C H A P T E R 7
Integration Techniques, L’Hôpital’s Rule,
and Improper Integrals
Section 7.1
Basic Integration Rules
Section 7.2
Integration by Parts
Section 7.3
Trigonometric Integrals . . . . . . . . . . . . . . . . . . . 65
Section 7.4
Trigonometric Substitution
Section 7.5
Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . 84
Section 7.6
Integration by Tables and Other Integration Techniques
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule . . . . . . . . . 96
Section 7.8
Improper Integrals
Review Exercises
. . . . . . . . . . . . . . . . . . . 50
. . . . . . . . . . . . . . . . . . . . . 55
. . . . . . . . . . . . . . . . . 74
. . 90
. . . . . . . . . . . . . . . . . . . . . 102
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
C H A P T E R 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.1
Basic Integration Rules
Solutions to Odd-Numbered Exercises
1. (a)
(b)
d
x2 1 C 21x2 1122x 2x
dx
x 1
(c)
1 1 2
x
d 1 2
x 1 C x 1122x dx 2
2 2
2x2 1
(d)
2x
d
lnx2 1 C 2
dx
x 1
x
x2 1
3. (a)
dx matches (b).
d
2x
x2 122 2x2x2 12x 21 3x2
2
2
2 C dx x 1
x2 14
x 13
(c)
1
d
arctan x C dx
1 x2
(d)
2x
d
lnx2 1 C 2
dx
x 1
1
dx matches (c).
x2 1
3x 24 dx
u 3x 2, du 3 dx, n 4
Use
11.
t sin t 2 dt
Use
7.
sin u du.
1
x 1 2x
dx
9.
u 1 2x, du un du.
u t 2, du 2t dt
50
x
d
2
lnx2 1 C 21 x2 2x
dx
1
x 1
(b)
5.
d
2x2 1 C 2 12 x2 1122x 2x
dx
x2 1
Use
1
x
dx
3
1 t2
u t, du dt, a 1
Use
du
.
u
13.
cos xesin x dx
u sin x, du cos x dx
Use
eu du.
dt
du
a 2 u 2
Section 7.1
17. Let u z 4, du dz
15. Let u 2x 5, du 2 dx.
2x 532 dx 1
2x 5322 dx
2
5
z 44
C
dz 5 z 45 dx 5
5
z 4
4
1
2x 552 C
5
19. Let u t3 1, du 3t2 dt.
3 t3 1 dt t2
1
3
Basic Integration Rules
21.
v
5
C
4z 44
1
dv 3v 13
t3 1133t2 dt
1 t3 143
C
3
43
t3 143
C
4
v dv 1
3v 133dv
3
1
1
v2 C
2
63v 12
23. Let u t3 9t 1, du 3t2 9 dt 3t2 3 dt.
25.
t2 3
1 3t 2 3
1
dt dt ln t 3 9t 1 C
t 3 9t 1
3 t 3 9t 1
3
x2
dx x1
29.
x 1 dx 1
dx
x1
1 2
x x ln x 1 C
2
1 2x22 dx 27. Let u 1 ex, du ex dx.
ex
dx ln1 ex C
1 ex
4
4
x
4x 4 4x2 1dx x5 x3 x C 12x 4 20x2 15 C
5
3
15
31. Let u 2 x2, du 4 x dx.
xcos 2 x2 dx 1
cos 2 x24 x dx
4
1
sin 2 x 2 C
4
33. Let u x, du dx.
cscx cot x dx 1
csc x cot x dx 1 cscx C
35. Let u 5x, du 5 dx.
e5x dx 1 5x
1
e 5 dx e5x C
5
5
37. Let u 1 e x, du e x dx.
2
dx 2
ex 1
2
1
ex 1
ee dx
x
x
ex
dx 2 ln1 e x C
1 ex
51
52
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
39.
ln x2
1
ln x
2
dx 2 ln x dx 2
C ln x C
x
x
2
41.
1 sin x
dx cos x
2
sec x tan x dx lnsec x tan x lnsec x C lnsec xsec x tan x C
1
1
cos 1 cos 1
43.
csc 1
d cos 1
cos 1
cos 1
cos 1 cos2 1 cos 1
sin2 cot csc2 csc cot csc2 d
csc cot C
45.
1
cos C
sin sin 1 cos C
sin 3z 2
3
2z
dz
dz dz 2 2
z2 9
2 z2 9
z 9
47. Let u 2t 1, du 2 dt.
3
2
z
lnz2 9 arctan
C
2
3
3
1
1 2t 12
dt 1
2
dt
2 1 2t 12
1
arcsin2t 1 C
2
49. Let u cos
2t , du 2 sint 2t dt.
2
1
tan2t
1
2 sin2t
dt dt
t2
2 cos2t
t2
1
2
ln cos
2
t
51.
6x x2
53.
4
dx 4x2 4x 65
55.
ds
t
1
, 0, dt
2
1 t 4
3
dx 3
C
1
9 x 32
dx 3 arcsin
(b) u t 2, du 2t dt
1
t
−1
x 3 3 C
1
1
1
x 12
2x 1
dx arctan
C arctan
C
x 122 16
4
4
4
8
s
(a)
1
0, t
1 t 4
dt 1
1
2t
dt arcsin t 2 C
2
2 1 t22
0.8
1
1 1
1
: arcsin 0 C ⇒ C 2
2 2
2
−1.2
−1
1.2
1
1
s arcsin t 2 2
2
− 0.8
Section 7.1
57.
59. y 10
Basic Integration Rules
1 ex2 dx 53
e2x 2ex 1 dx
1
e2x 2ex x C
2
−10
10
−2
y 3e0.2x
61.
dy
sec2 x
dx 4 tan2 x
63. Let u 2x, du 2 dx.
4
Let u tan x, du sec2 x dx.
y
0
sec2 x
1
tan x
dx arctan
C
4 tan2 x
2
2
1
2
x
xe
dx 0
1
2
1
0
1 2
2x dx ex
2
2
x
e
4
1
23
0
1
1
dx 4 9x2
3
23
0
4
cos 2x2 dx
0
4
2 sin 2x
0
0
2x
dx x2 9
1
1 e1 0.316
2
69. Let u 3x, du 3 dx.
1
2
1
0
1
2
67. Let u x2 9, du 2x dx.
65. Let u x2, du 2x dx.
cos 2x dx 3
4 3x2
x2 9122x dx
4
0
2x2 9
71.
1
3x
arctan
6
2
dx
4
0
4
1
1
x2
dx arctan
C
x2 4x 13
3
3
The antiderivatives are vertical translations of each other.
1
23
0
−7
0.175
18
5
−1
73.
1
2
d tan sec C or
1 sin 1 tan2
The antiderivatives are vertical translations of each other.
75. Power Rule:
un du un1
C, n 1.
n1
u x2 1, n 3
6
−
2
7
2
−6
77. Log Rule:
du
ln u C, u x2 1.
u
79. The are equivalent because
exC1 e x eC1 Ce x, C eC1
54
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
81. sin x cos x a sinx b
sin x cos x a sin x cos b a cos x sin b
sin x cos x a cos b sin x a sin b cos x
Equate coefficients of like terms to obtain the following.
1 a cos b
1 a sin b
and
Thus, a 1cos b. Now, substitute for a in 1 a sin b.
cos1 bsin b
1
1 tan b ⇒ b Since b 2
83.
0
4
1
,a
2. Thus, sin x cos x 2 sin x .
4
cos4
4
dx
sin x cos x
dx
2 sinx 4
1
2
csc x 1
dx ln csc x cot x 4
4
4
2
C
1a
4x
dx 3
x2 1
85. Let u 1 x2, du 2x dx.
0
12x
1
6a2
x1 x2 dx
A4
y
x ax2 dx 0
1
Matches (a).
87.
2
a
x3
3
1
2
3
1 x2122x dx
0
2
4
1 x232
3
1
1
0
Let
4
3
1
2
1
, 12a 2 3, a .
6a 2 3
2
y
( a1 , a1)
2
y
y=x
x
1
2
3
1
1
y = ax 2
1
2
x
x
− 12
1
1
2
− 12
2
−1
89. (a) Shell Method:
Let u x2, du 2x dx.
1
V 2
(b) Shell Method:
y
V 2
dx
1
2
ex
0
1
2
x
2x dx
e
0
xex dx
2
0
2
x
xe
b
1
x
1
2
2
b
0
1
1 eb 2
e
x2
1
0
1 e 1.986
1
eb 2
b
4
3
3 4
3
ln
33 4
0.743
1a
0
Section 7.2
4
91. A 0
x
1
A
x
5
dx 5 arcsin
5
25 x2
4
x
0
4
y
4
3
25 x 4
2
2 12
2x dx
(2.157, y )
1
0
x
1
5 25 x212
5 arcsin45
4
5
5
dx
25 x2
1
5
5 arcsin45
2
5 arcsin
0
Integration by Parts
1
4
2
3
4
0
1
3 5
arcsin45
2
2.157
arcsin45
y tan x
93.
y sec2 x
1 y 2 1 2 sec4 x
14
s
1 2 sec4x dx
0
1.0320
Section 7.2
Integration by Parts
1.
d
sin x x cos x cos x x sin x cos x x sin x. Matches (b)
dx
3.
d 2 x
x e 2xex 2ex x2ex 2xex 2xex 2ex 2ex x2ex. Matches (c)
dx
5.
xe2x dx
7.
ux
ln x2 dx
u ln x 2, dv dx
u x, dv e2x dx
11. dv e2x dx ⇒
v
1
e2x dx e2x
2
⇒ du dx
1
xe2x dx xe2x 2
1
e2x dx
2
1
1
1
xe2x e2x C 2x 2x 1 C
2
4
4e
9.
x sec2 x dx
u x, dv sec2 x dx
55
56
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
13. Use integration by parts three times.
(1) dv ex dx ⇒
v
(2) dv ex dx ⇒
ex dx ex
⇒ du 3x2 dx
u x3
u x2
v
(3) dv ex dx ⇒
ex dx ex
⇒ du 2x dx
v
⇒ du dx
ux
x3ex dx x3ex 3 x2ex dx x3ex 3x2ex 6 xex dx
x3ex 3x2ex 6xex 6ex C exx3 3x2 6x 6 C
15.
x2 ex dx 3
17. dv t dt
1
3
1 3
3
ex 3x2dx ex C
3
⇒
v
t dt t2
2
1
dt
t1
u lnt 1 ⇒ du t lnt 1 dt t2
1
lnt 1 2
2
t2
dt
t1
t2
1
lnt 1 2
2
t1
t2
2
lnt 1 1
dt
t1
t2
1
t lnt 1 C
2 2
1
2t 2 1 ln t 1 t 2 2t C
4
19. Let u ln x, du ln x2
dx x
1
dx.
x
21. dv ln x2
1x dx ln3x
3
1
dx ⇒
2x 12
v
u x2
xe2x
xe2x
dx 2x 12
22x 1
⇒ du 2x dx
x2 1ex dx x2ex dx ux
e2x
dx
2
e2x
xe2x
C
22x 1
4
e2x
C
42x 1
v
ex dx ex
⇒ du dx
exdx x2ex 2 xex dx ex
x2ex 2 xex (2) dv ex dx ⇒
ex dx ex
1
22x 1
e2x2x 1 dx
23. Use integration by parts twice.
v
2x 12 dx
⇒ du 2xe2x e2x dx
u xe2x
(1) dv ex dx ⇒
C
ex dx ex x2ex 2xex ex C x 12ex C
ex dx ex
Section 7.2
25. dv x 1 dx ⇒
2
x 112dx x 132
3
v
27. dv cos x dx ⇒
⇒ du dx
ux
2
2
xx 1 dx xx 132 3
3
x 132 dx
cos x dx sin x
x cos x dx x sin x sin x dx x sin x cos x C
4
2
xx 132 x 152 C
3
15
2x 132
3x 2 C
15
29. Use integration by parts three times.
(1) u x3, du 3x2, dv sin x dx, v cos x
x3 sin dx x3 cos x 3 x2 cos x dx
(2) u x2, du 2x dx, dv cos x dx, v sin x
x3 sin x dx x3 cos x 3 x2 sin x 2 x sin x dx
x3 cos x 3x2 sin x 6 x sin x dx
(3) u x, du dx, dv sin x dx, v cos x
x3 sin x dx x3 cos x 3x2 sin x 6 x cos x cos x dx
x3 cos x 3x2 sin x 6x cos x 6 sin x C
t csc t cot t dt t csc t csc t dt
⇒
33. dv dx
31. u t, du dt, dv csc t cot dt, v csc t
v
u arctan x ⇒ du t csc t ln csc t cot t C
dx x
1
dx
1 x2
arctan x dx x arctan x x arctan x 35. Use integration by parts twice.
