( ) 1 ( )
1 0
0 1 1 1
n
t
dv x t n− = −
u = F t = ∫ f tn dtn , ( ) 2
1
1
1
v x t n
− = − −
( ) du = f t1 dt1 , ( ) 1
1
n
−
t =
x
⎧⎪⎡ ⎤ ⎪⎫ = ⎨⎢− − ⎥ + − ⎬ − ⎩⎪⎣ − ⎦ − ⎪⎭
= −
1 1 1
2 ! 1 1
1 ( )( )
( 1)!
( ) ( ) ( ) ( )( ) 1
n t x n
− −
1 1
∫ 1
∫
I x t f t dt f t x t dt
n n
1 0 0 1 1 1
n n n
t
1
0
−
1
x n
f t x t dt
∫
0 1 1 1
n
=
−
.
(note: that the quantity in square brackets equals 0 when evaluated at the given limits)
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23. 86. Proof by induction.
n = 1:
u = P1(x) dv = exdx
du x) = dP1(dx
v = ex
dx
exP (x)dx exP (x) – ex dP (x) dx
exP (x) – dP (x) exdx
∫ = ∫ 1
1
1 1
dx
exP (x) – ex dP (x)
= ∫ 1
1
dx
1
dx
=
Note that dP1(x)
dx
is a constant.
Suppose the formula is true for n. By using integration by parts with u = Pn+1(x) and dv = exdx,
dP x
1
x ( ) x ( ) – x n
n n
+
∫ + = + ∫
Note that n 1( ) dP x
e P x dx e P x e dx
1 1
( )
dx
+ is a polynomial of degree n, so
dx
⎡ ⎛ ⎞⎤ = − ⎢ − ⎜ ⎟⎥ = − − ⎢⎣ ⎝ ⎠⎥⎦
n j n j
+
1
( ) ( ) ( ) e P x dx e P x e d dP ( x )
d P x e P x e
x x x j n x x j n
n n j n j
∫ Σ Σ
1 1
+ +
( ) ( ) ( 1) 1
1 1 1 1
+ + + +
dx dx dx
j j
0 0
= =
n j
1
d P x
+
= + Σ −
x x j n
n j
1
e P x e
1
( ) ( 1)
1
( )
j
+
dx
+
=
n 1
j
x j n
d P x
1
+
= Σ −
0
( )
( 1)
+
j
j
e
dx
=
87.
j
4 4 2
x x e dx e d x x
(3 2 ) (–1) (3 2 )
∫ 4 + 2
x = x Σ
j
= 0
dx
= ex[3x4 + 2x2 –12x3 – 4x + 36x2 + 4 – 72x + 72]
= ex (3x4 –12x3 + 38x2 – 76x + 76)
+
j
j
7.3 Concepts Review
1. 1 cos2
2
x dx
+ ∫
2. ∫(1– sin2 x) cos x dx
3. ∫sin2 x(1– sin2 x) cos x dx
4. cos mx cos nx = 1 [cos( m+ n ) x + cos( m− n ) x
]
2
Problem Set 7.3
∫ x dx = ∫ x dx
1 – 1 cos 2
2 2
1. sin2 1– cos 2
2
= ∫ dx ∫ x dx
= 1 x – 1 sin 2
x +C
2 4
2. u = 6x, du = 6 dx
sin4 6 1 sin4
∫ x dx = ∫ u du
6
1 1– cos 2 2
6 2
= ⎛ u ⎞ du ⎜ ⎟
⎝ ⎠ ∫
1 (1– 2cos 2 cos2 2 )
24
= ∫ u + u du
1 – 1 2cos 2 1 (1 cos 4 )
24 24 48
= ∫ du ∫ u du + ∫ + u du
3 – 1 2cos 2 1 4cos 4
48 24 192
= ∫ du ∫ u du + ∫ u du
3 (6 ) – 1 sin12 1 sin 24
48 24 192
= x x + x +C
= 3 x – 1 sin12 x + 1 sin 24
x +C
8 24 192
434 Section 7.2 Instructor Solutions Manual
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24. 3. ∫sin3 x dx = ∫sin x(1− cos2 x)dx
= ∫sin x dx − ∫sin x cos2 x dx
cos 1 cos3
= − x + x +C
3
4. ∫cos3 x dx =
= ∫cos x(1− sin2 x)dx
= ∫cos x dx − ∫cos x sin2 x dx
= sin x − 1 sin3
x + C
3
5. / 2 5 / 2 2 2
π π
∫ = ∫
cos θ dθ (1– sin θ ) cosθ dθ
0 0
/ 2 2 4
0
= ∫ +
(1– 2sin θ sin θ ) cosθ dθ
π
/ 2
⎡ π = sin – 2 sin 3 + 1 sin
5
⎤ ⎢⎣ ⎥⎦
1– 2 1 – 0 8
θ θ θ
0
3 5
= ⎛ + ⎞ = ⎜ ⎟
⎝ ⎠
3 5 15
6.
π / 2 π sin 6
= / 2 ⎛ 1– cos 2
θ
⎞ 3 0 0
⎜ ⎟
⎝ ⎠ ∫ ∫
d d
θ θ θ
2
1 / 2 (1– 3cos 2 3cos 2 2 – cos 3
2 )
8
0
= ∫ +
θ θ θ dθ
π
1 π / 2 3 π / 2 3 π / 2 2 1 π
– 2cos 2 cos 2 – / 2 cos 3
2
8 0 16 0 8 0 8
0
= ∫ ∫ + ∫ ∫
dθ θ dθ θ θ dθ
1[ 3 3 1 cos 4 θ
] – [sin 2 ] – 1 (1– sin 2 ) cos 2
8 16 8 2 8
π π π ⎛ + ⎞ π = + ⎜ ⎟
/ 2 / 2 / 2 / 2 2
0 0 0 0
⎝ ⎠ ∫ ∫
d d
θ θ θ θ θ θ
1 π 3 π / 2 3 π / 2 π 4cos 4 – 1 / 2 π
2cos 2 1 / 2 sin 2
2 2cos 2
8 2 16 0 64 0 16 0 16
0
= ⋅ + ∫ + ∫ ∫ + ∫ ⋅
dθ θ dθ θ dθ θ θ dθ
θ π θ π θ π π π
= + + + 5
3 3 [sin 4 ] – 1 [sin 2 ] 1 [sin 2 ]
/ 2 / 2 3 / 2
0 0 0
16 32 64 16 48
π
32
=
7. ∫sin5 4x cos2 4x dx = ∫(1– cos2 4x)2 cos2 4x sin 4x dx = ∫(1– 2cos2 4x + cos4 4x) cos2 4x sin 4x dx
= ∫ x x + x x dx – 1 cos3 4 1 cos5 4 – 1 cos7 4
– 1 (cos2 4 – 2cos4 4 cos6 4 )(–4sin 4 )
4
= x + x x +C
12 10 28
8. ∫(sin3 2t) cos 2tdt = ∫(1– cos2 2t)(cos 2t)1/ 2 sin 2t dt – 1 [(cos 2 )1/ 2 – (cos 2 )5/ 2 ](–2sin 2 )
= ∫ t t t dt
2
– 1 (cos 2 )3/ 2 1 (cos 2 )7 / 2
3 7
= t + t +C
9. ∫cos3 3θ sin–2 3θ dθ = ∫(1– sin2 3θ )sin–2 3θ cos3θ dθ 1 (sin 2 3 1)3cos3
= ∫ − θ − θ dθ
3
– 1 csc3 – 1 sin 3
3 3
= θ θ +C
10. ∫sin1/ 2 2z cos3 2z dz = ∫(1– sin2 2z)sin1/ 2 2z cos 2z dz
= ∫ z z z dz 1 sin3/ 2 2 – 1 sin7 / 2 2
1 (sin1/ 2 2 – sin5/ 2 2 )2cos 2
2
= z z +C
3 7
Instructor’s Resource Manual Section 7.3 435
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
25. 11.
2 2
t t dt t t dt
∫ sin4 + 3 cos4 3 = ∫ ⎛ 1– cos 6 ⎞ ⎛ 1 cos 6
⎞ 1 ⎜ ⎟ ⎜ ⎟
(1– 2cos2 6 cos4 6 )
⎝ 2 ⎠ ⎝ 2
⎠ = ∫ t + t dt
16
= 1 ∫ ⎡⎢ 1– (1 + cos12 t ) + 1 (1 + cos12 t )2
⎤⎥ dt – 1 cos12 1 (1 2cos12 cos2 12 )
16 ⎣ 4
⎦ = ∫ t dt + ∫ + t + t dt
16 64
– 1 12cos12 1 1 12cos12 1 (1 cos 24 )
192 64 384 128
= ∫ t dt + ∫ dt + ∫ t dt + ∫ + t dt
– 1 sin12 1 1 sin12 1 1 sin 24
= t + t + t + t + t +C 3 – 1 sin12 1 sin 24
192 64 384 128 3072
= t t + t +C
128 384 3072
12.
