C H A P T E R 7
Integration Techniques, L’Hôpital’s Rule,
and Improper Integrals
Section 7.1
Basic Integration Rules . . . . . . . . . . . . . . . . . . . 308
Section 7.2
Integration by Parts . . . . . . . . . . . . . . . . . . . . . 312
Section 7.3
Trigonometric Integrals . . . . . . . . . . . . . . . . . . . 321
Section 7.4
Trigonometric Substitution . . . . . . . . . . . . . . . . . 328
Section 7.5
Partial Fractions
Section 7.6
Integration by Tables and Other Integration Techniques . . 343
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule
Section 7.8
Improper Integrals
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . 336
. . . . . . . . 348
. . . . . . . . . . . . . . . . . . . . . 353
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
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 Even-Numbered Exercises
2. (a)
d
x
2
lnx2 1 C 21 x2 2x
dx
1
x 1
(b)
d
2x
x2 122 2x2x2 12x_ 21 3x2
C 2
dx x2 12
x2 14
x 13
(c)
1
d
arctan x C dx
1 x2
(d)
2x
d
lnx2 1 C 2
dx
x 1
x
dx matches (a).
x2 1
4. (a)
d
2x sinx2 1 C 2xcosx2 12x 2 sinx2 1 22x2 cosx2 1 sinx2 1
dx
(b)
d
1
1
sinx2 1 C cosx2 12x x cosx2 1
dx 2
2
(c)
d 1
1
sinx2 1 C cosx2 12x x cosx2 1
dx 2
2
(d)
d
2x sinx2 1 C 2xcosx2 12x 2 sinx2 1 22x2 cosx2 1 sinx2 1
dx
x cosx2 1 dx matches (c).
6.
2t 1
dt
t2 t 2
u t 2 t 2, du 2t 1 dt
Use
12.
du
.
u
sec 3x tan 3x dx
u 3x, du 3 dx
Use
308
sec u tan u du.
8.
2
dt
2t 12 4
10.
u 2t 1, du 2dt, a 2
Use
2x
x2 4
u x2 4, du 2x dx, n du
.
u2 a2
Use
14.
1
dx
xx2 4
u x, du dx, a 2
Use
dx
du
.
u u2 a2
un du.
1
2
Section 7.1
309
18. Let u t 9, du dt.
16. Let u x 4, du dx.
Basic Integration Rules
6x 45 dx 6 x 45 dx 6
x 46
C
6
2
2
dt 2 t 92 dt C
t 92
t9
x 46 C
20. Let u 4 2x2, du 4x dx.
x4 2x2 dx 22.
x
3
dx 2x 32
1
4 2x2124xdx
4
1
4 2x232 C
6
x dx 3
2
x2 3 2x 31
C
2
2
1
3
x2
C
2
22x 3
24. Let u x2 2x 4, du 2x 1 dx.
x1
x2 2x 4
dx 1
x2 2x 4122x 1 dx
2
x2 2x 4 C
26.
28.
2x
dx x4
2 dx 8
dx 2x 8 ln x 4 C
x4
1
1
1
1
1
1
dx 3 dx 3 dx
3x 1 3x 1
3 3x 1
3 3x 1
30.
32.
x 1
1
x
3
sec 4x dx 1
4
x 1
1
1
1 3x 1
ln 3x 1 ln 3x 1 C ln
C
3
3
3 3x 1
3
1
3
2 3 dx x
x
x
x3
3
1
1
1
2 dx x2 3x 3 ln x C
x
x
2
x
sec4x4 dx
34. Let u cos x, du sin x dx.
1
ln sec 4x tan 4x C
4
sin x
cos x
dx cos x12sin x dx
2cos x C
2
36. Let u cot x, du csc x dx.
csc2 xecot x dx ecot xcsc2 x dx ecot x C
38.
3ex
5
dx 5
2
5
3ex
1
2
x
ee dx
x
ex
dx
3 2ex
5
1
2ex dx
2 3 2ex
5
ln 3 2ex C
2
2
2x 3 2 dx
310
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sin x
dx
cos x
40. Let u lncos x, du tan xln cos x dx 1 cos d sin csc d cot d
ln cos xtan x dx
2
1
2
dx 3sec x 1
3 sec x 1
sec x 1
sec x 1 dx
46.