(1) dv e2xdx ⇒
v
1
e2x dx e2x
2
1
1
e2x sin x dx e2x sin x 2
2
5
4
(2) dv e2x dx ⇒
1
1 1 2x
1
e2x cos x dx e2x sin x e cos x 2
2 2
2
1
1
e2x sin x dx e2x sin x e2x cos x
2
4
1
e2x sin x dx e2x2 sin x cos x C
5
v
x
dx
1 x2
1
ln1 x2 C
2
1
e2x dx e2x
2
u cos x ⇒ du sin x dx
u sin x ⇒ du cos x dx
57
⇒ du dx
ux
v
Integration by Parts
e2x sin x dx
58
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
37. y xex
2
y
1 2
2
xe x dx ex C
2
39. Use integration by parts twice.
(1) dv 1
2 3t
dt ⇒
v
(2) dv 2 3t dt ⇒
41.
v
2
2 3t12 dt 2 3t32
9
⇒ du dt
ut
2
2 3t12 dt 2 3t
3
⇒ du 2t dt
u t2
y
2t 22 3t 4
t2
dt 3
3
2 3t
t2 3t dt
2
2t 22 3t 4 2t
2 3t32 3
3 9
9
8t
16
2t 22 3t
2 3t32 2 3t52 C
3
27
405
22 3t
27t 2 24t 32 C
405
2 3t32 dt
cos yy 2x
cos y dy 2x dx
sin y x 2 C
43. (a)
dy
xy cos x, 0, 4
dx
(b)
y
8
6
dy
y
y12 dy 2
2
2
x cos x dx
4
u x, du dx, dv cos x dx, v sin x
x cos x dx
2y12 x sin x x
4
sin x dx
6
x sin x cos x C
0, 4: 2412 0 1 C ⇒ C 3
2y x sin x cos x 3
45.
dy x x8
e , y0 2
dx y
10
−10
10
−2
−6
6
−2
Section 7.2
Integration by Parts
47. u x, du dx, dv ex2 dx, v 2ex2
xex2 dx 2xex2 4
Thus,
2ex2 dx 2xex2 4ex2 C
xex2 dx 2xex2 4ex2
0
8e2 4e2 4
4
0
12e2 4 2.376.
49. See Exercise 27.
2
0
1
51. u arccos x, du 1 x2
0
1
2
dx, dv dx, v x
arccos x dx x arccos x x
1 x2
dx x arccos x 1 x2 C
12
Thus,
2
x cos x dx x sin x cos x
arccos x x arccos x 1 x2
0
12
0
34 1
1
1
arccos
2
2
3
1 0.658.
6
2
53. Use integration by parts twice.
(1) dv e x dx ⇒
v
(2) dv ex dx ⇒
e x dx e x
u sin x ⇒ du cos x dx
e x sin x dx e x sin x e x cos x dx e x sin x e x cos x v
ex dx ex
u cos x ⇒ du sin x dx
e x sin x dx
2 e x sin x dx e xsin x cos x
e x sin x dx ex
sin x cos x C
2
1
Thus,
e x sin x dx 0
55. dv x2 dx, v x2 ln x dx x
1
0
e
1 esin 1 cos 1 1
sin 1 cos 1 0.909.
2
2
2
1
x3
, u ln x, du dx
3
x
x3
ln x 3
x3 1
dx
3 x
1 2
x3
ln x x dx
3
3
2
Hence,
e2 sin x cos x
x2 ln x dx x3 ln x 9x 3
1
2
3
1
1
8 1 8
7
8
ln 2 ln 2 1.071.
3
9 9 3
2
59
60
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
57. dv x dx, v x2
1
dx
, u arcsec x, du 2
x x2 1
x2
x arcsec x dx arcsec x 2
Hence,
4
x22
dx
xx2 1
2 arcsec x 2x
x2
x arcsec x dx 2
1
x2
2x
1
dx
arcsec x 2
4 x2 1
8 arcsec 4 x2
1
arcsec x x2 1 C
2
2
8 arcsec 4 2
15
2
15
2
1
4
2
23 23 3
2
2
3
7.380.
59.
x2e2x dx x2
12 e 2x
14 e 2
18 e C
2x
2x
2x
u and its
derivatives
x2
e2x
2x
1 2x
2e
2
1 2x
4e
0
1 2x
8e
1
1
1
x2e2x xe2x e2x C
2
2
4
1
e2x2x2 2x 1 C
4
61.
63.
u and its
derivatives
x3 cos x 3x2 sin x 6x cos x 6 sin x C
x3
sin x
3x2 6 sin x x3 6x cos x C
3x2
cos x
6x
sin x
6
cos x
0
sin x
x sec2 x dx x tan x ln cos x C
71. Yes. Let u x and du v and its
antiderivatives
Alternate
signs
u and its
derivatives
x
sec2 x
1
tan x
0
ln cos x
67. No. Substitution.
1
, dx.
x 1
73.
t 3e4t dt Substitution also works. Let u x 1
2
0
v and its
antiderivatives
Alternate
signs
x3 sin x dx x3cos x 3x2sin x 6x cos x 6 sin x C
65. Integration by parts is based on the
product rule.
75.
v and its
antiderivatives
Alternate
signs
e2x sin 3x dx e2x2 sin 3x 3 cos 3x
13
2
0
1
2e 3 0.2374
13
69. Yes. u x2, dv e2x dx
e4t
32t3 24t2 12t 3 C
128
Section 7.2
77. (a) dv 2x 3 dx ⇒
v
Integration by Parts
1
2x 312 dx 2x 332
3
⇒ du 2 dx
u 2x
2
2
2x2x 3 dx x2x 332 3
3
2x 332 dx
2
2
x2x 332 2x 352 C
3
15
2
2
2x 3323x 3 C 2x 332x 1 C
15
5
1
u3
and dx du
2
2
(b) u 2x 3 ⇒ x 2x2x 3 dx 2
1
u 3 12 1
u
du 2
2
2
u32 3u12 du 1 2 52
u 2u32 C
2 5
1
1
2
u32u 5 C 2x 3322x 3 5 C 2x 332x 1 C
5
5
5
x
79. (a) dv 4 x2
dx ⇒ v 4 x212x dx 4 x2
u x2 ⇒ du 2x dx
x3
dx x24 x2 2 x4 x2 dx
4 x2
2
1
x24 x2 4 x232 C 4 x2 x2 8 C
3
3
(b) u 4 x2 ⇒ x2 u 4 and 2x dx du ⇒ x dx x3
dx 4 x2
1
2
x2
x dx 4 x2
1
du
2
u41
du
u 2
u12 4u12 du 1 2 32
u 8u12 C
2 3
1
1
1
u12u 12 C 4 x2 4 x2 12 C 4 x2 x2 8 C
3
3
3
81. n 0:
n 1:
n 2:
n 3:
n 4:
ln x dx xln x 1 C
x ln x dx x2
2 ln x 1 C
4
x2 ln x dx x3
3 ln x 1 C
9
x3 ln x dx x4
4 ln x 1 C
16
x 4 ln x dx x5
5 ln x 1 C
25
In general,
xn ln x dx xn1
n 1ln x 1 C. (See Exercise 85.)
n 12
61
62
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
83. dv sin x dx ⇒
⇒ du nx n1 dx
u xn
x n sin
85. dv x n dx ⇒
v cos x
x dx x n
cos x n
u ln x
x n1
cos x dx
v
1
dx
x
⇒ du x n ln x dx x n1
n1
x n1
ln x n1
xn
dx
n1
x n1
x n1
C
ln x n1
n 12
x n1
n 1ln x 1 C
n 12
87. Use integration by parts twice.
1
v eax
a
(1) dv eax dx ⇒
u cos bx ⇒ du b sin bx dx
u sin bx ⇒ du b cos bx dx
eax cos bx dx
eax sin bx b eax cos bx b
a
a
a
a
Therefore, 1 eax sin bx b
a
a
eax sin bx dx b2
a2
e
ax
sin bx dx eax sin bx dx eax sin bx dx x3 ln x dx eax sin bx b2
2
a
a
eax sin bx dx
eaxa sin bx b cos bx
a2
eaxa sin bx b cos bx
C.
a2 b2
91. a 2, b 3 (Use formula in Exercise 88.)
89. n 3 (Use formula in Exercise 85.)
1
v eax
a
(2) dv eax dx ⇒
x4
4 ln x 1 C
16
e2x cos 3x dx e2x2 cos 3x 3 sin 3x
C
13
1
93. dv ex dx ⇒
4
A
xex dx xex
0
1
4
0
4
0
ex dx 4
ex
e4
4
0
5
0.908
e4
exsin x cos x
1 2
1
1
1
1 2 e
1 2 e
0.395 See Exercise 87.
3
−1
ex sin x dx
0
⇒ du dx
ux
95. A v ex
1
7
−1
0
1.5
0
1
0
Section 7.2
e
97. (a) A e
ln x dx x x ln x
1
1
1 See Exercise 4.
y
(b) Rx ln x, rx 0
Integration by Parts
2
(e, 1)
e
V
ln x2 dx
1
1
e
xln x2 2x ln x 2x
1
x
Use integration by parts twice, see Exercise 7.
1
2
3
e 2 2.257
e
(c) px x, hx ln x
(d)
e
y
e 2 1
13.177 See Exercise 85.
2
x, y x ln x dx 2
1
e
V 2
99. Average value 1
x
x2
1 2 ln x
4
1
4 sin 2t 2 cos 2t
1 4t 4 cos 2t 2 sin 2t
e
5e4t
20
20
7
1 e4 0.223
10
10
100,000 4000te0.05t dt 4000
0
25 te0.05t dt
0
Let u 25 t, dv e0.05t dt, du dt, v 100 0.05t
e
5
e
dt
100
e
e
4000 25 t
10,000
$931,265
5
25
25 t 100 0.05t
e
5
0.05t
10
0
10
cos n cos n
n
n
2
cos n
n
22nn, , ifif nn isis even
odd
10
0.05t
0
10
0
x
1
x sin nx dx cos nx 2 sin nx
n
n
100
5
0.05t
0
103.
ln x2dx e 2
0.359
1
2
e
1 e2
,
2.097, 0.359
4
2
2
e4t cos 2t 5 sin 2t dt
10
P 4000
1 e
2 1
0
101. ct 100,000 4000t, r 5%, t1 10
P
1 x ln x dx e2 1
2.097
1
4
0
From Exercises 87 and 88
63
64
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
105. Let u x, dv sin
1
I1 x sin
0
n2 xdx, du dx, v n2 cos
n2 x.
n
2x
n
x dx cos
x
2
n
2
2
x 2 sin
1
2
n
0
2
n
cos
n
2
n
2
2
cos
n
2
n
Let u x 2, dv sin
I2 1
1
cos
0
n2 x dx
n2 sin
n2 x
1
2
0
sin
n2
2
n2 xdx, du dx, v n2 cos
n2 x.
n
2x 2
n
x dx cos
x
2
n
2
2
n
cos
n
2
2
n
2
cos
n
2
n
2
1
2
n
2
1
n2 sin
n2 x
cos
n2 xdx
2
2
1
sin
n2
2
n2 sin
n2 n2 sin
n2 n sin
n2
2
hI1 I2 bn h
8h
2
2
107. Shell Method:
b
V 2
x f x dx
y
a
y = f ( x)
f (a )
dv x dx ⇒
x2
v
2
f (b )
u f x ⇒ du fx dx
V 2
x2 f x x2 fxdx
2
b
2
x
a
a
b
b2f b a2f a b
x2 fx dx
a
Disk Method:
V
f a
b2 a2 dy 0
f b
f a
b2 f 1 y 2 dy
b2 a2 f a b2 f b f a b2f b a2f a f b
f a
f b
f a
f 1y2 dy
f 1 y2 dy
Since x f 1 y, we have f x y and fxdx dy. When y f a, x a. When y f b, x b. Thus,
f b
f a
f 1 y 2 dy b
x 2fx dx
a
and the volumes are the same.
Section 7.3
Trigonometric Integrals
109. fx xex
(a) f x xex dx xex ex C
(b)
1
Parts: u x, dv ex dx
f 0 0 1 C ⇒ C 1
0
f x xex ex 1
(c) You obtain the points
n
xn
4
0
(d) You obtain the points
yn
n
xn
yn
0
0
0
0
0
0
1
0.05
0
1
0.1
0
2
0.10
2.378 103
2
0.2
0.0090484
3
0.15
0.0069
3
0.3
0.025423
4
0.20
0.0134
4
0.4
0.047648
80
0.9064
40
0.9039
4.0
1
4.0
1
0
4
0
0
4
0
(e) f 4 0.9084
The approximations are tangent line approximations. The results in (c) are better because x is smaller.
Section 7.3
Trigonometric Integrals
1. f x sin4 x cos4 x
(a) sin4 x cos4 x 2x
2x
1 cos
1 cos
2
2
2
2
1
1 2 cos 2x cos2 2x 1 2 cos 2x cos2 2x
4
1
1 cos 4x
22
4
2
1
3 cos 4x
4
(b) sin4 x cos4 x sin2 x2 cos4 x
1 cos2 x2 cos4 x
1 2 cos 2 x 2 cos4 x
(c) sin4 x cos4 x sin4 x 2 sin2 x cos2 x cos4 x 2 sin2 x cos2 x
sin2 x cos2 x2 2 sin2 x cos2 x
1 2 sin2 x cos2 x
—CONTINUED—
65
66
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1. —CONTINUED—
(d) 1 2 sin2 x cos2 x 1 2 sin x cos xsin x cos x
1 sin 2x
1
12 sin 2x
1 2
sin 2x
2
(e) Four ways. There is often more than one way to rewrite a trigonometric expression.
3. Let u cos x, du sin x dx.
5. Let u sin 2x, du 2 cos 2x dx.
cos3 x sin x dx cos3 xsin x dx
sin5 2x cos 2x dx 1
cos4 x C
4
1
2
sin5 2x2 cos 2xdx
1
sin6 2x C
12
7. Let u cos x, du sin x dx.
sin x1 cos2 x2 cos2 x dx
sin5 x cos2 x dx cos2 x 2 cos4 x cos6 xsin x dx 9.
cos3 sin d cos 1 sin2 sin 12 d
sin 12 sin 52cos d
2
2
sin 32 sin 72 C
3
7
13.
sin2 cos2 d 1 cos 2
2
1 cos 2
d
2
1
1 cos2 2 d
4
1
4
1
1 cos 4
d
2
1
1 cos 4 d
8
1
1
sin 4 C
8
4
1
4 sin 4 C
32
11.