3
∫ cos6 sin2 d = ∫ ⎛ 1 + cos 2 θ ⎞ ⎛ 1– cos 2
θ
⎞ ⎜ ⎟ ⎜ ⎟
d
1 (1 2cos 2 – 2cos3 2 – cos4 2 )
⎝ ⎠ ⎝ ⎠ = ∫ + θ θ θ dθ
θ θ θ θ
2 2
16
1 1 2cos 2 – 1 (1– sin2 2 ) cos 2 – 1 (1 cos 4 )2
16 16 8 64
= ∫ dθ + ∫ θ dθ ∫ θ θ dθ ∫ + θ dθ
1 1 2cos 2 – 1 2cos 2 1 2sin2 2 cos 2 – 1 (1 2cos 4 cos2 4 )
16 16 16 16 64
= ∫ dθ + ∫ θ dθ ∫ θ dθ + ∫ θ θ dθ ∫ + θ + θ dθ
1 1 sin2 2 2cos 2 – 1 – 1 4cos 4 – 1 (1 cos8 )
16 16 64 128 128
= ∫ dθ + ∫ θ ⋅ θ dθ ∫ dθ ∫ θ dθ ∫ + θ dθ
= 1 1 sin3 2 1 1 sin 4 1 1 sin8
16 48 64 128 128 1024
θ + θ − θ − θ − θ − θ +C
5 1 sin3 2 – 1 sin 4 – 1 sin8
128 48 128 1024
= θ + θ θ θ + C
13. sin 4 cos5 1 [sin 9 sin( )] 1 (sin 9 sin )
∫ y y dy = ∫ y + −y dy = ∫ y − y dy
2 2
1 1 cos9 cos 1 cos 1 cos9
2 9 2 18
= ⎛⎜ − y + y ⎞⎟ +C = y − y +C
⎝ ⎠
∫ y y dy = ∫ y + − y dy 1 sin 5 1 sin( 3 )
14. cos cos 4 1 [cos5 cos( 3 )]
2
= y − − y +C 1 sin 5 1 sin 3
10 6
= y + y +C
10 6
15.
2
w w dw w w dw
∫ sin4 ⎛ ⎞ cos2 ⎛ ⎞ = ∫ ⎛ 1– cos ⎞ ⎛ 1 + cos
⎞ 1 ⎜ (1– cos – cos2 cos3 )
⎝ 2 ⎟ ⎜ ⎠ ⎝ 2 ⎟ ⎜ ⎟ ⎜ ⎟
⎠ ⎝ 2 ⎠ ⎝ 2
⎠ = ∫ w w+ w dw
8
= 1 ∫ ⎡⎢ 1– cos w – 1 (1 + cos 2 w ) + (1– sin2 w ) cos
w⎤⎥ dw 1 1 – 1 cos 2 – sin2 cos
8 ⎣ 2
⎦ = ∫
⎡⎢ w w w⎤⎥dw 8 ⎣ 2 2
⎦ 1 – 1 sin 2 – 1 sin3
16 32 24
= w w w+C
sin 3 sin 1 cos 4 cos 2
16. [ ]
∫ ∫
t t dt = − t −
t dt
2
( )
1 ∫ cos 4 ∫
cos 2
2
1 1 sin 4 1 sin 2
2 4 2
1 sin 4 1 sin 2
8 4
tdt tdt
= − −
= − ⎛ − ⎞ + ⎜ ⎟
t t C
⎝ ⎠
t tC
= − + +
436 Section 7.3 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.
26. 17. 2
∫
x cos x sin
xdx
u = x du =
1
dx
dv =
2
x x dx
v = − x − x dx = −
x
cos sin
(cos ) ( sin ) 1 cos
2 3
cos
t =
x 3
∫
Thus
2
∫
x x xdx
x x xdx
cos sin
=
( 1 cos 3 ) ∫
(1)( 1 cos 3
)
3 3
1 cos ∫
cos
3
1 cos ∫
cos (1 sin )
3
− − − =
⎡− x 3 x + 3
⎣ xdx
⎤ ⎦
= ⎡− x 3 ⎣ x + x − 2
xdx
⎤ ⎦
= ⎡ ⎤
⎢− + − ⎥ =
⎣ ⎦
⎡− + − ⎤ + ⎢⎣ ⎥⎦
1 cos 3 ∫
(cos cos sin 2
)
3
1 cos sin 1 sin
3 3
x x x x xdx
t sin
x
3 3
=
x x x x C
18. 3
∫
x sin x cos
xdx
u = x du =
1
dx
dv =
sin 3
x cos
x dx
v = (sin x ) (cos xdx ) =
1 sin
x
3 4
sin
t =
x 4
∫
Thus
3
∫
x x xdx
x x xdx
sin cos
=
(1 sin 4 ) ∫
(1)(1 sin 4
)
4 4
1 sin ∫
(sin )
4
1 sin 1 ∫
(1 cos 2 )
4 4
− =
4 2 2
⎡ ⎣ x x − x dx
⎤ ⎦
= ⎡ 4 2
⎤ ⎢⎣ x x − − x dx
⎥⎦
= 1 sin 1 (1 2cos 2 cos 2 )
4 4
1 sin 1 1 sin 2 1 (1 cos 4 )
4 4 4 8
1 sin 3 1 sin 2 1 sin 4
4 8 4 32
⎡ ⎢⎣ x 4 x − ∫
− x + 2
xdx
⎤ ⎥⎦
= ⎡ ⎢⎣ x 4
x − x + x − ∫
+ xdx
⎤ ⎥⎦
= ⎡ 4
⎤ ⎢⎣ x x − x + x − x ⎥⎦
+ C
19. 4 ( 2 )( 2
)
∫ ∫
x dx x x dx
tan tan tan
=
= ∫
( 2 )
2
−
= ∫
( −
)
= ∫ − ∫
−
= − + +
x x dx
x x xdx
x x dx x dx
x x x C
tan (sec 1)
tan 2 sec 2 tan
2
tan 2 sec 2 (sec 2
1)
1 tan 3
tan
3
20. 4 ( 2 )( 2
)
∫ ∫
∫
( 2 2 2 )
x dx x x dx
x x dx
cot =
cot cot
cot (csc 1)
( )
2 2
= −
∫
∫ ∫
x x xdx
x x dx x dx
x x x C
cot csc cot
cot csc (csc 1)
1 cot cot
3
= −
= 2 2 − 2
−
= − 3
+ + +
21. tan 3 x ( )( 2
)
x x dx
x x dx
x x C
∫
∫
tan tan
tan sec 1
1 tan ln cos
2
=
= −
( )( 2
)
2
= + +
22. 3 ( )( 2
)
∫ ∫
∫
∫ ∫
t dt t t dt
t t dt
t tdt tdt
cot 2 =
cot 2 cot 2
( cot 2 )( csc 2
2 1
)
cot 2 csc 2 cot 2
1 cot 2 1 ln sin 2
4 2
= −
= 2
−
= − 2
− +
t t C
Instructor’s Resource Manual Section 7.3 437
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
27. θ
⎛ ⎞ θ
⎜ ⎟
⎝ ⎠ ∫
23. tan5
2
d
u du d
⎛θ ⎞ θ = ⎜ 2 ⎟ ;
=
⎝ ⎠
2
θ
⎛ ⎞ = ⎜ ⎟
⎝ ⎠
5 5
∫ ∫
d udu
tan 2 tan
( )( )
∫
∫ ∫
∫ ∫
∫ ∫ ∫
u u du
u udu udu
u u du u u du
u udu u udu u
2 tan sec 1
2 tan sec 2 tan
2 tan sec 2 tan sec 1
2 tan sec 2 tan sec 2 tan du
1 tan tan 2ln cos
2 2 2 2
= −
= −
= − ( −
)
= − +
3 2
3 2 3
3 2 2
3 2 2
θ θ θ
= 4 ⎛ ⎞ − 2
⎛ ⎞ ⎜ ⎟ ⎜ ⎟
− + 2
C
θ
⎝ ⎠ ⎝ ⎠
24. ∫cot5 2t dt
u = 2t;du = 2dt
cot 2 1 cot
5 5
∫ ∫
t dt u du
2
1 ∫ ( cot 3 u )( cot 2 udu ) 1 ∫
( cot 3 u )( csc 2
1
)
du
2 2
1 ∫ ( cot )( csc )
1 ∫
cot
2 2
1 ∫ ( cot )( csc ) 1 ∫
( cot )( csc 1
)
2 2
1 ∫ ( cot )( csc ) 1 ∫ ( cot )( csc )
1 ∫
cot
2 2 2
1 cot 1 cot 1 ln sin
8 4 2
1 cot 2 1 c
8 4
= = −
3 2 3
u udu udu
= −
3 2 2
u u du u u du
= − −
3 2 2
u u du u u du u
= − +
4 2
= − + + +
4
u u u C
t
=
= − +
ot2 2 1 ln sin 2
t + t +C
2
25. − 3 4 ( −
3 )( 2 )( 2
)
∫ ∫
x xdx x x x dx
tan sec tan sec sec
=
= +
= +
= − + +
( 3 )( 2 )( 2
)
∫
∫ ∫
x x x dx
x x x x dx
x x C
−
tan 1 tan sec
tan sec dx tan sec
1 tan ln tan
2
3 2 1 2
− −
( )
2
−
26. − 3/ 2 4 ( −
3/ 2 )( 2 )( 2
)
∫ ∫
x xdx x x x
tan sec tan sec sec
=
= +
= +
= − + +
( 3/ 2 )( 2 )( 2
)
∫
∫ ∫
x x x dx
x x dx x x dx
x x C
−
tan 1 tan sec
tan sec tan sec
2 tan 2 tan
3/2 2 1/2 2
1/ 2 3/ 2
3
−
−
438 Section 7.3 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.
28. 27. ∫ tan3 x sec2 x dx
Let u = tan x . Then du = sec2 x dx .
tan3 sec2 3 1 4 1 tan4
∫ x x dx = ∫ u du = u +C = x +C
4 4
28.
3 − 1/2 2 −
3/2
∫ ∫
x xdx x x x xdx
tan sec tan sec (sec tan )
=
= −
= −
= + +
( 2 )( −
3/ 2
)( )
∫
∫ ∫
x x x x dx
x x x dx x x x dx
x x C
sec 1 sec sec tan
sec 1/ 2 sec tan sec −
3/ 2
sec tan
2 sec 2sec
3
( ) ( )
3/ 2 −
1/ 2
29.