2 sec x 1
dx
3
tan2 x
2 sec x
2
dx cot2 x dx
3 tan2 x
3
2 cos x
2
dx csc2 x 1 dx
3 sin2 x
3
2
1
2
2
cot x x C
3
sin x
3
3
3
dt 3 arctan t C
t2 1
2
csc x cot x x C
3
1
1
50. Let u , du 2 dt.
t
t
48. Let u 3 x, du 3 dx.
3
1
1
dx
dx 4 3x2
3 4 3x 2
3 x
1
arctan
C
2
23
52.
54.
56.
dy
tan22x, 0, 0
dx
1
x 14x2 8x 3
1
1 4x x2
dx dx 2
2x 12x 12 1
1
5 x2 4x 4
y
(a)
(b)
1
dx tan22xdx e1t
1
dt e1t 2 dt e1t C
t2
t
dx arcsec 2x 1 C
1
5 x 22
dx arcsin
sec22x 1 dx x 52 C
a 5
1
tan2x x C
2
1.2
0, 0: 0 C
x
−0.6
0.6
y
1
tan2x x
2
− 1.2
1.2
−1
58.
− 1.2
60. r (0, 1) 5
−2
2
−2
ln csc cot ln sin C
lncos x2
C
2
tan x dx
44.
42.
1 et2
dt et
1 2et e2t
dt
et
et 2 et dt et 2t et C
Section 7.1
62. Let u 2x, du 2 dx.
y
1
dx x4x2 1
Basic Integration Rules
64. Let u sin t, du cos t dt.
2
dx
2x2x2 1
sin2 t cos t dt 0
13 sin t
3
0
0
arcsec 2x C
66. Let u 1 ln x, du e
1
1 ln x
dx x
68.
1
e
1 ln x
1
4
0
72.
1
25 x2
dx arcsin
x
5
4
0
x2
dx x
e
1
arcsin
2
1
1
2
dx
x
1x dx
1
1 ln x2
2
70.
2
1
dx.
x
2
x 2 ln x
1
1 ln 4 0.386
1
2
4
0.927
5
6
4
x2
1
x2
dx lnx2 4x 13 arctan
C
x2 4x 13
2
3
3
`
−10
10
The antiderivatives are vertical translations of each other.
−6
74.
ex ex
2
3
dx 1 3x
e 9ex 9ex e3x C
24
sec u tan u du sec u C
76.
The antiderivatives are vertical translations of each other.
5
−5
5
−5
78. Arctan Rule:
du
4
1
arctan
C
a2 u2 a
a
1
82. f x x3 7x2 10x
5
5
f x dx < 0 because
80. They differ by a constant:
sec2 x C1 tan2 x 1 C1 tan2 x C.
2
84.
0
4
dx 4
x2 1
86. A sin 2x dx
0
Matches (d).
0
2
y
more area is below the x-axis
than above.
2
1
cos 2x
2
0
1
y
3
1
5
2
1
2
1
0
5
x
1
−5
2
3
4
x
π
4
π
2
311
312
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
2
88.
2 x2 dx
0
2
A
(c) Let f x x over the interval
0, 2. Revolve this region
about the y-axis.
(b) Let f x 2 x over the
interval 0, 2. Revolve this
region about the x-axis.
(a) Let f x 2 x2 over the
interval 0, 2.
2 x2 dx
V
V 2
2
2 x dx
0
y
f ( x) =
2π x 2
20
xx dx
0
2
2
25
2
2
0
2 x2
dx
2 x2 dx
0
0
y
15
y
10
3
f ( x) =
2x
2
f ( x) = x
5
x
1
2
2
1
1
x
1
x
1
90. (a)
(b)
1
n 0
n
sinnx dx 0
3
1
n
sinnxn dx 0
1
1
dx arctan x
3 3 31 x2
6
1
3
3
y n
1
cosnx
94.
1
0
2
y x 23
y x
1 y 2 1 1 x1
x
x
2x
0
2
3x13
1 y 2 1 x x 1 dx
9
S 2
3
1
arctan 3
3
y 2x
92.