1
2
1
cos3 x cos5 x cos7 x C
3
5
7
cos2 3x dx 1 cos 6x
dx
2
1
1
x sin 6x C
2
6
1
6x sin 6x C
12
Section 7.3
Trigonometric Integrals
15. Integration by parts.
1 cos 2x
x
sin 2x 1
⇒ v 2x sin 2x
2
2
4
4
dv sin2 x dx u x ⇒ du dx
x sin2 x dx 1
1
x2x sin 2x 4
4
2x sin 2x dx
1
1
1
1
x2x sin 2x x2 cos 2x C 2x2 2x sin 2x cos 2x C
4
4
2
8
17. Let u sin x, du cos x dx.
2
2
cos3 x dx 0
1 sin2 x cos x dx
0
sin x 2
1 3
sin x
3
0
2
3
19. Let u sin x, du cos x dx.
2
2
cos7 x dx 0
1 sin2 x3 cos x dx 0
2
1 3 sin2 x 3 sin4 x sin6 x cos x dx
0
sin x sin3 x 21.
sec3x dx 1
ln sec 3x tan 3x C
3
25. dv sec2 x dx ⇒
sec3 x dx sec3x dx 27.
tan5
x
dx 4
sec4 5x dx 0
16
35
1 tan2 5x sec2 5x dx
tan3 5x
1
tan 5x C
5
3
tan 5x
3 tan2 5x C
15
1
tan x
1
sec x tan x 2 sec3 x dx ⇒ du sec x tan x dx
u sec x
v
23.
2
3 5
1
sin x sin7 x
5
7
sec x tan2 x dx 1
sec x tan x sec xsec2 x 1 dx
1
sec x tan x lnsec x tan x C1
1
sec x tanx ln sec x tan x C
2
sec2
tan3
x
x
1 tan3 dx
4
4
x
x
sec2 dx 4
4
29. u tan x, du sec2 x dx
tan3
x
dx
4
tan4
x
4
tan4
x
x
x
2 tan2 4 ln cos C
4
4
4
sec2
x
x
1 tan dx
4
4
sec2 x tan x dx 1 2
tan x C
2
67
68
31.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
tan2 x sec2 x dx tan3 x
C
3
33.
sec6 4x tan 4x dx 35. Let u sec x, du sec x tan x dx.
sec3 x tan x dx 39. r 37.
sec2 xsec x tan x dx
1
4
sec5 4x4 sec 4x tan 4x dx
sec6 4x
C
24
sec2 x 1
dx
sec x
sec x cos x dx
ln sec x tan x sin x C
1 cos22 d
41. y 1
4
1
4
1
1
1
sin2 sin4 C
4
2 8
1
12 8 sin2 sin4 C
32
1 2 cos2 cos22 d
1 2 cos2 1
sec3 x C
3
sin4 d 1 cos4
d
2
43. (a)
tan2 x
dx sec x
1
4
(b)
y
4
dy
sin2 x, 0, 0
dx
y
sin2 x dx tan3 3x sec 3x dx
sec2 3x 1sec 3x tan 3x dx
1
1
sec2 3x3 sec 3x tan 3x dx 3 sec 3x tan 3x dx
3
3
1
1
sec3 3x sec 3x C
9
3
4
1 cos 2x
dx
2
−6
6
x
sin 2x
1
C
x
2
4
4
−4
45.
−4
1
sin 2x
0, 0: 0 C, y x 2
4
dy 3 sin x
, y0 2
dx
y
47.
sin 3x cos 2x dx 8
−9
9
1
2
sin 5x sin x dx
1 1
cos 5x cos x C
2 5
1
cos 5x 5 cos x C
10
−4
49.
sin sin 3 d 1
2
cos 2 cos 4 d
1
1 1
sin 2 sin 4 C
2 2
4
1
2 sin 2 sin 4 C
8
51.
cot3 2x dx csc2 2x 1 cot 2x dx
1 2 cos 2x
1
cot 2x2csc2 2x dx dx
2
2 sin 2x
1
1
cot2 2x ln sin 2x C
4
2
1
ln csc2 2x cot2 2x C
4
Section 7.3
53. Let u cot , du csc2 d.
csc4 d 1
dx sec x tan x
cot2 t
dt csc t
csc2 1 cot2 d
csc2 d cot 57.
55.
Trigonometric Integrals
csc2 t 1
dt
csc t
csc t sin tdt
ln csc t cot t cos t C
csc2 cot2 d
1 3
cot C
3
cos2 x
dx sin x
1 sin2 x
dx
sin x
csc x sin x dx
ln csc x cot x cos x C
59.
tan4 t sec4 t dt tan2 t sec2 ttan2 t sec2 t dt
tan2 t sec2 t 1
tan2 t sec2 t dt 2 sec2 t 1 dt 2 tan t t C
61.
sin2 x dx 2
0
x
1
sin 2x
2
0
4
1 cos 2x
dx
2
63.
tan3 x dx 0
4
sec2 x 1 tan x dx
0
4
sec2 x tan x dx 0
4
0
sin x
dx
cos x
4
0
1 2
tan x ln cos x
2
1
1 ln 2
2
65. Let u 1 sin t, du cos t dt.
2
0
67. Let u sin x, du cos x dx.
2
0
cos t
dt ln 1 sin t
1 sin t
2
ln 2
2
2
cos3 x dx 2
1 sin2 x cos x dx
0
2 sin x 69.
2
1 3
sin x
3
0
1
x
cos4 dx 6x 8 sin x sin 2x C
2
16
6
−9
9
−6
71.
sec5 x dx 3
1
3
sec3 x tan x sec x tan x ln sec x tan x C
4
2
−3
3
−3
4
3
69
70
73.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sec5 x tan x dx 4
1
sec5 x C
5
75.
sin 2 sin 3 d 0
1
1
sin sin 5
2
5
4
0
32
10
5
−2
2
−5
2
77.
sin4 x dx 0
2
1 3x
1
sin 2x sin 4x
4 2
8
79. (a) Save one sine factor and convert the remaining sine
factors to cosine. Then expand and integrate.
0
3
16
(b) Save one cosine factor and convert the remaining
cosine factors to sine. Then expand and integrate.
(c) Make repeated use of the power reducing formula to
convert the integrand to odd powers of the cosine.
81. (a) Let u tan 3x, du 3 sec2 3x dx.
(b)
sec4 3x tan3 3x dx 1
3
1
3
tan6 3x tan4 3x
C1
18
12
sec2 3x tan3 3x sec2 3x dx
0.05
− 0.5
tan2 3x 1 tan3 3x3 sec2 3x dx
0.5
− 0.05
tan5 3x tan3 3x3 sec2 3x dx
Or let u sec 3x, du 3 sec 3x tan 3x dx.
(c)
sec4 3x tan3 3x dx 1
3
sec6 3x sec4 3x
C
18
12
sec3 3x tan2 3x sec 3x tan 3x dx
sec3 3xsec2 3x 13 sec 3x tan 3x dx
sec6 3x sec4 3x
1 tan2 3x3 1 tan2 3x2
C
C
18
12
18
12
1
1
1
1
1
1
1
tan6 3x tan4 3x tan2 3x tan4 3x tan2 3x C
18
6
6
18 12
6
12
tan6 3x tan4 3x
1
1
C
18
12
18 12
tan6 3x tan4 3x
C2
18
12
1
83. A sin2 x dx
y
0
1
0
1 cos2 x
dx
2
x
1
sin2 x
2 4
1
2
1
1
1
2
0
x
1
2
1
Section 7.3
85. (a) V 2
sin2 x dx 0
(b) A 0
sin x dx cos x
0
1
x sin 2x
2
2
1 cos 2x dx 0
0
Trigonometric Integrals
2
2
112
Let u x, dv sin x dx, du dx, v cos x.
x
1
A
x sin x dx 0
y
0
0
sin2 x dx
0
2
y
1
1 cos 2x dx
0
1
2
( π2 , π8 (
8
0
π
2
2 , 8 87. dv sin x dx ⇒
1
x cos x sin x
2
cos x dx 0
1
1
x sin 2x
8
2
x, y x cos x
1
2A
1
8
1
2
x
π
v cos x
u sinn1 x ⇒ du n 1sinn2 x cos x dx
sinn x dx sinn1 x cos x n 1 sinn2 x cos2 x dx
sinn1 x cos x n 1 sinn2 x1 sin2 x dx
sinn1 x cos x n 1 sinn2 x dx n 1 sinn x dx
Therefore, n sinn x dx sinn1 x cos x n 1 sinn2 x dx
sinn x dx sinn1 x cos x n 1
n
n
sinn2 x dx.
89. Let u sinn1 x, du n 1sinn2 x cos x dx, dv cosm x sin x dx, v mn
m1
cosm1 x
.
m1
cosm x sinn x dx sinn1 x cosm1 x
n1
m1
m1
sinn2 x cosm2 x dx
n1
sinn1 x cosm1 x
m1
m1
sinn2 x cosm x1 sin2 x dx
n1
sinn1 x cosm1 x
m1
m1
sinn2 x cosm x dx cosm x sinn x dx sinn1 x cosm1 x
n1
m1
m1
sinn2 x cosm x dx
cosm x sinn x dx cosm1 x sinn1
n1
mn
mn
cosm x sinn2 x dx
n1
m1
sinn x cosm x dx
71
72
91.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sin5 x dx sin4 x cos x 4
5
5
sin3 x dx
sin2 x cos x 2
sin4 x cos x 4
5
5
3
3
sin x dx
4
8
1
sin2 x cos x cos x C
sin4 x cos x 5
15
15
93.
sec4
cos x
3 sin4 x 4 sin2 x 8 C
15
25 x
dx 25 sec 25 x
25 dx
4
2
2 x
2 x
5 1
sec2
tan
2 3
5
5
3
2 x
2 x
2 x
5
sec2
tan
2 tan
6
5
5
5
2 x
5
tan
6
5
a1 1
b1 6
a0 sec2
25 x
25 dx
C
sec 25 x
2 C
2
t
t
b1 sin where:
6
6
12
1
12
1
6
95. (a) f t a0 a1 cos
a0 f t dt
0
12
f t cos
0
12
f t sin
0
t
dt
6
t
dt
6
12 0
30.9 432.2 241.1 453.7 264.6 474.0 278.2 477.0 271.0 3122
460.1 247.1 435.7 30.9 55.46
a1 12 0
2
30.9 cos 0 4 32.2 cos
2 41.1 cos
4 53.7 cos
2 64.6 cos
6312
6
3
2
3
5
7
4
278.2 cos 4 77.0 cos
2 71.0 cos
6
6
3
3
5
11
2 47.1 cos
4 35.7 cos
30.9 cos 2 23.88
2
3
6
4 74.0 cos
4 60.1 cos
b1 12 0
2
30.9 sin 0 4 32.2 sin
2 41.1 sin
4 53.7 sin
2 64.6 sin
6312
6
3
2
3
5
7
4
278.2 sin 4 77.0 sin
2 71.0 sin
6
6
3
3
5
11
2 47.1 sin
4 35.7 sin
30.9 sin 2 3.34
2
3
6
4 74.0 sin
4 60.1 sin
Ht 55.46 23.88 cos
—CONTINUED—
t
t
3.34 sin
6
6
Section 7.3
Trigonometric Integrals
95. —CONTINUED—
(b) a0 12 0
18.0 417.7 225.8 436.1 245.4 455.2 259.9 459.4 253.1 3122
443.2 234.3 424.2 18.0 39.34
a1 12 0
2
18.0 cos 0 4 17.7 cos
2 25.8 cos
4 36.1 cos
2 45.4 cos
6312
6
3
2
3
5
7
4
259.9 cos 4 59.4 cos
2 53.1 cos
6
6
3
3
5
11
2 34.3 cos
4 24.2 cos
18 cos 2 20.78
2
3
6
4 55.2 cos
4 43.2 cos
b1 12 0
2
18.0 sin 0 4 17.7 sin
2 25.8 sin
4 36.1 sin
2 45.4 sin
6312
6
3
2
3
5
7
4
259.9 sin 4 59.4 sin
2 53.1 sin
6
6
3
3
5
11
2 34.3 sin
4 24.2 sin
18 sin 2 4.33
2
3
6
4 55.2 sin
4 43.2 sin
Lt 39.34 20.78 cos
t
t
4.33 sin
6
6
(c) The difference between the maximum and minimum temperatures is greatest in the summer.
85
H
L
0
12
15
97.
cosmx cosnx dx sinmx sinnx dx sinmx cosnx dx 1 sinm nx sinm nx
2
mn
mn
1
2
0, m n
sinm nx sinm nx dx
1 cosm nx cosm nx
2
mn
mn
1
2
, m n
cosm n cosm n
cosm n cosm n mn
mn
mn
mn
0, since cos cos.
sinmx cosmx dx 0, m n
cosm nx cosm nx dx
1 sinm nx sinm nx
2
mn
mn
1
2
1 sin2mx
m
2
0
73
74
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.4
1.