∫ cos cos = 1 ∫ (cos[( + ) ] + cos[( − ) ])
1 1 sin[( ) ] 1 sin[( ) ]
mx nx dx m n x m n x dx π π
π π
2
– –
= ⎡ m + n x + m − n x
⎤ 2
⎢⎣ m + n m − n
⎥⎦
π
−π
= 0 for m ≠ n, since sin kπ = 0 for all integers k.
30. If we let u x
π
= then du dx
L
π
= . Making the substitution and changing the limits as necessary, we get
L
m π x n π
x dx L mu nu du
L L
L
π
cos cos cos cos 0 L
π
∫ = ∫ = (See Problem 29
− π −
(x sin x) dx π
∫ π + 2 2
31. 2
0
(x 2x sin x sin x) dx π
x dx x x dx x dx π π π π
= π∫ + + 2
0
= π∫ + π∫ + ∫ −
2 sin (1 cos2)
2
0 0 0
π π
1 2 sin cos 1 sin 2
3 2 2
⎡ 3 ⎤ π π = π ⎡ ⎤ ⎢⎣ x ⎥⎦ + π [ x − x x ]
+ x − x
0
⎢⎣ ⎥⎦
0 0
= π + π + π − + π − − 1 4 5 2 57.1437
1 4 π
2 (0 0) ( 0 0)
3 2
= π + π ≈
3 2
Use Formula 40 with u = x for ∫ x sin x dx
32. / 2 2 2
V 2 x sin (x )dx π
= π∫
0
u = x2 , du = 2x dx
/ 2 / 2 / 2 2 2
0 0 0
π π ⎡ ⎤π π = π = π = π ⎢ ⎥ = ≈ ⎣ ⎦ ∫ ∫
V u du u du u u
sin 1– cos 2 1 – 1 sin 2 2.4674
2 2 4 4
33. a. 1 f (x)sin(mx)dx π
π −π ∫
N
⎛ ⎞
1 sin( ) sin( )
a nx mxdx π
−π
∫ Σ
= ⎜⎜ ⎟⎟ π ⎝ ⎠
1
n
n
=
N
1 sin( )sin( )
=
π Σ ∫
1
a nx mxdx π
n
n
−π
=
From Example 6,
0 if
∫ ⎧ ≠
sin( nx )sin( mx )
dx
= ⎨so every term in the sum is 0 except for when n = m.
⎩ π if
= If m N, there is no term where n = m, while if m ≤ N, then n = m occurs. When n = m
an π
sin(nx)sin(mx) dx am n m
n m
π
−π
∫ = π so when m ≤ N,
−π
1 π
f ( x )sin( mx ) dx 1 am am ∫ = ⋅ ⋅π =
.
π −π
π Instructor’s Resource Manual Section 7.3 439
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
29. b. 1 f 2 (x)dx π
π −π ∫
N N
⎛ ⎞⎛ ⎞
1 sin( ) sin( )
a nx a mx dx π
−π
∫ Σ Σ
= π ⎜⎜ n ⎟⎟⎜⎜ m
⎟⎟ ⎝ ⎠⎝ ⎠
n m
1 1
= =
N N
1 sin( )sin( )
a a nx mxdx π
=
Σ n Σ ∫
πm
n m
1 1
−π
= =
From Example 6, the integral is 0 except when m = n. When m = n, we obtain
Σ Σ 2
.
πN N
1 ( )
a a π =
a
= =
n n n
n n
1 1
34. a. Proof by induction
x = x
n = 1: cos cos
2 2
Assume true for k ≤ n.
n
⎡ ⎤
x x x x x x x x + +
cos cos cos cos cos 1 cos 3 cos 2 –1 1 cos
⋅ = ⎢ + + + ⎥
n n n n n n n
1 –1 1
2 4 2 2 2 2 2 2 2
⎢⎣ ⎥⎦
Note that
k x x k x k x + + +
cos cos 1 1 cos 2 1 cos 2 –1 ,
2n 2n 2 2n 2n
⎛ ⎞⎛ ⎞ ⎡ + ⎤
⎜ ⎟⎜ = 1 ⎟ ⎢ + ⎥
⎝ ⎠⎝ ⎠ ⎣ 1 1
⎦
so
1
n n
+
⎡ ⎤ ⎛ ⎞ ⎡ ⎤
⎢ + + + ⎥ ⎜ ⎟ = ⎢ + + + ⎥
⎢⎣ ⎥⎦ ⎝ ⎠ ⎢⎣ ⎥⎦
cos 1 cos 3 cos 2 –1 cos 1 1 cos 1 cos 3 cos 2 –1 1
x x x x x x x
n n n n n n n
n n 1 –1 1 1 1
+ + + +
2 2 2 2 2 2 2 2 2
n n
⎡ ⎤ ⎡ ⎤
⎢ + + + ⎥ = ⎢ + + + ⎥
⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦
x x x x x x x
lim cos 1 cos 3 cos 2 –1 1 1 lim cos 1 cos 3 cos 2 –1
b.
–1 –1
n n n n n n n n n n
→∞ x →∞
2 2 2 2 2 2 2 2
1 cos x t dt
x
= ∫
0
1 ∫ x cos t dt = 1 [sin t ] x =
sin x
x x x
c. 0 0
x + x
= we see that since
35. Using the half-angle identity cos 1 cos ,
2 2
π π
= =
cos cos 2 2
4 2 2
2
π π + +
= = =
2 2 1 2 2 cos cos ,
8 4 2 2
2 2
π π + + + +
= = = etc.
2 2 1 2 2 2 cos cos ,
16 8 2 2
Thus, 2 2 + 2 2 + 2 +
2
⎛ π ⎞ ⎛ π ⎞ ⎛ π ⎞
⋅ ⋅ cos 2 cos 2 cos 2
2 2 2
= ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ ⎝ 2 ⎟ ⎜ ⎟ ⎜ ⎟
⎠ ⎝ 4 ⎠ ⎝ 8
⎠
( ) 2 2 2 2
π π π π
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
sin 2 lim cos cos cos
n 2 4 2n
= ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ π ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
→∞ π
2
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
30. π π
∫ π − = ∫ − +
36. Since (k − sin x)2 = (sin x − k)2 , the volume of S is 2 2 2
(k sin x) π (k 2k sin x sin x) dx
0 0
k dx k xdx xdx π π π π
x k x 2 k cos x x 1 sin 2
π
= π π + π π + π ⎡ − ⎤ ⎢⎣ ⎥⎦
2
= π ∫ − π∫ + ∫ − 2 [ ] [ ]
2
0 0 0
2 sin (1 cos2)
2
0 0
0
2 2
π π
2 2 2 ( 1 1) ( 0) 2 2 4
k k k k
= π + π − − + π − = π − π +
2 2
Let
2
π
= π − π + then f ′(k) = 2π2k − 4π and f ′(k) = 0 when k = 2 .
( ) 2 2 4 ,
2
f k k k
π
The critical points of f(k) on 0 ≤ k ≤ 1 are 0, 2 ,
π
1.
2 2 2
(0) 4.93, 2 4 8 0.93, (1) 2 4 2.24
π ⎛ ⎞ π π = ≈ ⎜ ⎟ = − + ≈ = π − π + ≈ ⎝ π ⎠
f f f
2 2 2
a. S has minimum volume when k = 2 .
π
b. S has maximum volume when k = 0.