2
2
4
9x 23
8
s
1
1
4
dx 7.6337
9x23
9
2
2x 1 dx
0
3x 1 4
2
32
9
0
8
1010 1 256.545
3
y
12
9
6
3
x
3
6
Section 7.2
2.
9
12
Integration by Parts
d 2
x sin x 2x cos x 2 sin x x2 cos x 2x sin x 2x sin x 2 cos x 2 cos x x2 cos x. Matches (d)
dx
Section 7.2
Integration by Parts
4.
1
d
x x ln x 1 x
ln x ln x. Matches (a)
dx
x
6.
x2 e2x dx
8.
u x2, dv e2x dx
12. dv ex dx ⇒
ln 3x dx
10.
u ln 3x, dv dx
v
ex dx ex
ux
⇒ du dx
2
x
dx 2 xex dx
ex
14.
x2 cos x dx
u x2, dv cos x dx
e1t
1
dt e1t 2 dt e1t C
t2
t
ex dx 2xex ex C
2 xex 2xex 2ex C
16. dv x 4 dx ⇒
⇒ du u ln x
x 4 ln x dx 20. dv v
⇒ du 2
1
dx x
ln x3
x5
x5 1
1 4
dx ln x x dx
5 x
5
5
x5
5 ln x 1 C
25
1
1
dx x2
x
1
dx
x
1
ln x 1
C
dx x2
x
x
v
x2 12 x dx 1
2
x2 1
⇒ du 2x3ex 2xex dx 2xex x2 1 dx
2
2
2
x3ex
x2ex
dx 2
2
1
2
x 1
x2
24. dv 1
dx ⇒ v x2
u ln 2x ⇒ du x
dx ⇒
x2 12
u x2ex
x5
1
ln x x5 C
5
25
ln x
ln x
dx x2
x
22. dv 18. Let u ln x, du 1
dx
x
x5
ln x 5
1
dx ⇒
x2
u ln x
x5
5
v
2
2
2
xex dx 2
2
1
1
dx x2
x
1
dx
x
ln
2x
ln
2x
dx x2
x
2
x2ex
ex
ex
C
C
2
2
2
x 1
2
2
x 1
ln
2x 1
1
ln
2x 1
dx C
C
x2
x
x
x
1
dx.