Trigonometric Substitution
Indefinite integral:
x2 16
x
dx
2
x
x2 16
2
4x
x2 16 x2 16 4
4x2
xx x 16 164 4x 4
3.
x2 16 4
d
d
4 ln
x2 16 C 4 ln x2 16 4 4 ln x x2 16 C
dx
x
dx
4
x
x
x2 16
xx2 16 x2 16 4
4x2 4x2 16 16x2 16 x2x2 16 4x2
xx2 16 x2 16 4
x2 16x2 16 4x2 16
xx2 16 x2 16 4
x2 16
x2 16 x2 16 4
2
2
x
x x 16 x 16 4
Matches (b)
x
x16 x2
14
x1216 x2122x 16 x2
d
C 8
8 arcsin dx
4
2
2
1 x42
16 x2
8
x2
2
2
2
16 x
216 x
16
x2
16 x2
x2
2
2
2
16 x2
216 x
216 x
216 x
Matches (a)
5. Let x 5 sin , dx 5 cos d, 25 x2 5 cos .
1
dx 25 x232
5 cos d
5 cos 3
1
25
5
x
sec d
2
θ
25 − x 2
1
tan C
25
x
2525 x2
7. Same substitution as in Exercise 5
25 x2
x
dx C
25 cos2 d
1 sin2 5
d 5 csc sin d
5 sin sin 4x2 16 x2 16 4 x2 x2 16 4
5
ln csc cot cos C 5 ln
5 25 x2
25 x2 C
x
Section 7.4
Trigonometric Substitution
9. Let x 2 sec , dx 2 sec tan d, x2 4 2 tan .
1
dx x2 4
2 sec tan d
2 tan ln
x
2
x2 4
2
x
x2 − 4
sec d ln sec tan C1
θ
2
C1
ln x x 4 ln 2 C1 ln x x2 4 C
2
11. Same substitution as in Exercise 9
x3x2 4 dx 8 sec3 2 tan 2 sec tan d 32 tan2 sec4 d
32 tan2 1 tan2 sec2 d 32
tan3 tan5 C
3
5
32 x2 432
x2 4
32 3
tan 5 3 tan2 C 53
C
15
15
8
4
1
1 2
x 432
20 3x2 4 C x2 4323x2 8 C
15
15
13. Let x tan , dx sec2 d, 1 x2 sec .
x1 x2 dx 1 + x2
sec3 1
C 1 x232 C
3
3
tan sec sec2 d x
θ
Note: This integral could have been evaluated with the Power Rule.
1
15. Same substitution as in Exercise 13
1
dx 1 x22
1
1 x2
4
dx
1 + x2
x
d
sec4 sec2
θ
cos2 d sin 2
1
2
2
1
sin cos C
2
x
1
arctan x 2
1 x2
1
x
arctan x C
2
1 x2
1
1
1 cos 2 d
2
1
1 x2
C
17. Let u 3x, a 2, and du 3 dx.
4 9x2 dx 1
22 3x2 3 dx
3
1 1
3x4 9x2 4 ln 3x 4 9x2 C
3 2
2
1
x4 9x2 ln 3x 4 9x2 C
2
3
75
76
19.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x
1
dx 9
2
x2
x2 9122x dx
x2
21.
1
16 x2
dx arcsin
4 C
x
9C
(Power Rule)
23. Let x 2 sin , dx 2 cos d, 4 x2 2 cos .
16 4x2 dx 2 4 x2 dx
2
x
θ
2 2 cos 2 cos d
4 − x2
8 cos2 d
4 1 cos 2 d
4 1
sin 2 C
2
4 4 sin cos C
4 arcsin
2x x4 x
2
C
27. Let x sin , dx cos d, 1 x2 cos .
25. Let x 3 sec , dx 3 sec tan d,
x2 9 3 tan .
1
x2 9
dx 3 sec tan d
3 tan 1 x2
x4
dx sec d
cos cos d
sin4 cot2 csc2 d
1
cot3 C
3
ln sec tan C1
x2 9
x
ln C1
3
3
ln x x2 9 C
1
1 x232
C
3x3
x
θ
x
x2 − 9
1 − x2
θ
3
29. Same substitutions as in Exercise 28
1
dx x4x2 9
1
3
32 sec2 d
32 tan 3sec 1
1 4x2 9 3
csc d ln csc cot C ln
C
3
3
2x
Section 7.4
31. Let x 5 tan , dx 5 sec2 d, x2 5 5 sec2 .
5x
dx x2 532
55 tan 5 sec2 d
5 sec2 32
5
x2 + 5
x
tan d
sec θ
5
5 sin d
5 cos C
5
5
x2 5
C
5
C
5
x2
33. Let u 1 e2x, du 2e2x dx.
e2x1 e2x dx 1
2
1
1 e2x122e2x dx 1 e2x32 C
3
35. Let ex sin , ex dx cos d, 1 e2x cos .
ex1 e2x dx cos2 d
1
2
1
1 cos 2 d
sin 2
1
2
2
ex
θ
1 − e 2x
1
1
sin cos C arcsin ex ex1 e2x C
2
2
37. Let x 2 tan , dx 2 sec2 d, x2 2 2 sec2 .
1
dx 4 4x2 x4
1
dx
x2 22
2
2
4
sec2 4 sec4 cos2 x2 + 2
x
d
θ
2
d
21 cos 2 d
2 1
4
2
8
2
8
21 sin 2 C
sin cos C
1
x
1
x
arctan
C
4 x2 2 2
2
Trigonometric Substitution
77
78
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1
39. Since x > ,
2
1
dx, dv dx ⇒ v x
x4x2 1
u arcsec 2x, ⇒ du arcsec 2x dx x arcsec 2x 2x sec , dx arcsec 2x dx x arcsec 2x 1
4x x2
dx
1
sec tan d, 4x2 1 tan 2
x arcsec 2x 41.
1
4x2 1
dx 1
12 sec tan d
x arcsec 2x tan 2
sec d
1
1
ln sec tan C x arcsec 2x ln 2x 4x2 1 C.
2
2
1
4 x 22
dx arcsin
x 2 2 C
43. Let x 2 2 tan , dx 2 sec2 d, x 22 4 2 sec .
x
dx x 4x 8
2
x
dx x 22 4
2 tan 22 sec2 d
2 sec 2 tan 1sec d
2
sec ln sec tan C1
2
x 22 4
2
ln
x 4x 8 2 ln
x 22 4
2
x2
2
C1
x 4x 8 x 2 C
x2 4x 8 2
ln x2 4x 8 x 2 ln 2 C1
2
2
45. Let t sin , dt cos d, 1 t 2 cos2 .
(a)
t2
dt 1 t232
sin2 cos d
cos3 1
tan2 d
t
θ
1 − t2
sec2 1 d
tan C
32
Thus,
0
t
1 t2
arcsin t C
32
t2
t
dt arcsin t
1 t232
1 t 2
0
32
14
(b) When t 0, 0. When t 32, 3. Thus,
32
0
t2
dt tan 1 t232
3
0
3 0.685.
3
arcsin
3
2
3 0.685.
3
Section 7.4
Trigonometric Substitution
47. (a) Let x 3 tan , dx 3 sec2 d, x2 9 3 sec .
x3
x2 9
dx 27 tan3 3 sec2 d
3 sec 27 sec2 1 sec tan d
13 sec
27
3
9
3
Thus,
0
sec C 9
sec3 3 sec C
3
x2 9
3
3
C 31 x 9
x2 9
2
3
13 54
9x2 9 C
3
x3
1
dx x2 932 9x2 9
3
x2 9
32
0
9 27
2 272
18 92 9 2 2 5.272.
(b) When x 0, 0. When x 3, 4. Thus,
3
0
x3
dx 9 sec3 3 sec 9
x2
4
0
9 22 32 91 3 9 2 2 5.272.
49. (a) Let x 3 sec , dx 3 sec tan d, x2 9 3 tan .
x2
x2 9
dx 9 sec2 3 sec tan d
3 tan 9
sec3
x
d
θ
3
12 sec tan 21sec d
9
(7.3 Exercise 90)
9
sec tan ln sec tan 2
9 x
2 3
x2 9
3
Hence,
6
4
ln
x2 9
x
3
3
x2 9
x2
9 xx2 9
x
dx ln x2 9
2
9
3
3
9
2
27
627
ln 2 9
3
6
4
4 9 7 ln43 37
93 27 6 27
4 7
9
ln
ln
2
3
3
93 27 9
6 33
ln
2
4 7
93 27 9
4 7 2 3 12.644.
ln
2
3
—CONTINUED—
x2 − 9
79
80
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
49. —CONTINUED—
(b) When x 4, arcsec
43.
When x 6, arcsec2 6
x2
9
dx sec tan ln sec tan 9
2
x2
4
.
3
9
2
2
x2
53.
x2
9 4 7
4 7
ln 3 3
3
3
3 ln 2 3 2
93 27 51.
3
arcsec43
9
6 33
12.644
ln
2
4 7
1
x2
dx x2 10x 9 x 15 33 ln x 5 x2 10x 9 C
2
10x 9
x2
1
dx xx2 1 ln x x2 1 C
2
1
a
57. A 4
0
55. (a) u a sin b
a2 x2 dx
a
y
a
4b
a
b
y= a
(b) u a tan a2 − x2
b
a2 x2 dx
0
4b 1
a 2
a2 arcsin
x
xa2 x2
a
−a
a
x
a
0
−b
2b 2 a
a
2
ab
Note: See Theorem 7.2 for a2 x2 dx.
59. x2 y2 a2
x ± a2 y2
a
A2
a2 y2 dy a2 arcsin
h
ay ya
2
2 a
a2
2
arcsin
y2
a
(Theorem 7.2)
h
ha ha
2
h2
h
a2
a2 arcsin
ha2 h2
2
a
61. Let x 3 sin , dx cos d, 1 x 32 cos .
y
Shell Method:
2
4
1
V 4
x1 x 32 dx
x
2
4
1
2
−1
3 sin cos 2 d
2
−2
4
32 4
2 2 sin 2 3 cos 2
1 cos 2 d 2
3
1
1
2
cos2 sin d
2
2
3
2
6 2
3
(c) u a sec Section 7.4
Trigonometric Substitution
1
1
x2 1
63. y ln x, y , 1 y 2 1 2 x
x
x2
Let x tan , dx sec2 d, x2 1 sec .
5
s
1
sec sec2 d tan b
x2 1
dx x2
a
5
x2 1
x
1
b
a
x2 + 1
x
dx
θ
sec 1 tan2 d
tan 1
b
csc sec tan d
a
ln csc cot sec b
a
x
26 1
ln
26 ln 2 1 2
5
5 2 1
26 1
26 2 4.367 or ln
ln
26 1 5 2 1
x2 1
ln
1
x2 1
x
1
5
26 2
65. Length of one arch of sine curve: y sin x, y cos x
L1 1 cos 2 x dx
0
Length of one arch of cosine curve: y cos x, y sin x
L2 2
1 sin2 x dx
2
2
1 cos x 2 dx
ux
2
2
, du dx
2
0
1 cos 2 u du
1 cos 2 u du L1
0
67. (a)
(b) y 0 for x 200 (range)
60
−25
250
−10
(c) y x 0.005x 2, y 1 0.01x, 1 y2 1 1 0.01x 2
Let u 1 0.01x, du 0.01 dx, a 1. (See Theorem 7.2.)
200
s
200
1 1 0.01x2 dx 100
0
1 0.01x2 1 0.01 dx
0
50 1 0.01x1 0.01x2 1 ln 1 0.01x 1 0.01x2 1
50
2 ln 1 2 2 ln 1 2 1002 50 ln
2 1
2 1
229.559
200
0
81
82
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
69. Let x 3 tan , dx 3 sec2 d, x2 9 3 sec .
4
A2
0
4
3
dx 6
2
x 9
b
0
sec d 6 ln sec tan 6
a
b
dx
6
x2 9
a
b
a
3 sec2 d
3 sec 6 ln
x 0 (by symmetry)
y
4
1 1
2 A
4
3
9
x2
3
4
6 ln 3
0
y
dx
2
3
4
1
2
4
9
12 ln 3
3
1
x
arctan
4 ln 3 3
3
1
dx
2
4 x 9
2
4
arctan 0.422
4 ln 3
3
x2 9 x
(0, 0.422)
1
4
x
4
−4
−2
2
4
4
1
4
arctan
0, 0.422
2 ln 3
3
x, y 0,
1 y 1 4x 2
y 2x,
71. y x2,
2x tan , dx 1
sec2 d, 1 4x2 sec 2
(For sec5 d and sec3 d, see Exercise 80 in Section 7.3)
2
S 2
a
0
4
b
x21 4x2 d x 2
b
sec3 tan2 d a
4
tan 2
1
sec sec2 d
2
2
sec d sec d
b
b
5
3
a
a
3
1
1
sec3 tan sec tan lnsec tan sec tan lnsec tan 4 4
2
2
1
1
1 4x 2322x 1 4x 2122x ln 1 4x2 2x 4 4
8
51 2 ln 3 2 2 32 102
4
4
8
b
a
2
0
542 62 1
ln 3 22 4
4
6
8
2 ln 3 22
13.989
73. (a) Area of representative rectangle: 21 y2 y
y
Pressure: 262.43 y1 y 2 y
2
1
F 124.8
1
x=
3 y1 y d y
1
124.8 3
1
1
1 y 2 dy x
−2
y1 y 2 dy
1
32 arcsin y y1 y 21 231 y 1
124.8
2
2 32
1
62.43
arcsin 1 arcsin1 187.2 lb
1
(b) F 124.8
1
1 − y2
2
1
d y1 y 2 dy 124.8 d
124.8
1
1
1 y 2 dy 124.8
1
d2arcsin y y1 y y1 y 2 dy
1
2
1
124.80 62.4 d lb
2
Section 7.4
75. (a) m dy y y 144 x2 dx
x0
y
( 0, y +
Trigonometric Substitution
144 − x 2 (
12
144 x2
x
12
144 − x 2
( x, y )
x
y
x
2
(b) y 144 x2
x
4
6
8
10 12
dx
12
Let x 12 sin , dx 12 cos d, 144 x2 12 cos .
y
x
θ
12 cos 1 sin2 12 cos d 12
d
12 sin sin 144 − x 2
12 csc sin d 12 ln csc cot 12 cos C
C
12 ln
144 x2
12 144 x2
12
x
x
12
12 ln
12 144 x2
144 x2 C
x
0
12 144 x2
x
144 x2.