7.4 Concepts Review
1. x – 3
2. 2 sin t
3. 2 tan t
4. 2 sec t
Problem Set 7.4
1. u = x +1, u2 = x +1, 2u du = dx
∫ x x +1dx = ∫(u2 –1)u(2u du)
= ∫(2u4 – 2u2 )du 2 5 – 2 3
= u u +C
5 3
2 ( 1)5 / 2 – 2 ( 1)3/ 2
5 3
= x + x + + C
2. u = 3 x + π, u3 = x + π, 3u2du = dx
∫ x3 x + πdx = ∫(u3 – π)u(3u2du)
= ∫(3u6 – 3πu3)du 3 7 – 3 π
4
u u C
= +
7 4
3 7 / 3 3 π
( )– ( )4 / 3
7 4
x x C
= +π +π +
3. u = 3t + 4, u2 = 3t + 4, 2u du = 3 dt
1 2 2
3 3 2 ( 4) 2 ( –4)
t dt u −
u du u du
t u
∫ = ∫ =
∫
3 + 4 9
2 3 – 8
27 9
= u u +C
2 (3 4)3/ 2 – 8 (3 4)1/ 2
27 9
= t + t + + C
4. u = x + 4, u2 = x + 4, 2u du = dx
2 2 2 2 3 ( – 4) 3( – 4)2
x + x u +
dx u u du
x u
∫ =
∫
+ 4
= 2∫(u4 – 5u2 + 4)du 2 5 – 10 3 8
= u u + u + C
5 3
2 ( 4)5/ 2 – 10 ( 4)3/ 2 8( 4)1/ 2
5 3
= x + x + + x + +C
5. u = t , u2 = t, 2u du = dt
dt 2u du 2 u e e du
t e u e u e
2 2 2
1 1 1
+ −
∫ = ∫ =
∫
+ + + 2 –2 e
du du
2 2
1 1
+ ∫ ∫
u e
=
= 2[u] 2 1 – 2e ⎡⎣ln u + e ⎤⎦
2
1 = 2( 2 –1) – 2e[ln( 2 + e) – ln(1+ e)]
⎛ 2
+ ⎞
2 2 – 2 – 2 ln e e
= ⎜⎜ ⎝ 1
+ e
⎟⎟ ⎠
6. u = t , u2 = t, 2u du = dt
t dt u u du
t u
1 1
0 0 2
(2 )
∫ =
∫
+ + 1 1
1 2 1 2
0 2 0 2
u du u du
u u
+ −
2 2 1 1
+ + ∫ ∫
1 1
0 0 2
= =
1 1
2 – 2 1
∫ du ∫ du
1 –1 1
+ 1
u
=
= 2[u]0 – 2[tan u]0
π
2 – 2 tan–11 2 – 0.4292
= = ≈
2
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31. 7. u = (3t + 2)1/ 2 , u2 = 3t + 2, 2u du = 3dt
(3 2)3/ 2 1 ( 2 – 2) 3 2
t t + dt = u u ⎛⎜ u du ⎞⎟
∫ ∫
3 ⎝ 3
⎠ 2 ( 6 – 2 4 ) 2 7 4 5
9 63 45
= ∫ u u du = u − u +C
= 2 (3 t + 2)7 / 2 – 4 (3 t + 2)5 / 2
+C
63 45
8. u = (1– x)1/ 3, u3 = 1– x, 3u2du = –dx
∫ x(1– x)2 / 3dx = ∫(1– u3)u2 (–3u2 )du
3 ( 7 – 4 ) 3 8 3 5
= ∫ u u du = u − u +C
8 5
3 (1– )8/ 3 – 3 (1– )5 / 3
8 5
= x x +C
9. x = 2 sin t, dx = 2 cos t dt
4 – 2 2cos (2cos )
x dx t t dt
x 2sin
t
∫ = ∫
1– sin2 2
= ∫ = 2∫csct dt – 2∫sin t dt
sin
= 2ln csct − cot t + 2cos t + C
t dt
t
2
2ln 2 4 – x 4 – x2 C
= + +
x
−
10. x = 4sin t, dx = 4cos t dt
2 2
x dx t t dt
∫ = ∫
16 – x 2
cos
t
= 16∫sin2 t dt = 8∫(1– cos 2t)dt
= 8t – 4sin 2t +C = 8t −8sin t cos t +C
16 sin cos
2
= 8sin–1 ⎛ x ⎞ – x 16 –
x ⎜ ⎟
+C 4 2
⎝ ⎠
11. x = 2 tan t, dx = 2sec2 t dt
2
dx 2sec t dt 1 cos
t dt
∫ = =
( x 2 + 4) 3/2 ∫ 2 ∫
(4sec t
) 3/2
4
1 sin
4
= t + C
x C
x
= +
4 2 +
4
12. t = sec x, dt = sec x tan x dx
x
π
Note that 0 .
2
≤
t2 –1 = tan x = tan x
3 sec–1(3)
2 2 2 / 3 2
dt sec x tan
x dx
∫ = ∫
t t –1 π
sec x tan
x sec–1(3)
/ 3
= ∫
cos x dx π
sec–1(3) 1
x −
π
[sin ] / 3 sin[sec (3)] sin
π
3
= = −
sin cos 1 1 3 2 2 – 3 0.0768
⎡ = − ⎛ ⎞⎤ ⎢ ⎜ − = ≈ ⎣ ⎝ 3 ⎟⎥ ⎠⎦
2 3 2
13. t = sec x, dt = sec x tan x dx
π
≤ π
x
Note that .
2
t2 –1 = tan x = – tan x
–3 2 sec–1(–3)
–2 3 2 / 3 3
t –1 dt – tan x sec x tan
x dx
t π sec
x
∫ = ∫
sec–1(–3) sec–1(–3) 2
2 /3 2 /3
– sin 1 cos 2 – 1
= = ⎛ ⎞ ⎜ ⎟
⎝ ⎠ ∫ ∫
x dx x dx π π
2 2
sec–1(–3)
π
2 /3
1 sin 2 – 1
4 2
= ⎡ ⎤ ⎢⎣ x x
⎥⎦
sec–1(–3)
= ⎡ ⎢⎣ x x ⎤ ⎥⎦
2 /3
– 2 – 1 sec–1(–3) π
3 x0.151252
1 sin cos – 1
2 2
π
= + + ≈
9 2 8 3
14. t = sin x, dt = cos x dx
t dt x dx
t
∫ = ∫ = –cos x + C
2
sin
1–
= – 1– t2 +C
15. z = sin t, dz = cos t dt
z dz t dt
z
2 –3 (2sin – 3)
1–
∫ = ∫
2
= –2 cos t – 3t + C
= –2 1– z2 – 3sin–1 z +C
442 Section 7.4 Instructor’s Resource Manual
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32. 16. x = π tan t, dx = πsec2 t dt
π
x –1 dx ( 2
tan t –1)sect dt
x
∫ = ∫
π
2 + π
2
= π2 ∫ tan t sect dt – ∫sect dt
= π2 sect – ln sect + tan t +C
= π x2 + π2 − ln 1 x2 + π2 + x + C
π π
x 1 dx
x
π π −
∫
0 2 2
+ π
⎡ π x 2 + π 2
⎤ = ⎢π x 2 + π 2
– ln + x
⎥
⎢ π π ⎥
⎣ ⎦
0
= [ 2π2 – ln( 2 +1)] – [π2 − ln1]
= ( 2 –1)π2 – ln( 2 +1) ≈ 3.207
17. x2 + 2x + 5 = x2 + 2x +1+ 4 = (x +1)2 + 4
u = x + 1, du = dx
dx du
∫ =
∫
u = 2 tan t, du = 2sec2 t dt
2 2 5 2 4
x x u
+ + +
du sec t dt ln sec t tan
t C
u
∫ ∫
= = + +
2 1
4
+
2
u u C
1
+
ln 4
= + +
2 2
2
x x x C
1
+ 2 + 5 + +
ln 1
= +
2
= ln x2 + 2x + 5 + x +1 +C
18. x2 + 4x + 5 = x2 + 4x + 4 +1 = (x + 2)2 +1
u = x + 2, du = dx
dx du
∫ =
∫
u = tan t, du = sec2 t dt
2 4 5 2 1
x x u
+ + +
du t dt t t C
u
∫ ∫
2
sec ln sec tan
1
= = + +
+
dx u 2
u C
2
ln 1
x x
4 5
= + + +
+ +
∫
= ln x2 + 4x + 5 + x + 2 +C
19. x2 + 2x + 5 = x2 + 2x +1+ 4 = (x +1)2 + 4
u = x + 1, du = dx
x dx u du
3 3– 3
2 5 4
∫ =
∫
2 2
x x u
+ + +
u du du
u u
∫ ∫
3 –3
2 2
4 4
=
+ +
(Use the result of Problem 17.)
= 3 u2 + 4 – 3ln u2 + 4 + u +C
= 3 x2 + 2x + 5 – 3ln x2 + 2x + 5 + x +1 +C
20. x2 + 4x + 5 = x2 + 4x + 4 +1 = (x + 2)2 +1
u = x + 2, du = dx
2 x –1 2 u −
dx 5
du
x 4 x 5 u
1
∫ =
∫
2 2
+ + +
u du du
u u
2 – 5
=
∫ ∫
2 + 1 2
+
1
(Use the result of Problem 18.)
= 2 u2 +1 – 5ln u2 +1 + u +C
= 2 x2 + 4x + 5 – 5ln x2 + 4x + 5 + x + 2 +C
21. 5 − 4x − x2 = 9 − (4 + 4x + x2 ) = 9 − (x + 2)2
u = x + 2, du = dx
∫ 5 – 4x – x2 dx =∫ 9 – u2 du
u = 3 sin t, du = 3 cos t dt
9 2 9 cos2 9 (1 cos 2 )
∫ − u du = ∫ t dt = ∫ + t dt
2
9 1sin 2
2 2
= ⎛⎜ t + t ⎞⎟ +C
⎝ ⎠
9( sin cos )
2
= t + t t +C
= ⎛ u ⎞ + u u +C ⎜ ⎟
9 sin–1 1 9 – 2
2 3 2
⎝ ⎠
x x x x C
9 sin–1 2 2 5 – 4 – 2
2 3 2
⎛ + ⎞ + = ⎜ ⎟ + +
⎝ ⎠
22. 16 + 6x – x2 = 25 − (9 − 6x + x2 ) = 25 – (x – 3)2
u = x – 3, du = dx
dx du
x x u
∫ =
∫
u = 5 sin t, du = 5 cos t
16 +
6 – 2 25 – 2
du dt t C
u
= ⎛ u ⎞ +C ⎜ ⎟
∫ ∫ sin–1
25 2
= = +
−
5
⎝ ⎠
= sin–1 ⎛ x – 3
⎞ ⎜ ⎟
+C 5
⎝ ⎠
Instructor’s Resource Manual Section 7.4 443
© 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.
33. 23. 4x – x2 = 4 − (4 − 4x + x2 ) = 4 – (x – 2)2
u = x – 2, du = dx
dx du
x x u
∫ = ∫
4 – 2 4 – 2
u = 2 sin t, du = 2 cos t dt
du dt t C
u
= ⎛ u ⎞ +C ⎜ ⎟
∫ ∫ sin–1
4 2
= = +
−
2
⎝ ⎠
= sin–1 ⎛ x – 2
⎞ ⎜ ⎟
+C 2
⎝ ⎠
24. 4x – x2 = 4 − (4 − 4x + x2 ) = 4 – (x – 2)2
u = x – 2, du = dx
x u +
dx 2
du
x x u
∫ = ∫
2 2
4 – 4–
u du du
u u
– – 2
= ∫ + ∫
4 – 2 4 –
2
(Use the result of Problem 23.)
