Techniques of Integration Chapter

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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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
+
∫
Instructor’s Resource Manual Section 7.1 413

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
∫ex sec exdx = ∫secu du
= ln secu + tan u + C
= ln sec ex + tan ex +C
∫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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

∫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

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

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

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.

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)

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

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

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

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

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
−
−

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

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

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

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

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
= 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
⎝ ⎠

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
– 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

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.

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

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

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

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
−

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
⎝ ⎠

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

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
+

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
+

Techniques of Integration Chapter

Techniques of Integration Chapter

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