12 144 x2
> 0 for 0 < x ≤ 12
x
(c) Vertical asymptote: x 0
(d) y 144 x2 12 ⇒ y 12 144 x2
Thus,
12 144 x2 12 ln
12 12 1 ln
144 x2
x
144 x2
x
144 x2
xe1 12 144 x2
xe1 122 144 x2 2
x2e2 24xe1 144 144 x2
x2e2 1 24xe1 0
x
xe2 1 24e1 0
x 0 or x 24e1
7.77665.
e2 1
Therefore,
12
s
7.77665
12
7.77665
1 144 x2
x
2
12
dx 7.77665
x
2
144 x2
dx
x2
7.77665 12ln 12 ln 7.77665 5.2 meters.
12
dx 12 ln x
x
12
12
0
When x 12, y 0 ⇒ C 0. Thus, y 12 ln
Note:
30
83
84
Chapter 7
77. True
dx
1 x2
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
cos d
cos 79. False
3
d
0
dx
1 x2 3
3
0
sec2 d
sec3 3
cos d
0
81. Let u a sin , du a cos d, a 2 u 2 a cos .
a2 u2 du a2 cos2 d a2
1 cos 2
d
2
1
a2
a2
sin 2 C sin cos C
2
2
2
a2
u
u
arcsin 2
a
a
a2 u2
a
C 21 a arcsin ua u
a2 u2 C
2
Let u a sec , du a sec tan d, u2 a2 a tan .
u 2 a 2 du a tan a sec tan d a 2 tan2 sec d
a 2 sec2 1 sec d a2 sec3 sec d
12 sec tan 21 sec d a sec d a 12 sec tan 21 lnsec tan u a
a u
ln 2 a
a
ua u a a C
a2
2
2
2
2
2
2
2
1
1
uu2 a2 a2 ln u u2 a2 C
2
Let u a tan , du a sec2 d, u2 a2 a sec d.
u2 a2 du a sec a sec2 d
12 sec tan 21 lnsec tan C
a2 sec3 d a2
Section 7.5
a2 u2 a2
2
a
u
1
a ln
u2 a2
a
u
a
C1 1
uu2 a2 a2 ln u u2 a2 C
2
Partial Fractions
1.
5
A
B
5
x 2 10x xx 10
x
x 10
3.
2x 3
2x 3
A
Bx C
2
x3 10x xx 2 10
x
x 10
5.
16x
16x
A
C
B
2
x3 10x 2 x 2x 10
x
x
x 10
7.
1
1
A
B
x 2 1 x 1x 1 x 1 x 1
1 Ax 1 Bx 1
When x 1, 1 2A, A 12 .
When x 1, 1 2B, B 12 .
1
1
dx x2 1
2
1
1
dx x1
2
1
dx
x1
1
1
ln x 1 ln x 1 C
2
2
1 x1
ln
C
2 x1
Section 7.5
9.
3
3
A
B
x 2 x 2 x 1x 2 x 1 x 2
5 x Ax 1 B2x 1
When x 1, 3 3A, A 1.
When x 2, 3 3B, B 1.
3
dx x2 x 2
1 9
3
When x 2 , 2 2 A, A 3.
1
dx x1
1
dx
x2
When x 1, 6 3B, B 2.
ln x 1 ln x 2 C
ln
13.
5x
dx 3
2x 2 x 1
x1
C
x2
1
1
dx 2
dx
2x 1
x1
3
ln 2x 1 2 ln x 1 C
2
B
C
x 2 12x 12
A
xx 2x 2
x
x2 x2
x 2 12x 12 Ax 2x 2 Bxx 2 Cxx 2
When x 0, 12 4A, A 3. When x 2, 8 8B, B 1. When x 2, 40 8C, C 5.
x 2 12x 12
dx 5
x3 4x
1
dx x2
1
dx 3
x2
1
dx
x
5 ln x 2 ln x 2 3 ln x C
15.
2x3 4x 2 15x 5
x5
A
B
2x 2x x 2 2x 8
x 4x 2
x4 x2
x 5 Ax 2 Bx 4
When x 4, 9 6A, A 32 . When x 2, 3 6B, B 12 .
2x3 4x 2 15x 5
dx x2 2x 8
2x x2 17.
12
32
dx
x4 x2
3
1
ln x 4 ln x 2 C
2
2
B
4x 2 2x 1 A
C
2
x 2x 1
x
x
x1
4x 2 2x 1 Axx 1 Bx 1 Cx 2
When x 0, B 1. When x 1, C 1. When x 1, A 3.
4x 2 2x 1
dx x3 x 2
1
3
1
1
2
dx 3 ln x ln x 1 C
x
x
x1
x
1
ln x 4 x 3 C
x
2
2
C
B
19. x 3x 4 x 3x 4 A 3
2
x 4x 4x
xx 22
x
x 2 x 22
x2 3x 4 Ax 22 Bxx 2 Cx
When x 0, 4 4A ⇒ A 1. When x 2, 6 2C ⇒ C 3. When x 1, 0 1 B 3 ⇒ B 2.
x2 3x 4
dx x3 4x2 4x
1
dx x
85
5x
5x
A
B
2x 2 x 1 2x 1x 1 2x 1 x 1
11.
3 x 2 Bx 1
Partial Fractions
2
dx x 2
ln x 2 ln x 2 3
dx
x 22
3
C
x 2
86
21.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x2 1
A Bx C
2
xx 2 1
x
x 1
x 2 1 Ax 2 1 Bx Cx
When x 0, A 1. When x 1, 0 2 B C. When x 1, 0 2 B C.
Solving these equations we have A 1, B 2, C 0.
x2 1
dx x3 x
1
dx x
2x
dx
x2 1
ln x 2 1 ln x C
ln
23.
x2
1
C
x
A
B
Cx D
x2
2
x 4 2x 2 8 x 2 x 2
x 2
x 2 Ax 2x 2 2 Bx 2x 2 2 Cx Dx 2x 2
When x 2, 4 24A. When x 2, 4 24B. When x 0, 0 4A 4B 4D, and when x 1,
1 9A 3B 3C 3D. Solving these equations we have A 16 , B 16 , C 0, D 13 .
x4
x 1 2 dx x 1 2 dx 2 x
1
x2
x
ln
2 arctan
C
6 x 2
2
x2
1
dx 2x 2 8
6
25.
2
1
dx
2
x
A
B
Cx D
2x 12x 14x 2 1 2x 1 2x 1 4x 2 1
x A2x 14x 2 1 B2x 14x 2 1 Cx D2x 12x 1
When x 12 , 12 4A. When x 12 , 12 4B. When x 0, 0 A B D, and when x 1,
1 15A 5B 3C 3D. Solving these equations we have A 18 , B 18 , C 12 , D 0.
x
1
dx 16x 4 1
8
27.
2x 1 1 dx 2x 1 1 dx 4 4x x 1 dx
2
1
4x 2 1
ln
C
16 4x 2 1
x2 5
A
Bx C
x 1x 2 2x 3 x 1 x 2 2x 3
x 2 5 Ax 2 2x 3 Bx Cx 1
A Bx 2 2A B Cx 3A C
When x 1, A 1. By equating coefficients of like terms, we have A B 1, 2A B C 0,
3A C 5. Solving these equations we have A 1, B 0, C 2.
x3
x2 5
dx x2 x 3
1
dx 2
x1
1
dx
x 12 2
ln x 1 2 arctan
x21 C
Section 7.5
29.
Partial Fractions
3
A
B
2x 1x 2 2x 1 x 2
3 Ax 2 B2x 1
When x 1
0
1
2 ,
A 2. When x 2, B 1.
3
dx 2x 5x 2
2
1
0
2
dx 2x 1
1
dx
x2
0
1
0
1
ln 2x 1 ln x 2
ln 2
31.
x1
A Bx C
2
xx 2 1
x
x 1
x 1 Ax 2 1 x Cx
When x 0, A 1. When x 1, 2 2A B C. When x 1, 0 2A B C. Solving these
equations we have A 1, B 1, C 1.
2
1
x1
dx xx 2 1
2
1
1
dx x
ln x 2
1
x2
x
dx 1
2
1
1
dx
x2 1
1
lnx 2 1 arctan x
2
2
1
1 8 ln arctan 2
2 5
4
0.557
33.
3x dx
9
3 ln x 3 C
x 2 6x 9
x3
4, 0: 3 ln4 3 35.
9
C0⇒C9
43
2
x2 x 2
x
1
arctan
dx C
x 2 22
2
2x 2 2
2
0, 1: 0 30
1
5
C1⇒C
4
4
3
(0, 1)
−6
(4, 0)
−3
10
3
−1
−10
37.
2x 2 2x 3
1
2x 1
dx ln x 2 ln x 2 x 1 3 arctan
C
x2 x 2
2
3
x3
3, 10: 0 39.
20
(3, 10)
7
1
7
1
C 10 ⇒ C 10 ln 13 3 arctan
ln 13 3 arctan
2
2
3
3
1
1 x2
dx ln
C
x2 4
4 x2
6, 4:
1
4
1
1 1
ln C 4 ⇒ C 4 ln 4 ln 2
4
8
4 2
4
10
(6, 4)
− 10
10
−3
−2
6
−5
87
88
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
41. Let u cos x du sin x dx.
43.
3 cos x
1
dx 3 2
du
sin2 x sin x 2
u u2
A
B
1
uu 1
u
u1
1 Au 1 Bu
When u 0, A 1. When u 1, B 1, u cos x,
du sin x dx.
u1
C
u2
ln
1 sin x
C
2 sin x
(From Exercise 9 with u sin x, du cos x dx)
sin x
1
dx du
cos xcos x 1
uu 1
ln
1
du u
1
du
u1
ln u ln u 1 C
u
C
ln
u1
ln
cos x
C
cos x 1
45. Let u ex, du ex dx.
47.
1
A
B
u 1u 4 u 1 u 4
1
A
B
xa bx
x
a bx
1 Aa bx Bx
When x 0, 1 a A ⇒ A 1a.
When x ab, 1 abB ⇒ B ba.
1 Au 4 Bu 1
1
1
When u 1, A 5 . When u 4, B 5 , u ex,
x dx.
du e
ex
dx x
e 1ex 4
1
5
49.
1
u 1u 4
1
du u1
du
1 u1
ln
C
5 u4
1 ex 1
ln
C
5 ex 4
1
du
u4
x
A
B
a bx2 a bx a bx2
51.
−2
1b
ab
dx
a bx a bx2
1
b
1
a
dx a bx
b
1
dx
a bx2
a
1
1
ln a bx 2
C
b2
b a bx
1
a
ln a bx
b2 a bx
1
x
ln
C
a a bx
C
2
−4
1
b
dx
x
a bx
1
ln x ln a bx C
a
10
When x ab, B ab.
When x 0, 0 aA B ⇒ A 1b.
6
dy
, y0 3
dx 4 x2
x Aa bx B
x
dx a bx2
1
1
dx xa bx
a
Section 7.5
53. Dividing x3 by x 5.
Partial Fractions
89
55. (a) Substitution: u x2 2x 8
(b) Partial fractions
(c) Trigonometric substitution (tan) or inverse tangent rule
57. Average Cost 1
80 75
1
5
80
75
80
75
124p
dp
10 p100 p
124
1240
dp
10 p11 100 p11
1 124
1240
ln10 p ln100 p
5 11
11
80
75
1
24.51 4.9
5
Approximately $490,000.
3
59. A 1
10
dx 3
xx2 1
Matches (c)
y
5
4
3
2
1
x
1
2
3
5
A
B
1
1
,AB
x 1n x x 1 n x
n1
61.
1
n1
4
1
1
dx kt C
x1 nx
x1
1
ln
kt C
n1 nx
When t 0, x 0, C 1
1
ln .
n1 n
x1
1
1
1
ln
kt ln
n1 nx
n1 n
x1
1
1
ln
ln
kt
n1
nx
n
ln
nx n
n 1k t
nx
nx n
en1kt
nx
x
nen1kt 1
n en1kt
Note: lim x n
t →
90
63.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Cx D
x
Ax B
1 x4 x2 2 x 1 x2 2 x 1
x Ax B x2 2 x 1 Cx D x2 2 x 1
A Cx3 B D 2 A 2 Cx2 A C 2 B 2 Dx B D
0 A C ⇒ C A
0 B D 2 A 2 C
1 A C 2 B 2 D
22 A 0 ⇒ A 0 and C 0
22 B 1 ⇒ B 2
and D 4
2
4
0 B D ⇒ D B
Thus,
1
0
x
dx 1 x4
1
0
1
2
4
2
4
24
24
2
dx
x 2 x 1 x 2 x 1
2
0
1
1
dx
2
2
x 22 12 x 22 12
1
x 22
x 22
arctan
arctan
12
12
12
1
0
1
1
arctan 2 x 1 arctan 2 x 1
2
1
arctan 2 1 arctan 2 1 arctan 1 arctan1
2
1
arctan 2 1 arctan 2 1 .