– 4 – 2 2sin–1
= u + ⎛ u ⎞ +C ⎜ ⎟
2
⎝ ⎠
= – 4 x – x 2 + 2sin–1 ⎛ x – 2
⎞ ⎜ ⎟
+ C 2
⎝ ⎠
25. x2 + 2x + 2 = x2 + 2x +1+1 = (x +1)2 +1
u = x + 1, du = dx
x dx u du
2 +
1 2 –1
2 2 1
∫ =
2 ∫
+ + 2
+ x x u
u du du
u u
2 –
1 1
=
∫ 2 ∫
+ 2
+ = ln u2 +1 – tan–1 u +C
= ln (x2 + 2x + 2)− tan−1(x +1) +C
26. x2 – 6x +18 = x2 – 6x + 9 + 9 = (x – 3)2 + 9
u = x – 3, du = dx
x dx u du
2 –1 2 5
–6 18 9
∫ =
2 ∫
+ 2
+ x x u
+
2 u du 5
du
u u
+ + ∫ ∫
ln ( 2 9) 5 tan 1
= +
2 2
9 9
= u + + − ⎛ u ⎞ +C ⎜ ⎟
3 3
⎝ ⎠
ln ( 2 6 18) 5 tan 1 3
= x − x + + − ⎛ x − ⎞ +C ⎜ ⎟
3 3
⎝ ⎠
27.
2
⎛ ⎞
1
0 2
1
2 5
⎝ + + ⎠ ∫
V dx
= π ⎜ ⎟
x x
2
⎡ ⎤
1
1
0 2
∫
= π ⎢ ⎥
x
( 1) 4
dx
⎢⎣ + + ⎥⎦
x + 1 = 2 tan t, dx = 2sec2 t dt
1 2sec
4sec
π ⎛ ⎞
/ 4 2
tan (1/ 2) 2
⎝ ⎠ ∫
V tdt
= π –1
⎜ ⎟
2
t
1
t dt π π
π π
= ∫ –1
/ 4
tan –1
(1/ 2) 2
8 sec
dt
t
/ 4 cos
2
tan (1/ 2)
= ∫
8
t dt π π ⎛ ⎞ = ⎜ + ⎟
/ 4
tan (1/ 2)
∫
–1
⎝ ⎠ 1 1cos 2
8 2 2
/ 4
–1
π ⎡ ⎤π = ⎢ + ⎥ ⎣ ⎦
tan (1/ 2)
1 1sin 2
8 2 4
t t
/ 4
t t t π
–1
tan (1/ 2)
1 1sin cos
8 2 2
π ⎡ ⎤ = ⎢ + ⎥ ⎣ ⎦
1 – 1 tan–1 1 1
π π = ⎡⎛ + ⎞ ⎛ ⎢⎜ ⎟ ⎜ + ⎞⎤ 8 ⎣⎝ 8 4 ⎠ ⎝ 2 2 5
⎠⎦
⎟⎥ 1 – tan–1 1 0.082811
π ⎛ π ⎞ = ⎜ + ⎟ ≈
16 10 4 2
⎝ ⎠
2 1
28. 1
+ + ∫
1
0 2
V xdx
0 2
x x
2 5
= π
2
x dx
∫
( x
+ 1) + 4
1 1
0 2 0 2
= π
x dx dx
x x
+
2 1 – 2 1
+ + + + ∫ ∫
= π π
( 1) 4 ( 1) 4
1 1
x x
2 1 ln[( 1) 4] – 2 1 tan 1
⎡ ⎡ = π + 2 + ⎤ π –1
⎛ + ⎞⎤ ⎜ ⎣ ⎢ 2 ⎥ ⎦ ⎢ ⎣ 2 ⎝ 2
⎠⎦
⎟⎥ [ln8 – ln 5] – tan–11– tan–1 1
0 0
= π π ⎡ ⎤ ⎢⎣ 2
⎥⎦
ln 8 – tan–1 1 0.465751
5 4 2
⎛ π ⎞ = π⎜ + ⎟ ≈
⎝ ⎠
29. a. u = x2 + 9, du = 2x dx
xdx du u C
x u
∫ 2
∫
+ 1 ln 2 9 1 ln ( 2 9)
2 2
1 1ln
= = +
9 2 2
= x + +C = x + +C
b. x = 3 tan t, dx = 3sec2 t dt
xdx t dt
x
∫ ∫ tan
= – ln cost +C
2 + 9
=
⎛ ⎞
ln 3 ln 3
= − + = − ⎜ ⎟ +
C C
2 1 2 1
+ ⎜ + ⎟ ⎝ ⎠
x x
9 9
= ln ⎛⎜ x 2
+ 9 ⎞⎟ − ln 3 +
C1 ⎝ ⎠
= ln ((x2 + 9)1/ 2 )+C 1 ln ( 2 9)
= x + +C
2
444 Section 7.4 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.
34. 30. u = 9 + x2 , u2 = 9 + x2 , 2u du = 2x dx
3 3 3 2 3 2 2
0 2 0 2 3
x dx x u −
x dx 9
udu
∫ = ∫ =
∫
x x u
9 9
+ +
3 2 ⎡ 3 2
⎤
3 2 u du u u
3
∫
≈ 5.272
( 9) – 9 18 – 9 2
= − = ⎢ ⎥ =
3
3
⎢⎣ ⎥⎦
31. a. u = 4 – x2 , u2 = 4 – x2 , 2u du = –2x dx
2 2 2
x dx x x dx u du
x x u
4 – 4 −
–
∫ = ∫ = ∫
2 2
4 –
2
2 2
u du du du
u u
4 4 4 1
4 4
− + −
− − ∫ ∫ ∫
4 1 ln 2
= =− +
u +
u C
u
= − ⋅ + +
4 2
−
2
x x 2
C
x
− +
ln 4 2 4
= − + − +
2
4 − −
2
b. x = 2 sin t, dx = 2 cos t dt
4 – 2 cos2 2
x dx t dt
x sin
t
∫ = ∫
(1– sin2 ) 2
= ∫
sin
= 2∫csct dt – 2∫sin t dt
= 2ln csct − cot t + 2cos t + C
t dt
t
2
−
2ln 2 4 x 4 x2 C
= − + − +
x x
2
− −
2ln 2 4 x 4 x2 C
= + − +
x
To reconcile the answers, note that
2 2
2 2
x x
x x
− + − −
ln 4 2 ln 4 2
− =
4 − − 2 4 − +
2
2 2
x
− −
ln ( 4 2)
( 4 x 2 2)( 4 x
2
2)
=
− + − −
2 2 2 2
x x
(2 − 4 − ln ) ln (2 − 4 −
)
= =
2 2
x x
4 4
− − −
2
⎛ − − ⎞ − − = ⎜ ⎟ =
2 4 2 2 4 2 ln x 2ln x
x x
⎜ ⎟
⎝ ⎠
32. The equation of the circle with center (–a, 0) is
(x + a)2 + y2 = b2 , so y = ± b2 – (x + a)2 . By
symmetry, the area of the overlap is four times
the area of the region bounded by x = 0, y = 0,
and y = b2 – (x + a)2 dx .
A = 4 ∫ b – a b 2 –( x + a ) 2
dx
0
x + a = b sin t, dx = b cos t dt
/ 2 2 2
sin ( / )
A b tdt π
= ∫
4 cos
–1
a b
b tdt π
2 / 2
sin ( / )
= ∫ +
2 (1 cos 2 )
–1
a b
b t t π
/ 2
= ⎡ + ⎤ ⎢⎣ ⎥⎦
2 –1
a b
2 2
1 sin 2
sin ( / )
b t t tπ
2 /2
–1
2 [ sin cos ]
a b
sin ( / )
= +
⎡π ⎛ ⎛ a ⎞ a b 2 –
a
2
⎞⎤ = 2 b 2 ⎢ – ⎜ sin–1 ⎢ 2
⎜ ⎜ ⎟ + ⎟⎥ ⎝ ⎝ b ⎠ b b
⎟⎥ ⎣ ⎠⎦
b2 – 2b2 sin–1 a – 2a b2 – a2
= π ⎛ ⎞ ⎜ ⎟
b
⎝ ⎠
33. a. The coordinate of C is (0, –a). The lower arc
of the lune lies on the circle given by the
equation x2 + ( y + a)2 = 2a2 or
y = ± 2a2 – x2 – a. The upper arc of the
lune lies on the circle given by the equation
x2 + y2 = a2 or y = ± a2 – x2 .
– – 2 – – a a
a a
A = a 2 x 2 dx ⎛⎜ a 2 x 2
a ⎞⎟ dx
∫ ∫
– –
⎝ ⎠ – – 2 – 2 a a
a a
2 2 2 2 2
= ∫ a x dx ∫ a x dx + a
– –
Note that 2 2
– a
a
∫ a x dx is the area of a
–
semicircle with radius a, so
2
a 2 x 2
dx a
a
a
∫ =
–
For 2 2
π
– .
2
2 – , a
a
∫ a x dx let
–
x = 2a sin t, dx = 2a cos t dt
2 – 2 cos a
a
a 2 x 2 dx π
/ 4 a 2 2
tdt π
∫ = ∫
– –/ 4
2 π / 4 1 π
/ 4 (1 cos 2 ) 2
sin 2
– π /4 – π
/ 4
= a ∫
+ tdt = a ⎡ ⎢⎣ t + t
⎤ 2
⎥⎦ 2
2
a a
π
= +
2
2 2
– 2 2 2 2
2 2
π ⎛ π ⎞
= ⎜⎜ + ⎟⎟ + =
A a a a a a
⎝ ⎠
Thus, the area of the lune is equal to the area
of the square.
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35. b. Without using calculus, consider the
following labels on the figure.
Area of the lune = Area of the semicircle of
radius a at O + Area (ΔABC) – Area of the
sector ABC.
1 2 2 – 1 ( 2 )2
2 2 2
⎛ π ⎞ = π + ⎜ ⎟
A a a a
⎝ ⎠
1 2 2 – 1 2 2
2 2
= πa + a πa = a
Note that since BC has length 2a, the
π
measure of angle OCB is ,
4
so the measure
π
of angle ACB is .
2
34. Using reasoning similar to Problem 33 b, the area is
a a b a a b
1 1 (2 ) – – 1 2sin
2 2 2
1 – – sin .
2
π + ⎛ ⎞ ⎜ ⎟
2 2 2 –1 2
⎝ ⎠
a 2 a b 2 a 2 b 2 –1
a
b
b
= π +
35.
2 – 2 dy – a x
dx x
= ;
2 – 2 y – a x dx
x
= ∫
x = a sin t, dx = a cos t dt
cos cos2 – cos –
y a t a t dt a t dt
= ∫ = ∫
a sin t sin
t
1– sin2 – (sin – csc )
a t dt a t t dt
= ∫ = sin
t
∫
= a (– cos t − ln csct − cot t ) +C
2 cos t = a – x 2 2 2 , csct = a , cot t =
a – x
a x x
⎛ − ⎞ = ⎜ − − ⎟ +
2 – 2 2 2 y a – a x ln a a x C
⎜ a x x
⎟
⎝ ⎠
2 2
a2 x2 a − a ln a −
x C
= − − − +
x
Since y = 0 when x = a,
0 = 0 – a ln 1 + C, so C = 0.
2 2
y – a2 x2 a ln a a – x
x
−
= − −
7.5 Concepts Review
1. proper
2. –1 5
1
x
x
+
+
3. a = 2; b = 3; c = –1
A B Cx +
D
x x x
4. + +
–1 ( –1)2 2 +
1
Problem Set 7.5
1. 1
A B
= +
xx x x
( + 1) +
1
1 = A(x + 1) + Bx
A = 1, B = –1
1 1 – 1
( 1) 1
+ + ∫ ∫ ∫
= ln x – ln x +1 + C
dx =
dx dx
x x x x
2. 2
A B
2 = 2
= +
3 ( 3) 3
x x x x x x
+ + +
2 = A(x + 3) + Bx
2 , – 2
3 3
A = B =
∫ 2
∫ ∫
+ + 2 ln – 2 ln 3
3 3
dx dx B dx
2 =
2 1 – 2
3 3 3 3
x x x x
= x x + + C
A B
3 3
–1 ( 1)( –1) 1 –1
3. 2
= = +
x x + x x +
x
3 = A(x – 1) + B(x + 1)
– 3 , 3
2 2
A = B =
3 – 3 1 3 1
–1 2 1 2 –1
∫ 2
∫ ∫
+ – 3 ln 1 3 ln –1
dx = dx +
dx
x x x
= x + + x + C
2 2
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36. x x
5 5 5
4. 3 2 2
= =
x x x x x x
2 + 6 2 ( + 3) 2 ( +
3)
3
A B
x x
= +
+
5 ( 3)
2
= A x + + Bx
5 , – 5
6 6
A = B =
x dx dx
5 5 1 – 5 1
∫ =
x 3 + x 2
∫ x ∫
x
+ 5 ln – 5 ln 3
6 6
2 6 6 6 3
= x x + +C
x x A B
x x x x x x
5. 2
11 11
3 4 ( 4)( 1) 4 1
− −
= = +
+ − + − + −
x – 11 = A(x – 1) + B(x + 4)
A =3, B = –2
x dx dx dx
x x x x
∫ 2
∫ ∫
+ − + − = 3ln x + 4 − 2ln x −1 + C
11 3 1 2 1
3 4 4 1
−
= −
x x A B
x x x x x x
6. 2
– 7 = – 7
= +
– –12 ( – 4)( + 3) – 4 +
3
x – 7 = A(x + 3) + B(x – 4)
– 3 , 10
7 7
A = B =
x dx dx dx
x x x x
∫ 2
∫ ∫
+ – 3 ln – 4 10 ln 3
– 7 = – 3 1 +
10 1
– –12 7 – 4 7 3
= x + x + +C
7 7
7. 2
x x A B
3 − 13 3 −
= 13
= +
3 10 ( 5)( 2) 5 2
x x x x x x
+ − + − + −
3x −13 = A(x − 2) + B(x + 5)
A = 4, B = –1
∫ x −
dx
4 1 1
2
+ − 3 13
3 10
x x
+ − ∫ ∫
dx dx
= −
x x
5 2
= 4ln x + 5 − ln x − 2 +C
x x
+ π + π
8. =
2 – 3 2 2 ( – 2 )( – )
x x x x
A B
x x
= +
π + π π π – 2 –
π π
x +π = A(x −π ) + B(x − 2π )
A = 3, B = –2
x + π
dx dx dx
∫ 2 π + π 2
∫ ∫
π π = 3ln x – 2π – 2ln x – π + C
3 – 2
=
x x x x
– 3 2 – 2 –
x x
x x x x
2 21 2 21
2 9 – 5 (2 –1)( 5)
9. 2
+ +
=
A B
x x
= +
+ + 2 –1 +
5
2x + 21 = A(x + 5) + B(2x – 1)
A = 4, B = –1
x dx dx dx
x x x x
2 21 4 – 1
2 9 – 5 2 –1 5
∫ 2
∫ ∫
+ + = 2ln 2x –1 – ln x + 5 +C
+
=
10.
2 2
2 2
x x x x x
x x x x
2 − − 20 2( + − 6)− 3 −
8
=
6 6
+ − + −
x
x x
2 3 8
2
6
+
= −
+ −
x x
x x x x
3 8 3 8
2
+ +
=
A B
x x
= +
+ − + − 3 2
6 ( 3)( 2)
+ −
3x + 8 = A(x – 2) + B(x + 3)
1 , 14
5 5
A = B =
2
2
x x dx
x x
2 20
6
− −
+ − ∫
2 1 1 14 1
+ − ∫ ∫ ∫
2 1 ln 3 14 ln 2
dx dx dx
= − −
x x
5 3 5 2
= x − x + − x − +C
5 5
x x
x x x x
17 – 3 17 – 3
3 – 2 (3 – 2)( 1)
11. 2
=
A B
x x
= +
+ + 3 – 2 +
1
17x – 3 = A(x + 1) + B(3x – 2)
A = 5, B = 4
x dx dx dx
x x x x
+ + ∫ ∫ ∫ 5 ln 3 – 2 4ln 1
17 – 3 = 5 +
4
3 2
– 2 3 – 2 1
= x + x + +C
3
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
37. 12. 2
x x
5 – 5 –
=
x x x x
– ( 4) 4 ( – )( – 4)
A B
x x
= +
π + + π π – π
– 4
5 – x = A(x – 4) + B(x – π )
5 – π
, 1
–4 4–
A B
= =
π π
∫ 2
∫ ∫ 5 – ln – 1 ln – 4
− π + + π π π π x dx dx dx
5 – 5 – π
= 1 +
1 1
( 4) 4 – 4 – 4 – – 4
x x x x
x x C
π
= π + +
π π
– 4 4 –
13.
2 2
3 2
x x x x
x x x x x x
2 + − 4 2 + −
4
=
A B C
x x x
= + +
− − + − 1 2
2 ( 1)( 2)
+ −
2x2 + x − 4 = A(x +1)(x − 2) + Bx(x − 2) + Cx(x +1)
A = 2, B = –1, C = 1
2
3 2
2 x + x −
4 dx 2 dx 1 dx 1
dx
x x x x x x
∫ = ∫ − ∫ +
∫ = 2ln x − ln x +1 + ln x − 2 + C
− − + − 2 1 2
14.
7 2 2 – 3
x +
x A B C
= + +
x x x x x x
(2 –1)(3 + 2)( – 3) 2 –1 3 +
2 – 3
7x2 + 2x – 3 = A(3x + 2)(x – 3) + B(2x –1)(x – 3) +C(2x –1)(3x + 2)
1 , – 1 , 6
35 7 5
A = B = C =
7 2 2 – 3 1 1 1 1 6 1 –
x +
x dx dx dx dx
∫ = ∫ ∫ +
∫
+ + 1 ln 2 –1 – 1 ln 3 2 6 ln – 3
70 21 5
x x x x x x
(2 –1)(3 2)( – 3) 35 2 –1 7 3 2 5 – 3
= x x + + x +C
15.
2 2
x x x x
x x x x x x
6 + 22 − 23 + −
=
6 22 23
(2 1)( 2
6) (2 1)( 3)( 2)
A B C
x x x
= + +
− + − − + − 2 − 1 + 3 −
2
6x2 + 22x − 23 = A(x + 3)(x − 2) + B(2x −1)(x − 2) + C(2x −1)(x + 3)
A = 2, B = –1, C = 3
2
x x dx dx dx dx
x x x x x x
6 + 22 −
23 2 1 3
(2 1)( 6) 2 1 3 2
∫ = ∫ − ∫ +
∫ = ln 2x −1 − ln x + 3 + 3ln x − 2 + C
− 2
+ − − + − 16.