2
4
4
0
Since arctan x arctan y arctan
x y1 xy, we have:
1
0
1
x
2 1 2 1
1
1
2
dx arctan
arctan
1 x4
2
1 2 1 2 1
2
2
2
2
2
4
2
8
Section 7.6
1. By Formula 6:
3. By Formula 26:
Integration by Tables and Other Integration Techniques
x
x2
dx 2 x ln 1 x C
1x
2
e x1 e 2x dx 1 x
e e 2x 1 ln e x e 2x 1 C
2
u e x, du e x dx
5. By Formula 44:
1 x 2
1
dx
C
x
x 21 x 2
Section 7.6
7. By Formulas 50 and 48:
sin42x dx Integration by Tables and Other Integration Techniques
1
sin42x2 dx
2
1 sin 2x cos2x 3
2x sin 2x cos 2x C
2
4
8
1 sin32x cos2x 3
sin22x2 dx
2
4
4
3
9. By Formula 57:
1
x 1 cos x
u x, du 1
6x 3 sin 2x cos 2x 2 sin3 2x cos 2x C
16
dx 2
2 cot x csc x C
1
dx
2x
11. By Formula 84:
1
1
dx
1 cos x 2x
13. By Formula 89:
1
1
dx x ln1 e 2x C
1 e 2x
2
x 3 ln x dx x4
4 ln x 1 C
16
x 2e x dx x 2e x 2 xe x dx
15. (a) By Formulas 83 and 82:
x 2e x 2
x 1e x C1
x 2e x 2xe x 2e x C
(b) Integration by parts: u x 2, du 2x dx, dv e x dx, v e x
x 2e x dx x 2ex 2xe x dx
Parts again: u 2x, du 2 dx, dv e x dx, v e x
x 2e x dx x 2e x 2 xe x 2e x dx x 2e x 2xe x 2e x C
17. (a) By Formula: 12, a b 1, u x, and
1
1 1 1
x
ln
dx x 2x 1
1 x
1 1x
(b) Partial fractions:
C
1
x
C
ln
x
1x
1
x1
ln
C
x
x
1
A
B
C
2
x 2x 1
x
x
x1
1 Axx 1 Bx 1 Cx 2
x 0: 1 B
x 1: 1 C
x 1: 1 2A 2 1 ⇒ A 1
1
dx x 2x 1
1
1
1
2
dx
x
x
x1
ln x 1
ln x 1 C
x
1
x
ln
C
x
x1
91
92
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
19. By Formula 81:
21. By Formula 79:
1 2
2
xex e x C
2
x arcsecx 2 1 dx 1
arcsecx 2 1(2x dx
2
1 2
x 1 arcsecx 2 1 ln x 2 1 x 4 2x 2 C
2
u x 2 1, du 2x dx
23. By Formula 89:
27. By Formula 4:
x 2 ln x dx x3
1 3 ln x C
9
25. By Formula 35:
31. By Formula 73:
e x arccos e x dx e x arccos e x 1 e 2x C
u e x, du e x dx
cos x
dx arctansin x C
1 sin2 x
2
cos 1 sin d arctan
C
3 2 sin sin2 2
2
u sin , du cos d
1
dx 3
x 22 9x 2
3
3x 2 3x2
2
2
32 9x 2
C
6x
2 9x 2
2x
C
39. By Formulas 54 and 55:
x 2 4
4x
C
x
1
2x
dx dx
1 sec x 2
2 1 sec x 2
1
x 2 cot x 2 csc x 2 C
2
u sin x, du cos x dx
37. By Formula 35:
4
dx 35. By Formula 14:
1
x 2x 2
2x
x
2
1
dx 2
dx ln1 3x C
1 3x2
1 3x2
9
1 3x
29. By Formula 76:
33. By Formula 23:
t3 cos t dt t3 sin t 3 t2 sin t dt
t3 sin t 3 t2 cos t 2 t cos t dt
t3 sin t 3t2 cos t 6 t sin t sin t dt
t3 sin t 3t2 cos t 6t sin t 6 cos t C
dx
Section 7.6
41. By Formula 3:
Integration by Tables and Other Integration Techniques
ln x
1
dx 2 ln x 3 ln 3 2 ln x C
x3 2 ln x
4
1
dx
x
u ln x, du 43. By Formulas 1, 25, and 33:
x
1
2x 6 6
dx dx
x 2 6x 102
2 x 2 6x 102
3
x3
1
arctanx 3 C
2x 2 6x 10 2 x 2 6x 10
45. By Formula 31:
x
x 4 6x 2 5
3x 10
3
arctanx 3 C
2x 2 6x 10 2
1
2x
dx
2 x 2 32 4
dx 1
1
x 2 6x 1022x 6 dx 3
dx
2
x 32 12
1
ln x 2 3 x 4 6x 2 5 C
2
u x 2 3, du 2x dx
47.
x3
dx 4 x 2
8 sin3 2 cos d
2 cos x
8 1 cos2
sin d
θ
4 − x2
8 sin cos 2
8 cos 8 cos3 C
3
2
sin d
4 x 2 2
x 8 C
3
x 2 sin , dx 2 cos d, 4 x 2 2 cos 49. By Formula 8:
e 3x
dx 1 e x3
e x2
e x dx
1 e x3
2
1
ln 1 e x C
1 e x 21 e x 2
u e x, du e x dx
51.
1
2abu a2b2
1
A
B
u2
2
2 2 a bu
b
a bu2
b
a bu a bu2
2a
a2
u 2 Aa bu B aA B bAu
b
b
Equating the coefficients of like terms we have aA B a2b2 and bA 2ab. Solving these equations
we have A 2ab2 and B a2b2.
a 1 bub du ba 1ba 1bu b du b1 u 2ab lna bu ba a 1 bu C
u2
1
2a 1
du 2 du 2
a bu2
b
b b
2
2
C
a2
1
bu 2a ln a bu
b3
a bu
2
2
2
3
3
93
94
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
53. When we have u2 a2:
When we have u2 a2:
u a tan du a
u2
a2
1
du u2 a232
sec2
a2
u a sec d
sec2
du a sec tan d
a sec2 d
a3 sec3 a2 a2 tan2 1
du u2 a232
a sec tan d
a3 tan3 1
cos d
a2
1
sin C
a2
u
C
a2u2 a2
u
u
θ
1 cos d
a2 sin2 u2 + a2
1
csc C
a2
u
C
a2u2 a2
u2 + a2
θ
a
a
55.
u2
arctan u du u arctan u u arctan u 1
2u
du
2 1 u2
1
ln1 u2 C
2
u arctan u ln1 u2 C
w arctan u, dv du, dw 57.
du
,vu
1 u2
21 x
1
dx C
x
x321 x
8
12, 5: 21212 C 5 ⇒ C 7
( 12, 5)
−0.5
1.5
21 x
7
y
x
59.
1
1
x3
dx tan1x 3 2
C
x2 6x 102
2
x 6x 10
3, 0:
y
61.
−2
2
1
0
0
C0⇒C0
2
10
1
x3
tan1x 3 2
2
x 6x 10
−8
−2
1
d csc C
sin tan 4 , 2: 22 C 2 ⇒ C 2 2
y csc 2 2
(3, 0)
10
(π4, 2)
−
2
2
−2
8
Section 7.6
63.
1
d 2 3 sin 2
2 du
1 u2
2u
23
1 u2
Integration by Tables and Other Integration Techniques
2
65.
0
1
d 1 sin cos 2 du
1 u2
2u
1 u2
1
1 u2 1 u2
1
du
1u
0
1
du
u 3u 1
0
ln 1 u
1
ln 2
1
du
3 2 5
u
2
4
u 23 25
u 23 25
u tan
1
ln
5
0
1
2
du
1 u2 6u
2
1
95
2
C
2u 3 5
C
2u 3 5
2 tan
3 5
2
1
ln
C
5
2 tan
3 5
2
1
5
ln
u tan
67.
2
1
sin d 3 2 cos 2
2 sin d
3 2 cos 69.
cos 1
d 2 cos d
2 2 sin C
1
ln u C
2
u , du 1
ln3 2 cos C
2
1
d
2
u 3 2 cos , du 2 sin d
8
71. A 0
x
dx
x 1
y
73. Arctangent Formula, Formula 23,
4
22 x
x 1
3
8
3
0
2
4
12 3
1
x
2
1
du, u ex
u2 1
4
6
8
40
13.333 square units
3
75. Substitution: u x2, du 2x dx
Then Formula 81.
77. Cannot be integrated.
79. Answers will vary. For example,
2xe2x dx
can be integrated by first letting
u 2x and then using Formula 82.
96
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
5
81. W 2000xex dx
0
5
2000
xex dx
0
5
2000
xex1 dx
0
2000 xex ex
2000 6
1
e5
5
0
1919.145 ft lbs
3
83. (a) V 202
2
1 y 2
0
80 ln y 1 y 2
80 ln 3 10
W 148 80 ln 3 10 dy
3
11,840 ln 3 10 0
21,530.4 lb
145.5 cubic feet
(b) By symmetry, x 0.
3
M 2
0
3
Mx 2
0
y
2
1 y 2
dy 4 ln y 1 y 2
2y
dy 4
1 y 2
1 y 2
3
0
0 4 ln 3 10 3
4 10 1
Mx
4 10 1
1.19
M
4 ln 3 10 Centroid: x, y 0, 1.19
4
85. (a)
0
k
4
0
4
k
dx 10
2 3x
(b)
0
15.417
dx
2 3x
8
10
10
1
0.6486
dx
2 3x
15.417
4
0
−1
ln307
87. False. You might need to convert your integral using substitution or algebra.
Section 7.7
1. lim
x →0
Indeterminate Forms and L’Hôpital’s Rule
sin 5x
5
2.5 exact:
sin 2x
2
x
f x
3
0.1
0.01
0.001
0.001
0.01
0.1
2.4132
2.4991
2.500
2.500
2.4991
2.4132
−1
1
−1
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule
3. lim x5ex100 0
10,000,000
x →
x
f x
1
10
102
103
104
105
0.9901
90,484
3.7 109
4.5 1010
0
0
0
100
0
5. (a) lim
x →3
(b) lim
x →3
7. (a) lim
1
2x 3
2x 3
2
lim
lim
x →3 x 3x 3
x →3 x 3
x2 9
3
2 1
2x 3
2
ddx2x 3
lim
lim
x →3 ddxx2 9
x →3 2x
x2 9
6 3
x 1 2
x3
x →3
(b) lim
x 1 2
x →3
x3
lim
x 1 2
x3
x →3
lim
x →3
x 1 2
x 1 2
lim
x →3
1
x 1 4
1
lim
x 3
x 1 2 x →3 x 1 2 4
1 2
x 1 1
ddx
x 1 2
lim
x →3
ddxx 3
1
4
5x2 3x 1
5 3x 1x2 5
lim
2
x →
x →
3x 5
3 5x2
3
9. (a) lim
5x2 3x 1
ddx5x2 3x 1
10x 3
ddx10x 3
10 5
lim
lim
lim
lim
2
x →
x →
x →
x →
x → 6
3x 5
ddx3x2 5
6x
ddx6x
3
(b) lim
x2 x 2
2x 1
lim
3
x →2
x →2
x2
1
11. lim
15. lim
x →0
13. lim
4 x2 2
x →0
x
x
4 x2
0
x →0
1
lim
ex 1 x
ex 1
lim
2
x →0
x
1
17. Case 1: n 1
lim
x →0
ex 1 x
ex 1
lim
0
x →0
x
1
Case 2: n 2
lim
x →0
ex 1 x
ex 1
ex 1
lim
lim 2
x →0
x →0 2
x
2x
2
Case 3: n ≥ 3
lim
x →0
ex 1 x
ex 1
ex
lim n1 lim
n
x →0 nx
x →0 nn 1xn2
x
sin 2x
2 cos 2x 2
lim
x→0 sin 3x
x→0 3 cos 3x
3
19. lim
arcsin x
1
1 x2
lim
1
x →0
x →0
x
1
3x2 2x 1
6x 2
lim
x→ x→ 2x2 3
4x
21. lim
23. lim
lim
x→ x2 2x 3
2x 2
lim
x→ x →
x1
1
25. lim
6 3
4 2
x3
3x2
lim
x2
x→ e
x→ 12e x2
27. lim
lim
x→ 6x
6
lim
0
14e x2 x→ 18e x2
97
98
Chapter 7
x
29. lim
x→ x2 1
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
lim
1
x→ 1 1x2
1
31. lim
x→ cosx x ≤ 1x Note: L’Hôpital’s Rule does not work on this limit.
See Exercise 79.
ln x
1x
1
lim
lim 2 0
x→ x2
x→ 2x
x→ 2x
cos x
0 by Squeeze Theorem
x
ex
ex
ex
lim
lim
2
x→ x
x→ 2x
x→ 2
33. lim
35. lim
37. (a) lim x ln x 0 0
39. (a) lim x sin
x →0
ln x
(b) lim x ln x lim
x →0
x →0 1x
lim
x →0
x→ (b) lim x sin
x→ 1x
1x2
1
sin1x
lim
x x→ 1x
1x2 cos1x
x→ 1x2
lim
lim x 0
x →0
(c)
1
0
x
lim cos
x→ 2
0
(c)
10
1.5
−1
−12
1x 1
1
− 0.5
41. (a) lim x1x 0 0, not indeterminant
x →0
43. (a) lim x1x 0
x →
(b) Let y lim x1x.