3 2
3 2
x x x
x x x
6 11 6
− + −
− + −
4 28 56 32
3 2
3 2
⎛ − + − ⎞
x x x
x x x
1 6 11 6
4 7 14 8
= ⎜⎜ ⎟⎟ ⎝ − + − ⎠
2
⎛ − + ⎞
x x
1 1 3 2
4 7 14 8
= ⎜⎜ + ⎝ x 3 ⎟⎟ − x 2
+ x
− ⎠
⎛ − − ⎞
x x
x x x
1 1 ( 1)( 2)
4 ( 1)( 2)( 4)
= ⎜ + ⎟ ⎝ − − − ⎠
1 1 1
4 x 4
= ⎛ + ⎞ ⎜ − ⎟ ⎝ ⎠
3 2
3 2
x x x dx
x x x
∫ – 6 +
11 – 6
1 1 1
+ 4 –28 56 –32
= ∫ dx + ∫ dx
1 1ln – 4
x
4 4 –4
= x + x +C
4 4
448 Section 7.5 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.
38. 17.
3
x x x
–1 3 – 2
= +
2 2
x x x x
–2 –2
+ +
3 x –2 = 3 x –2
= A +
B
x 2
x x x x x
– 2 ( 2)( –1) 2 –1
+ + +
3x – 2 = A(x – 1) + B(x + 2)
8 , 1
3 3
A = B =
3
x dx
∫ ( 1) 8 1 1 1
x 2 + x – 2
∫ x dx ∫ dx ∫ dx
1 2 – 8 ln 2 1 ln –1
+ − = − + +
x x
3 2 3 1
= x x + x + + x +C
2 3 3
18.
3 2
2
x x x x
x x x x
– 4 14 24
+ +
= +
5 6 ( 3)( 2)
+ + + +
14 x +
24
= A +
B
( x 3)( x 2) x 3 x
2
+ + + +
14x + 24 = A(x + 2) + B(x + 3)
A = 18, B = –4
3 2
2 5 6
x x dx
x x
+
+ + ∫ ( 4) 18 – 4
∫ x dx ∫ dx ∫ dx
1 2 4 18ln 3 – 4ln 2
+ + x x
3 2
= − +
= x − x + x + x + +C
2
19.
4 2 2
3
x x x x
x x x x x
8 8 12 8
4 ( 2)( – 2)
+ + +
= +
− +
12 x 2 +
8
= A + B +
C
( 2)( – 2) 2 – 2
x x + x x x +
x
12x2 + 8 = A(x + 2)(x – 2) + Bx(x – 2) + Cx(x + 2)
A = –2, B = 7, C = 7
4 2
3
x x dx x dx dx dx dx
x x x x x
+ ∫ ∫ ∫ ∫ ∫ 1 2 – 2ln 7ln 2 7 ln – 2
8 8 – 2 1 7 1 7 1
– 4 2 – 2
+ +
= + +
= x x + x + + x +C
2
20.
6 3 2
x x x x x x
x x x x
4 4 4 16 68 272 4
– 4 – 4
+ + +
3 2
= + + + +
3 2 3 2
2
x A B C
272 +
4
= + +
x 2 ( x – 4) x x 2
x
– 4
272x2 + 4 = Ax(x – 4) + B(x – 4) +Cx2
– 1 , –1, 1089
4 4
A = B = C =
6 3
3 2
x 4 x 4
dx
x – 4
x
( 4 16 68) – 1 1 – 1 1089 1
+ + ∫ 3 2
− ∫ ∫ ∫ ∫
x x x dx dx dx dx
= + + + +
2
x x x
4 4 4
1 4 4 3 8 2 68 – 1 ln 1 1089 ln – 4
4 3 4 4
x x x x x x C
= + + + + + +
x
x +
1
A B
x x x
21. = +
2 2
( 3) 3 ( 3)
− − −
x + 1 = A(x – 3) + B
A = 1, B = 4
x dx dx dx
x x x
∫ +
1 = ∫ 1 +
∫ 4
ln 3 4
− 2 − − 2
( 3) 3 ( 3)
x C
= − − +
3
x
−
Instructor’s Resource Manual Section 7.5 449
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
39. x x A B
5 + 7 5 +
7
4 4 ( 2) 2 ( 2)
22. = = +
2 2 2
x x x x x
+ + + + +
5x + 7 = A(x + 2) + B
A = 5, B = –3
x dx dx dx
+ + + + ∫ ∫ ∫ 5ln 2 3
5 +
7 = 5 −
3
4 4 2 ( 2)
2 2
x x x x
x C
= + + +
2
x
+
x x
3 + 2 3 +
2
3 3 1 ( 1)
23. =
3 2 3
A B C
x x x
= + +
+ + + + 1 ( 1)2 ( 1)3
x x x x
+ + +
3x + 2 = A(x +1)2 + B(x +1) +C
A = 0, B = 3, C = –1
x dx dx dx C
3 +
2 3 1 3 1
3 3 1 ( 1) ( 1) 1 2( 1)
+ + + + + + + ∫ ∫ ∫
= − = − + +
3 2 2 3 2
x x x x x x x
24.
6
x A B C D E F G
= + + + + + +
( x – 2)2 (1– x )5 x – 2 ( x – 2)2 1– x (1– x )2 (1– x )3 (1– x )4 (1– x
)5
A = 128, B = –64, C = 129, D = –72, E = 30, F = –8, G = 1
6
2 5 2 2 3 4 5
⎡ ⎤
x dx 128 – 64 129 – 72 30 8 1
dx
∫ ∫
= ⎢ + + − + ⎥
x x x x x x x x x
( – 2) (1– ) – 2 ( – 2) 1– (1– ) (1– ) (1– ) (1– )
⎢⎣ ⎥⎦
128ln – 2 64 –129ln 1– 72 15 8 1
x x C
= + + − + − +
2 3 4
x x x x x
– 2 1– (1– ) 3(1– ) 4(1– )
25.
2 2
3 2 2 2
x x x x A B C
x x x x x x x x
3 − 21 + 32 3 − 21 +
32
= = + +
8 16 ( 4) 4 ( 4)
− + − − −
3x2 − 21x + 32 = A(x − 4)2 + Bx(x − 4) + Cx
A = 2, B = 1, C = –1
2
3 2 2
x x dx dx dx dx
x x x x x
∫ 3 − 21 +
32 = ∫ 2 + ∫ 1 −
∫ 1
2ln ln 4 1
− + − − 8 16 4 ( 4)
x x C
= + − + +
4
x
−
26.
2 2
x x x x
x x x x
+ 19 + 10 + 19 +
=
10
2 4 5 3 3
(2 5)
A B C D
x x x x
= + + +
+ + 2 3 2 +
5
A = –1, B = 3, C = 2, D = 2
2
x + 19 x + 10 ⎛ 1 3 2 2
⎞
dx – dx
2 x 5 x x x x 2 x
5
– ln x – 3 – 1 ln 2x 5 C
∫ = 4 3 ∫ ⎜ + + + + ⎝ 2 3
+ ⎟ ⎠ 2
= + + +
x x
27.
2 2
3 2 2
x x x x A BxC
x x x x x x
2 –8 2 –8
+ + +
= = +
4 ( 4) 4
+ + +
A = –2, B = 4, C = 1
2
3 2
x x dx dx x dx
x x x x
2 + – 8 –2 1 4 +
1
dx x dx dx
x x x
2 1 2 2 1
+ + ∫ ∫ ∫
∫ = ∫ +
∫ + + 2 2
4 4
= − + +
4 4
= –2ln x + 2ln x 2 + 4 + 1 tan–1
⎛ x ⎞ ⎜ ⎟
+C 2 2
⎝ ⎠
450 Section 7.5 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.
40. x x
3 + 2 3 +
2
( 2) 16 ( 4 20)
28. =
2 2
A Bx +
C
x x x
= +
+ + + + 2 4 20
x x x x x x
+ +
1 , – 1 , 13
10 10 5
A = B = C =
x dx
3 +
2
( 2) 16
∫
+ 2
+ xx x
1 x
13
10 5
2
1 1 –
+
10 4 20
1 1 14 1
10 5 ( 2) 16
+ + ∫ ∫ 2
dx dx
x x x
= +
x dx
1 2 4
20 4 20
+ + ∫ ∫ 2
dx dx
x x
= +
+
+ + ∫
x x
−
x x
1 ln 7 tan–1 2
10 10 4
⎛ + ⎞ = + ⎜ ⎟
⎝ ⎠
– 1 ln 2 4 20
20
x + x + +C
29.
2
x x A Bx C
x x x x
2 –3 –36
+
= +
(2 − 1)( 2 + 9) 2 –1 2
+
9
A = –4, B = 3, C = 0
2
x x dx dx x dx
x x x x
2 – 3 – 36 –4 1 3
(2 –1)( 9) 2 –1 9
∫ = ∫ +
∫ –2ln 2 –1 3 ln 2 9
2 + 2
+ = x + x + +C
2
1 1
x –16 (x 2)(x 2)(x 4)
30. =
4 2
− + +
A B Cx +
D
x x x
= + +
– 2 + 2 2 +
4
1 , – 1 , 0, – 1
32 32 8
A = B = C = D =
= x x + ⎛ x ⎞ +C ⎜ ⎟
+ + ∫ ∫ ∫ ∫ 1 ln – 2 – 1 ln 2 – 1 tan–1
1 1 1 – 1 1 1 1
–16 32 – 2 32 2 8 4
dx = dx dx −
dx
4 2
x x x x
32 32 16 2
⎝ ⎠
1
A B C D
31. = + + +
2 2 2 2
x x x x x x
( –1) ( + 4) –1 ( –1) + 4 ( +
4)
– 2 , 1 , 2 , 1
125 25 125 25
A = B = C = D =
1 – 2 1 1 1 2 1 1 1
+ + + ∫ ∫ ∫ ∫ ∫
– 2 ln –1 – 1 2 ln 4 – 1
dx = dx + dx + dx +
dx
2 2 2 2
x x x x x x
( –1) ( 4) 125 –1 25 ( –1) 125 4 25 ( 4)
x x C
= + + +
x x
125 25( –1) 125 25( +
4)
32.
3 2 2
x x x x
x x x x x x
– 8 –1 +
= 1 +
–7 7 –16
2 2
( + 3)( –4 + 5) ( + 3)( − 4 +
5)
2
2 2
x x A Bx C
–7 + 7 –16
+
= +
( x 3)( x – 4 x 5) x 3 x – 4 x
5
+ + + +
– 50 , – 41, 14
A = B = C =
13 13 13
3 2 41 14
⎡ ⎛ ⎞ + ⎤ = ⎢ ⎜ ⎟ + ⎥ + + ⎢⎣ ⎝ + ⎠ + ⎥⎦
x – 8 x –1 – x dx 1–
50 1 13 13
dx
x x x x x x
∫ ∫
2 2
( 3)( – 4 5) 13 3 – 4 5
dx dx dx x dx
50 1 68 1 41 2 4
13 3 13 ( 2) 1 26 4 5
+ − + − + ∫ ∫ ∫ ∫
– 50 ln 3 – 68 tan–1( – 2) – 41 ln 2 – 4 5
= − − −
2 2
−
x x x x
= x x + x x x + +C
13 13 26
Instructor’s Resource Manual Section 7.5 451
© 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. 33. x = sin t, dx = cos t dt
3 2 3 2
t t t dt x x dx
t t t x x x
(sin − 8sin − 1) cos − 8 −
1
(sin 3)(sin 4sin 5) ( 3)( 4 5)
∫ =
2 ∫
+ − + + 2
− + 50 ln 3 68 tan 1( 2) 41 ln 2 4 5
13 13 26
= x − x + − − x − − x − x + + C
which is the result of Problem 32.
3 2
t t t dt t t t t t C
t t t
(sin – 8sin –1) cos sin – 50 ln sin 3 – 68 tan –1 (sin – 2) – 41 ln sin 2
– 4sin 5
(sin 3)(sin – 4sin 5) 13 13 26
∫
+ 2
+ = + + +
34. x = sin t, dx = cos t dt
t dt dx x x x C
t x
cos 1 1 ln 2 1 ln 2 1 tan
sin 16 16 32 32 16 2
= = − − + − − 1
⎛ ⎞ ⎜ ⎟
+ ∫ 4 ∫
− 4
− ⎝ ⎠ which is the result of Problem 30.
t dt t t t C
t
cos 1 ln sin – 2 – 1 ln sin 2 – 1 tan sin
sin –16 32 32 16 2
= + –1
⎛ ⎞ ⎜ ⎟
+ ∫
4
⎝ ⎠ 35.
3
2 2 2 2 2
x – 4
x + +
= Ax B +
Cx D
x x x
( + 1) + 1 ( +
1)
A = 1, B = 0, C = –5, D = 0
3
2 2 2 2 2
x – 4 x dx x dx 5
x dx
x x x
1 ln 1 5
2 2( 1)
∫ = ∫ −
∫ 2
+ + + ( 1) 1 ( 1)
x C
= + + +
2
x
+
36. x = cos t, dx = –sin t dt
2 2
2 4 2 4
t t dt x dx
(sin )(4cos –1) – 4 –1
∫ =
∫
+ + + + t t t x x x
(cos )(1 2cos cos ) (1 2 )
2 2
2 4 2 2 2 2 2
x x A BxC DxE
4 − 1 4 − 1
+ +
= = + +
(1 2 ) ( 1) 1 ( 1)
x x x x x x x x
+ + + + +
A = –1, B = 1, C = 0, D = 5, E = 0
⎡ ⎤
− ⎢− + + ⎥ = − + + +
⎢⎣ + + ⎥⎦ +
x x dx x x C
1 5 ln 1 ln 1 5
ln cos 1 ln cos 1 5
∫ 2
2
2 2 2 2
x x x x
1 ( 1) 2 2( 1)
t t C
= − + + +
2
2 2(cos t
+
1)
37.
3 2 2
5 3 4 2
x x x x x x
x x x xx x
2 + 5 + 16 (2 + 5 +
16)
=
8 16 ( 8 16)
+ + + +
2
2 2 2 2 2
x x Ax B Cx D
x x x
2 + 5 + 16
+ +
( 4) 4 ( 4)
= = +
+ + +
A = 0, B = 2, C = 5, D = 8
3 2
5 3 2 2 2
x x x dx dx x dx
x x x x x
2 + 5 + 16 2 5 +
8
dx x dx dx
2 5 8
4 ( 4) ( 4)
∫ = ∫ +
∫ + + + + 2 2 2 2 2
8 16 4 ( 4)
+ + + ∫ ∫ ∫
= + +
x x x
8 ,
x + ∫ let x = 2 tan θ, dx = 2sec2θ dθ .
To integrate 2 2
( 4)
dx
2
8 16sec
θ
∫ dx =
∫ d
cos2 1 1 cos 2
2 + 2 4
x
( 4) 16sec
θ
θ
= θ dθ = ⎛⎜ + θ ⎞⎟ dθ
∫ ∫
⎝ 2 2
⎠ 1 1 sin 2 1 1 sin cos
2 4 2 2
x x C
1 tan
2 2 4
= θ + θ +C = θ + θ θ +C –1
= + +
2
x
+
3 2
x x x dx x x x C
x x x x x
2 5 16 tan – 5 1 tan
x x C
3 tan 2 – 5
2 2 2( 4)
∫ –1 –1
–1
+ + + + + +
= + + +
5 3 2 2
8 16 2 2( 4) 2 2 4
= + +
2
x
+
452 Section 7.5 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.
42. x x A B
x x x x x x
38. 2
–17 = –17
= +
–12 ( 4)( – 3) 4 – 3
+ + +
A = 3, B = –2
6 6
4 2 4
x dx dx
x x x x
–17 3 – 2
–12 4 – 3
= ⎛ ⎞ ⎜ + ⎟ + ⎝ ⎠ ∫ ∫ 6
4 = ⎡⎣3ln x + 4 – 2ln x – 3 ⎤⎦ = (3ln10 – 2ln 3) – (3ln8 – 2ln1)
= 3ln10 – 2ln 3 – 3ln8 ≈ –1.53
39. u = sin θ, du = cos θ dθ
cos θ
1
∫ / 4 d ∫ 1/ 2
du
1/ 2
0 2 2 + 2 0 2 2 + 2
u u
θ
=
(1– sin θ )(sin θ
1) (1– )( 1)
π
1
+ + ∫
0 2 2
u u u
(1 – )(1 )( 1)
du
=
1
A B Cu + D Eu +
F
= + + +
2 2 2 2 2 2
u u u u u u
(1– )( + 1) 1– 1 + + 1 ( +
1)
1 , 1 , 0, 1 , 0, 1
8 8 4 2
A = B = C = D = E = F =
1/ 2 1 1 1/ 2 1 1 1/ 2 1 1 1/ 2 1 1 1/ 2
1
0 2 2 2 0 0 0 2 0 2 2
− + − + + + ∫ ∫ ∫ ∫ ∫
du = du + du + du +
du
u u u u u u
(1 )( 1) 8 1 8 1 4 1 2 ( 1)
1/ 2
⎡ ⎛ ⎞⎤
= ⎢ – 1 ln 1– u + 1 ln 1 + u + 1 tan –1 u + 1 ⎜ tan
–1
u + u
⎣ 8 8 4 4 ⎝ 2
⎟⎥
+ 1
⎠⎦
0
u
1/ 2
⎡ + = 1 ln 1 u 1 ⎢ + tan
–1
u + u
⎤
⎥
⎢⎣ 8 1 − u 2 4( u
2
+ 1)
⎥⎦
0
1 2 +
ln 1 1 tan–1 1 1 0.65
8 2 1 2 2 6 2
= + + ≈
−
1 ,
u + ∫ let u = tan t.)
(To integrate 2 2
( 1)
du
x x
x x x x
3 13 3 13
4 3 ( 3)( 1)
40. 2
+ +
=
A B
x x
= +
+ + + + + 3 +
1
A = –2, B = 5
5 5
1 2 1
x dx x x
x x
3 +
13 –2ln 3 5ln 1
4 3
∫ = ⎡⎣ + + + ⎤⎦
= –2 ln 8 + 5 ln 6 + 2 ln 4 – 5 ln 2 = 5ln 3− 2ln 2 ≈ 4.11
+ + 41. dy y(1 y)
dt
= − so that
∫ dy ∫
dt t C
− a. Using partial fractions:
1
1 1
(1 )
y y
= = +
A B A y By
1 (1 − )
+
(1 ) 1 (1 )
= + = ⇒
y y y y y y
− − −
( ) 1 0 1, 0 1, 1 1 1 1
+ − = + ⇒ = − = ⇒ = = ⇒ = +
y (1 y ) y 1
y
A B Ay y A B A A B
− −
⎛ ⎞
⎛ ⎞
y y y
+ = ⎜ + ⎟ = ⎝ − ⎠ ∫ ln ln(1 ) ln
t C dy
Thus: 1
1 1
y 1
y
− − = ⎜ ⎟ ⎝ 1
− y
⎠
so that
y e Ce
y
t C t
= 1
=
+
1 ( C1 )
C =
e
−
y t e
or 1 ( )
C
t
t
e
=
+
(0) 0.5, 0.5 1 or 1
y = = C =
Since 1
1
C
+
y t e
; thus ( )
=
+
1
t
t
e
b.
3
3 (3) 0.953
y e
= ≈
1
e
+
Instructor’s Resource Manual Section 7.5 453
© 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.