(See Exercise 95)
x→ y x1x
(b) Let
ln y ln
x1x
ln y lim
1
ln x.
x
x→ Thus, ln y 0 ⇒ y e0 1. Therefore,
1
ln x → . Hence,
x
Since x → 0 ,
ln y → ⇒ y → 0.
ln x
1x
lim
0
x→ x
1
lim x1x 1.
x →
(c)
2
Therefore, lim x1x 0.
x→0
(c)
−5
2
20
−0.5
− 0.5
2
− 0.5
45. (a) lim 1 x1x 1
(c)
x →0
6
(b) Let y lim 1 x1x.
x →0
ln1 x
x
11 x
1
lim
x →0
1
ln y lim
x →0
Thus, ln y 1 ⇒ y e1 e.
Therefore, lim 1 x1x e.
x →0
−1
4
−1
Section 7.7
47. (a) lim 3xx2 00
(c)
x→0
Indeterminate Forms and L’Hôpital’s Rule
99
7
(b) Let y lim 3xx2.
x→0
x
ln x
2
ln x
2x
ln y lim ln 3 x→0
lim ln 3 x→0
−6
lim ln 3 lim
1x
2x2
lim ln 3 lim
x
2
x→0
x→0
x→0
x→0
6
−1
ln 3
Hence, lim 3xx2 3.
x→0
49. (a) lim ln xx1 00
ln x
(b) Let y lim
51. (a) lim
x
(b) lim
x
2
x →2
x→1
x1
x1
lim x 1ln x 0
2
x →2
x→1
x
8
4 x2
8
8 xx 2
x
lim
x →2
4 x2
x2 4
Hence, lim ln xx1 1
x→1
(c)
lim
2 x4 x
x 2x 2
lim
x 4 3
x2
2
x →2
6
x →2
−4
8
(c)
4
−2
−7
5
−4
53. (a) lim
x →1
(b) lim
x →1
ln3x x 2 1 ln3x x 2 1 lim
x →1
lim
x →1
55. (a)
3x 3 2 ln x
x 1ln x
3
−1
7
3 2x
x 1x ln x −1
(b) lim
(c)
x →3
8
x3
1
lim
ln2x 5 x →3 22x 5
lim
x →3
−1
2x 5 1
2
2
4
−4
57. (a)
(b) lim x2 5x 2 x lim x2 5x 2 x
10
x→ x→ x2 5x 2 x
x2 5x 2 x
x2 5x 2 x2
x→ x2 5x 2 x
lim
−8
10
−2
lim
x→ lim
x→ 5x 2
x2 5x 2 x
5 2x
1 5x 2x2 1
5
2
100
59.
Chapter 7
0 , ,0
0 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
, 1, 00, 61. (a) Let f x x2 25 and gx x 5.
(b) Let f x x 52 and gx x2 25.
(c) Let f x x2 25 and gx x 53.
63. lim
x2
2x
2
lim
lim
0
e5x x → 5e5x x → 25e5x
65. lim
ln x3
3ln x21x
lim
x→ x
1
x →
x→ 67. lim
x→ ln xn
nln xn1x
lim
x→ xm
mxm1
3ln x2
x
lim
6ln x1x
x→ 1
lim
lim
x→ x→ nn 1ln xn2
x→ m2xm
lim
lim
x→ 69.
x
ln x4
x
6ln x
6
lim 0
x→ x
x
. . . lim
x→ 10
102
104
106
108
1010
2.811
4.498
0.720
0.036
0.001
0.000
71. y x1x, x > 0
lim
x →
1
ln x
x
2x
2
lim x 0
x → e
ex
Horizontal asymptote: y 0
1
1 dy 1 1
ln x 2
y dx
x x
x
1
dy
x1x 2 1 ln x x1x 21 ln x 0
dx
x
Critical number:
n!
0
mnxm
73. y 2xex
Horizontal asymptote: y 1 (See Exercise 37)
ln y nln xn1
mxm
xe
0, e
Sign of dydx:
y f x: Increasing
Relative maximum: e, e1e Intervals:
e, Decreasing
dy
2xex 2ex
dx
2ex 1 x 0
Critical number:
x1
Intervals: , 1
Sign of dydx:
y f x: Increasing
Relative maximum:
4
1, Decreasing
1, 2e 3
(1, 2e )
−2
(e, e1/e)
0
6
−5
0
e2x 1 0
0
x →0
ex
1
75. lim
Limit is not of the form 00 or .
L’Hôpital’s Rule does not apply.
10
77. lim x cos
x →
1
1 x Limit is not of the form 00 or .
L’Hôpital’s Rule does not apply.
Section 7.7
79. (a) lim
x→ x
x2 1
xx
lim
x→ x→ lim
x→ (c)
(b) lim
x2 1x
lim
Indeterminate Forms and L’Hôpital’s Rule
x→ x
x2 1
1
x 1
x2
1
x
x2 1
x2 1
x
x→ 1
lim
1 1x2
1
x→ lim
2
1 0
lim
x→ lim
Applying L’Hôpital’s rule twice results in the original limit,
so L’Hôpital’s rule fails.
1
−1.5
81. lim
v0kekt
32
k
k→0
321 ekt
lim v0ekt k→0
k→0
k
lim
lim
k→0
v0
320 tekt lim kt 32t v0
k→0 e
1
1
2x1 cos x x x cos x
2
Shaded area: Area of rectangle Area under curve
83. Area of triangle:
x
2x1 cos x 2
1 cos t dt 2x1 cos x 2 t sin t
0
x
0
2x1 cos x 2x sin x 2 sin x 2x cos x
Ratio: lim
x →0
x x cos x
1 x sin x cos x
lim
2 sin x 2x cos x x →0 2 cos x 2x sin x 2 cos x
lim
1 x sin x cos x
2x sin x
lim
x cos x sin x sin x
2x cos x 2 sin x
lim
x cos x 2 sin x
2x cos x 2 sin x
lim
x 2 tan x
2x 2 tan x
x →0
x →0
x →0
x →0
1cos x
1cos x
1 2 sec2 x 3
x →0 2 2 sec2 x
4
lim
85. f x x3, gx x2 1, 0, 1
fc
f b f a
gb ga
gc
f 1 f 0 3c2
g1 g0
2c
1 3c
1
2
c
2
3
x
x2 1
1
x
6
32 1 ekt x→ x2 1
1.5
−6
101
2 87. f x sin x, gx cos x, 0,
fc
f 2 f 0
g
2 g0
gc
1
cos c
1 sin c
1 cot c
c
4
102
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
89. False. L’Hôpital’s Rule does not apply since
91. True
lim x2 x 1 0.
x →0
x2 x 1
1
1
lim x 1 x →0
x →0
x
x
lim
93. (a) sin BD
cos DO ⇒ AD 1 cos 1
1
1
1
Area ABD bh 1 cos sin sin sin cos 2
2
2
2
(b) Area of sector:
1
1
1
1
1
Area OBD cos sin sin cos 2
2
2
2
2
Shaded area:
(c) R 1
2
12 sin 12 sin cos sin sin cos 12 12 sin cos sin cos sin 12 sin 2
12 sin 2
(d) lim R lim
→0
→0
cos cos 2
sin 2 sin 2
cos 4 cos 2 3
lim
lim
→0
→0
1 cos 2
2 sin 2
4 cos 2
4
lim
→0
95. lim f xgx
x →a
y f xgx
ln y gx ln f x
lim gx ln f x x →a
As x → a, ln y ⇒ , and hence y 0. Thus,
lim f xgx 0.
x →a
b
97. fab a f tt b dt fab a a
ftt b
b
a
b
fab a faa b f t
dv f tdt ⇒
v ft
u t b ⇒ du dt
Section 7.8
Improper Integrals
1. Infinite discontinuity at x 0.
4
0
1
x
4
dx lim
b →0
b
1
x
lim 2x
b →0
dx
4
b
lim 4 2b 4
b →0
Converges
ft dt
a
b
a
f b f a
Section 7.8
3. Infinite discontinuity at x 1.
2
0
1
dx x 12
1
1
dx x 12
0
lim
lim
b
b→1
b→1
0
2
1
b
lim 0
103
1
dx
x 12
1
dx lim
c→1
x 12
1
x1
Improper Integrals
c→1
2
c
1
dx
x 12
1
x1
2
1 1 c
Diverges
1
5. Infinite limit of integration.
7.
b
ex dx lim
b→ 0
ex dx
because the integrand is not defined at x 0.
Diverges
0
e lim
x
b→ 1
dx 2
x2
1
b
0
011
Converges
9.
1
1
dx lim
b→ x2
b
1
1x
b
lim
b→
1
dx
x2
11.
3
x
1
dx lim
b→ b→
3x13 dx
1
9x 2
b
lim
1
1
b
3
23
1
Diverges
0
13.
0
xe2x dx lim
b→
0
1
2x 1e2x
b→ 4
xe2x dx lim
b
b
lim
b→
1
1 2b 1e2b (Integration by parts)
4
Diverges
15.
b→ 0
Since lim b→ b
x2ex dx lim
b→
0
b
0
lim b→ b2 2b 2
2 2
eb
ex cos x dx lim
0
x2 2x 2
x
b2 2b 2
0 by L’Hôpital’s Rule.
eb
17.
e
x2ex dx lim
b→ 1 x
e cos x sin x
2
b
0
1
1
0 1 2
2
19.
4
1
dx lim
b→ xln x3
1
ln x3 dx
x
4
b
21.
2
2 dx 4 x
0
2
2 dx 4 x
0
2 ln x lim
1
1
ln b2 ln 42
2
2
lim
1
1
1
2 2 ln 22 8ln 22
0 2
lim
b→
1
2
b
4
b→
b→
b
0
2
dx
4 x2
2
dx lim
c→ 4 x2
c
2
dx
4
x2
0
arctan2 lim arctan2
x
0
b
x
c→
0 2
c
0
104
Chapter 7
23.
0
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1
1
dx lim
b →0
x2
x
cos x dx lim
b→ 0
sin x
b
1
0
Diverges since sin x does not approach a limit as x → .
arctane 0
0
25.
2
4
4
1
0
ex
dx
1 e2x
x
b→ b
b
lim
27.
1
dx lim
b→ ex ex
1
1 b
Diverges
8
29.
1
dx lim
3
b→8
0 8 x
1
31.
0
x ln x dx lim
b→0
b
0
tan d 0
x2 lnx 4 2
1
x2
b
2
33.
1
3
dx lim
8 x 23
b→8
2
8x
3
lnsec b
lim
b→ 2
lim
b→0
14 b
0
6
b2 ln b b2
1
since lim b2 ln b 0 by L’Hôpital’s Rule.
b→0
2
4
4
4
35.
2
0
2
dx lim
b→2
xx2 4
2
dx
xx2 4
x
2
4
b
lim arcsec
Diverges
b→2
4
b
lim arcsec 2 arcsec
b→2
4
37.
2
4
1
lim ln x x2 4
b→2
x 4
2
0
3
3
b
ln 4 23 ln 2
ln 2 3 1.317
2
39.
0
1
dx 3
x 1
1
0
1
dx 3
x 1
lim
b→1
41.
0
4
dx xx 6
4
xx 6
Thus,
0
dx 4
xx 6
1
23
0
1
0
1
dx
x1
3
b
3
2
2 x 1 lim
c→1
4
dx xx 6
Let u x, u2 x, 2u du dx.
2 x 1 2
3
23
c
4
xx 6
1
3 3
0
2
2
dx
C
x
42u du
du
4
8
8
8 2
arctan
C
arctan
2
uu 6
u 6
6
6
6
6
dx lim
b→0
8
6
6
arctan
x
86 arctan 16 2
6
8
.
26
3
1
lim
b
8
6
c→ 86 2 0 c
x
8
arctan
6
6
8
6
1
16
arctan
b2
Section 7.8
43. If p 1,
1
1
dx lim
b→ x
b
1
Improper Integrals
105
b
1
dx lim ln x .
b→ x
1
Diverges. For p 1,
1
1
x1p
lim
p dx b→
x
1p
This converges to
b
1
b 1 p.
1p
1p
lim
b→
1
1
if 1 p < 0 or p > 1.
p1
45. For n 1 we have
b
xex dx lim
b→ 0
lim
b→ xex dx
0
e
xx
b
ex
Parts: u x, dv ex dx
0
lim ebb eb 1
b→ lim
b→ b
e
b
1
1 1 (L’Hôpital’s Rule)
eb
Assume that
x nex dx converges. Then for n 1 we have
0
x n1ex dx x n1ex n 1 x nex dx
by parts u xn1, du n 1xn dx, dv ex dx, v ex.
Thus,
b→
0
1
47.
0
x
xn1ex dx lim
n1ex
b
0
n 1
0
1
dx diverges.
x3
53. Since
x2
49.
1
x3
1
1
x2
1
dx converges by Exercise 43,
1
1
≥ 3 2 on 2, and
3
x
xx 1
2
1
1
converges.
31 2
dx (See Exercise 43, p 3.
1
1
≤ 2 on 1, and
5
x
55. Since ex ≤ ex on 1, and
xnex dx, which converges.
0
(See Exercise 44, p 3 1.
51. Since
xnex dx 0 n 1
1
2
1
dx converges.
x2 5
1
dx diverges by Exercise 43,
3 2
x
ex dx converges (see Exercise 5),
0
2
1
3
xx 1
dx diverges.
ex dx converges.
2
0
1
57. Answers will vary. See pages 540, 543.
59.
1
3 dx x
1
0
1
3 dx x
1
1
0
1
dx
x3
These two integrals diverge by Exercise 44.
61. f t 1
Fs 63. f t t2
0
s e est dx lim
b→
1
st
b
1
,s > 0
s
0
Fs s
s
t 2est dx lim
b→
0
1
3
2
,s > 0
s3
2 t2
2st 2est
b
0
106
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
65. f t cos at
Fs est cos at dt
0
st
e s cos at a sin at
s a
lim
b→
b
2
2
0
s
s
,s > 0
s2 a2 s2 a2
0
67. f t cosh at
Fs est cosh at dt 0
est
0
e
at
etsa etsa dt
0
1
1
1
et sa et sa
b→ 2 s a
s a
lim
eat
1
dt 2
2
b
0
0
1
1
1
2 s a s a
1
1
s
1
2
,s > a
2 s a s a
s a2
69. (a) A ex dx
(b) Disk:
0
lim
b→ ex
b
V
0 1 1
ex 2 dx
0
0
1
lim e2x
b→ 2
(c) Shell:
V 2
xex dx
0
2
lim 2
exx 1
b→
b
0
x23 y23 4
71.
2 13 2 13
x
y
y 0
3
3
y 1 y 2 y13
x13
1 xy
8
s4
0
y
8
2
(− 8, 0)
−8
(0, 8)
−2
−8
(8, 0)
x
2
(0, −8)
8
2
13
x
23
23
x
y23
x 23
2 x 23b 48
23
dx lim 8
b→0
3
8
x 4
23
2
x1/3
b
0
2
Section 7.8
Improper Integrals
73. n xn1ex dx
0
(a) 1 b→
0
2 1
e
x 1
b→
0
x
(b) n 1 b→
x e
x 2ex dx lim
0
b
x
xex dx lim
0
3 e ex dx lim
b→
1
x e xnex dx lim
0
0
b
2xex 2ex
2 x
b
n x
b
0
0
2
b
lim n
b→ xn1ex dx 0 nn
u xn, dv ex dx
0
(c) n n 1!
1 t7
e
dt 7
75. (a)
0
1 t7
e
dt lim et7
b→ 7
b
0
4
1
(b)
0
1 t7
e
dt et7
7
4
0
et7 1
0.4353 43.53%
(c)
17 e dt lim te
t7
t
0
t7
b→
7et7
b
0
077
5
77. (a) C 650,000 25,000 e0.06t dt 650,000 0
e
25,000
0.06
5
0.06t
0
$757,992.41
10
(b) C 650,000 25,000e0.06t dt $837,995.15
0
(c) C 650,000 25,000e0.06t dt 650,000 lim
b→ 0
e
25,000
0.06
0.06t
b
0
$1,066,666.67
79. Let x a tan , dx a sec2 d, a2 x2 a sec .
1
dx a2 x232
a sec2 d
1
2 cos d
a3 sec3 a
Pk
1
81.
x
1
k
dx 2 lim
a x232
a b→ a2 x2
2
x
θ
a
1
1
x
2 sin 2
a
a a2 x2
Hence,
a2 + x 2
b
1
k
1
ka2 1 1
1
.
a2
a2 1
a2a2 1
10
10
⇒ x 0, 2.
x2 2x xx 2
You must analyze three improper integrals, and each must converge in order for the original integral to converge.
3
0
1
f x dx 0
2
f x dx 1
3
f x dx 2
f x dx
107
108
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
83. For n 1,
I1 x2
0
x
1
dx lim
b→ 2
14
b
0
1
x2 142x dx lim 1
b→ 6 x2 13
For n > 1,
In x2n1
x2n2
lim
n3 dx b→
2
n 2
x 1
2n 2x 1
2
0
0
(b)
0
(c)
0
1
x
dx lim 2
b→ x2 14
6x 13
3
1
x
dx x2 15
4
2
x5
x 16 5
2
0
0
0
n1
n2
0
b
0
1
.
6
n1
x2n3
dx 0 I x 1n2
n 2 n1
2
x
1
dx, v x2 1n3
2n 2x2 1n2
u x 2n2, du 2n 2x 2n3 dx, dv (a)
b
b
0
1
6
1 1
1
x
dx x2 14
4 6
24
2 1
1
x3
dx x 15
5 24
60
2
85. False. f x 1x 1 is continuous on 0, , lim 1x 1 0, but
x →
0
0 .
1
dx lim ln x 1
b→ x1
Diverges
87. True
Review Exercises for Chapter 7
1.
xx2 1 dx 1
x2 1122x dx
2
3.
x
1
2x
dx dx
x2 1
2 x2 1
1 x2 132
C
2
32
1
ln x2 1 C
2
1
x2 132 C
3
5.
ln2x
ln 2x2
dx C
x
2
1
2
e2x sin 3x dx e2x cos 3x 3
3
9.
13
9
7.
16
16 x2
dx 16 arcsin
e2x cos 3x dx
2 1 2x
2
1
e sin 3x e2x cos 3x 3
3 3
3
e2x sin 3x dx
1
2
e2x sin 3x dx e2x cos 3x e2x sin 3x
3
9
e2x sin 3x dx e2x
2 sin 3x 3 cos 3x C
13
(1) dv sin 3x dx ⇒
u e2x
1
v cos 3x
3
⇒ du 2e2x dx
(2) dv cos 3x dx ⇒
u e2x
v
1
sin 3x
3
⇒ du 2e2x dx
4x C
b
Review Exercises for Chapter 7
2
11. u x, du dx, dv x 512 dx, v x 532
3
2
xx 5 dx xx 532 3
1
x2 sin 2x dx x2 cos 2x 2
13.
2
x 532 dx
3
4
(1) dv sin 2x dx ⇒
156 x 34 C
u x2
x 532
x2
arcsin 2x 2
x arcsin 2x dx 19.
21.
dx
22x2
dx
1 2x2
1 1
x2
arcsin 2x 2x1 4x2 arcsin 2x C (by Formula 43 of Integration Tables)
2
8 2
1
8x2 1 arcsin 2x 2x1 4x2 C
16
v
cos3 x 1 dx sec4
⇒ du dx
1
sin 2x
2
1
x2
arcsin 2x 2
8
u arcsin 2x ⇒ du 17.
x2
1 4x2
v
⇒
dv x dx
ux
1
v cos 2x
2
⇒ du 2x dx
(2) dv cos 2x dx ⇒
2
x 532
3x 10 C
15
x2
2
2
dx
1 4x2
1 sin2 x 1 cos x 1 dx
1
1
sin x 1 sin3 x 1 C
3
1
sin x 1
3 sin2 x 1 C
3
1
sin x 1
3 1 cos 2 x 1 C
3
1
sin x 1
2 cos 2 x 1 C
3
2x dx tan 2x 1 sec 2x dx
2
2
x
x
x
2 3 x
2
tan
2 tan
C tan3
3 tan
3
2
2
3
2
2
tan2
1
d 1 sin 2x sec 2x dx sec 2x dx
2
2
1 sin d cos2 sin 2x dx
1
x
1
x2 cos 2x sin 2x cos 2x C
2
2
4
23x 15 x 5 C
15.
x cos 2x dx
1
1
1
x2 cos 2x x sin 2x 2
2
2
2
4
xx 532 x 552 C
3
15
x 532
109
C
sec2 sec tan d tan sec C
110
23.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
12
dx x24 x2
24 cos d
4 sin2 2 cos x
csc2 3
2
d
θ
4 − x2
3 cot C
34 x2
C
x
x 2 sin , dx 2 cos d, 4 x2 2 cos x 2 tan 25.
x2 + 4
dx 2 sec2 d
x
θ
4 x2 4 sec2 x3
dx 4 x2
2
8 tan3 2 sec2 d
2 sec 8 tan3 sec d
8 sec2 1tan sec d
sec3 sec C
3
8
x2 432 x2 4
C
24
2
8
x2 4
13 x
2
4 4 C
1
8
x2x2 4 x2 4 C
3
3
1
x2 412x2 8 C
3
27.
4 x2 dx 2 cos 2 cos d
2
x
2 1 cos 2 d
θ
4 − x2
1
2 sin 2 C
2
2 sin cos C
2 arcsin
2x 2x
4 x2
2
C
1
x
4 arcsin
x4 x2 C
2
2
x 2 sin , dx 2 cos d, 4 x2 2 cos Review Exercises for Chapter 7
29. (a)
x3
4 x2
dx 8
sin3 d
cos4 (b)
x3
4 x2
8
sec sec2 3 C
3
4 x2
3
(c)
x3
dx x24 x2 4 x2
x
dx ⇒
4 x2
2x4 x2 dx
31.
v 4 x2
⇒ du 2x dx
u x2
x 28
A
B
x2 x 6 x 3 x 2
x 28 Ax 2 Bx 3
x 2 ⇒ 30 B5 ⇒ B 6
⇒ 25 A5
x3
33.
x 28
dx x2 6 6
⇒ A 5
5
6
dx 5 ln x 3 6 ln x 2 C
x3 x2
x2 2x
A
Bx C
2
x 1x2 1 x 1
x 1
x2 2x Ax2 1 Bx Cx 1
Let x 1: 3 2A ⇒ A 3
2
3
2
Let x 0: 0 A C ⇒ C Let x 2: 8 5A 2B C ⇒ B x2 2x
3
dx x x2 x 1
2
3
1
2
1
1
dx x1
2
1
1
dx x1
4
x3
dx
x2 1
2x
3
dx x2 1
2
1
dx
x2 1
3
2
1
3
3
ln x 1 ln x2 1 arctan x C
2
4
2
1
6 ln x 1 lnx2 1 6 arctan x C
4
4 x2
3
x2 8 C
u2 4 x2, 2u du 2x dx
4 x2 2
2
x24 x2 4 x232 C x 8 C
3
3
dv u2 4 du
u
u2 12 C
3
x2 8 C
x 2 tan , dx 2 sec2 d
1
u3 4u C
3
8 cos4 cos2 sin d
dx 111
112
35.
Chapter 7
x2
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
15 2x
x2
1 2
2x 15
x 2x 15
B
15 2x
A
x 3x 5 x 3 x 5
15 2x Ax 5 Bx 3
Let x 3:
9 8A ⇒ A 9
8
Let x 5: 25 8B ⇒ B x2
dx x2 2x 15
dx x
37.
9
8
25
8
25
1
dx x3
8
1
dx
x5
9
25
ln x 3 ln x 5 C
8
8
x
1
2
dx ln2 3x C
2 3x2
9 2 3x
39.
x
1
1
dx du
1 sin x2
2 1 sin u
1
tan u sec u C
2
(Formula 4)
1
tan x2 sec x2 C
2
41.
x
1
1
dx ln x2 4x 8 4 2
dx
x2 4x 8
2
x 4x 8
(Formula 15)
1
2
2x 4
ln x2 4x 8 2
arctan
32 16
32 16
2
43.
C
(Formula 14)
1
x
ln x2 4x 8 arctan 1 C
2
2
1
1
1
dx dx
sin x cos x
sin x cos x
1
ln tan x C
u x
(Formula 58)
45. dv dx
⇒
vx
1
u ln xn ⇒ du nln xn1 dx
x
ln xn dx xln xn n ln xn1 d x
47.
sin cos d 1
2
sin 2 d
1
1
cos 2 4
4
dv sin 2 d ⇒
u
1
v cos 2
2
⇒ du d
1
1
1
cos 2 d cos 2 sin 2 C sin 2 2 cos 2 C
4
8
8
u x2
(Formula 56)
Review Exercises for Chapter 7
49.
x14
uu3
dx 4
du
1 x12
1 u2
4
4
u2 1 13 u
3
51.
1 cos x dx 1
du
u2 1
sin x
1 cos x
113
dx
1 cos x12sin x dx
21 cos x C
u arctan u C
u 1 cos x, du sin x dx
4
x34 3x14 3 arctanx14 C
3
4
x , x u4, dx 4u3 du
y 53.
cos x lnsin x dx sin x lnsin x 3 x3
9
dx ln
C
x2 9
2 x3
(by Formula 24 of Integration Tables)
sin x lnsin x sin x C
dv cos x dx ⇒
v sin x
u lnsin x ⇒ du 57. y 55. y cos x dx
cos x
dx
sin x
lnx2 xdx x ln x2 x x ln x 2 x x ln x2 x 5
2x2 x
dx
x2 x
59.
xx2 432 dx 2
15 x
2
452
5
2
1
5
2x 1
dx
x1
2dx 1
dx
x1
x ln x2 x 2x ln x 1 C
⇒
dv dx
vx
u lnx2 x ⇒ du 4
61.
1
ln x
1
dx ln x2
x
2
4
65. A 4
1
2x 1
dx
x2 x
1
ln 42 2ln 22 0.961
2
0
x4 x dx 0
4 u2u2u du
2
63.
0
2u4 4u2 du
y
2
u5 4u3 2
5
3
0
2
128
15
1
1
x x3
3
0
1
67. By symmetry, x 0, A .
2
0
x sin x dx x cos x sin x
1
2 1
2
1
x, y 0,
4
3
1 x22 dx
u 4 x, x 4 u2, dx 2u du
69. s 1 cos2 x dx 3.82
0
e2x
2e2x
4e2x
lim
lim
2
x → x
x → 2x
x → 2
73. lim
71. lim
x →1
ln x2
x 1 lim x →1
21xln x
0
1
75. y lim ln x2x
x →
ln y lim
x →
2x ln x
2 lnln x
0
lim
x →
x
1
Since ln y 0, y 1.
1
1
4
3