x
ln x3
1x dx 2
1
ln x
2
C
313
314
Chapter 7
26. dv Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1
2 3x
dx ⇒
2
2 3x12 dx 2 3x
3
⇒ du dx
ux
v
x
2 3x
2x2 3x 2
2 3x dx
3
3
dx 2x2 3x
4
22 3x
22 3x
2 3x32 C 9x 2
2 3x C 3x 4 C
3
27
27
27
28. dv sin x dx ⇒
v cos x
⇒ du dx
ux
x sin dx x cos x cos x dx x cos x sin x C
30. Use integration by parts twice.
(2) u x, du dx, dv sin x dx, v cos x
(1) u x2, du 2x dx, dv cos x dx, v sin x
x2 cos x dx x2 sin x 2 x cos x x2 cos x dx x2 sin x 2 x sin x dx
x2 sin x 2x cos x 2 sin x C
32. dv sec tan d ⇒
v
sec tan d sec
34. dv dx
⇒ du d
u
sec tan d sec v
sec d
v
(2) dv e x dx ⇒
e x dx e x
v
ex cos 2x dx ex cos 2x 2 ex sin 2x dx ex cos 2x 2 ex sin 2x 2 ex cos 2x dx
5 e x cos 2x dx ex cos 2x 2ex sin 2x
e x cos 2x dx 38. dv dx
⇒
ex
cos 2x 2 sin 2x C
5
vx
u ln x ⇒ du 1
dx
x
y ln x
y
ln x dx x ln x x
x
1 x2
e x dx e x
u sin 2x ⇒ du 2 cos 2x dx
u cos 2x ⇒ du 2 sin 2x dx
dx
dx
4 x arccos x 1 x2 C
36. Use integration by parts twice.
⇒
1
1 x2
4 arccos x dx 4 x arccos x sec ln sec tan C
(1) dv e x dx
dx x
u arccos x ⇒ du ⇒
1
dx x ln x x C x
1 ln x C
x
cos x dx
Section 7.2
40. Use integration by parts twice.
(1) dv x 1 dx ⇒
v
315
2
x 112 dx x 132
3
⇒ du 2x dx
u x2
(2) dv x 132dx ⇒
v
2
x 132 dx x 152
5
⇒ du dx
ux
y
Integration by Parts
x 2x 1 dx
2
4
4 2
2
2
x2
x 132 x
x 132 dx x 2
x 132 x
x 152 3
3
3
3 5
5
x 152 dx
2
x 132
2
8
16
x 2
x 132 x
x 152 x 172 C 15x2 12x 8 C
3
15
105
105
⇒
42. dv dx
u arctan
y
v
dx x
1
2
1
x
⇒ du dx dx
2
1 x22 2
4 x2
arctan
x
x
dx x arctan 2
2
y
44. (a)
(b)
4
x
2x
dx x arctan ln
4 x 2 C
4 x2
2
y
x
−6
dy
18
ex3 sin 2x, 0, dx
37
ex3 sin 2x dx
Use integration by parts twice.
4
(1) u sin 2x, du 2 cos 2x
−4
dv ex3 dx, v 3ex3
ex3 sin 2x dx 3ex3 sin 2x 6ex3 cos 2x dx
(2) u cos 2x, du 2 sin 2x
dv ex3 dx, v 3ex3
ex3 sin 2x dx 3ex3 sin 2x 6 3ex3 cos 2x 37 ex3 sin 2x dx 3ex3 sin 2x 18ex3 cos 2x C
y
ex3 sin 2x dx 1
3ex3 sin 2x 18ex3 cos 2x C
37
18
1
:
0 18 C ⇒ C 0
0, 18
37 37
37
y
1 x3
3e
sin 2x 18ex3 cos 2x
37
6ex3 sin 2x dx C
316
46.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
dy x
sin x, y
0 4
dx y
48. See Exercise 3.
(0, 4)
8
1
0
−5
x2e x dx x2e x 2xe x 2e x
1
0
e 2 0.718
10
−2
50. dv sin 2x dx ⇒
1
sin 2x dx cos 2x
2
v
⇒ du dx
ux
1
1
x sin 2x dx x cos 2x 2
2
52. dv x dx
Thus,
cos 2x dx
x arcsin x2 dx 1
1
x cos 2x sin 2x C
2
4
x sin 2x dx 0
v
u arcsin x2 ⇒ du 1
sin 2x 2x cos 2x C
4
⇒
14 sin 2x 2x cos 2x
0
.
2
2x
dx
1 x4
x3
dx
1 x 4
1
x2
arcsin x2 2
1 x 412 C
2
4
1 2
x arcsin x2 1 x 4 C
2
x arcsin x2dx 0
x2
2
x dx x2
arcsin x2 2
1
Thus,
1
2.
4
54. Use integration by parts twice.
(1) dv ex, v ex, u cos x, du sin x dx
ex cos x dx ex cos x ex sin x dx
(2) dv ex dx, v ex, u sin x, du cos x dx
ex cos x dx ex cos x ex sin x ex cos x dx ⇒ 2 ex cos x dx ex sin x ex cos x
Thus,
2
ex cos x dx 0
⇒
56. dv dx
x
e
e
2
sin x ex cos x
2
2
v
u ln
1 x2 ⇒ du 1
0
0
1
2
dx x
2x
dx
1 x2
x ln
1 x2 2
2
ln
1 x2dx x ln
1 x2 Thus,
sin 2 cos 2 2x2
dx
1 x2
1
1
dx x ln
1 x2 2x 2 arctan x C
1 x2
ln
1 x2 dx x ln
1 x2 2x 2 arctan x
1
0
ln 2 2 .
2
1 2
x arcsin x2 1 x 4
2
1
0
Section 7.2
58. u x, du dx, dv sec2 x dx, v tan x
x sec2 x dx x tan x 60.
4
tan x dx
2
4 ln 22 0
1
ln 2
4
2
x3 cos 2x dx x3
u and its
derivatives
x3
e2x
3x2
2e2x
6x
1 2x
4e
6
8e2x
0
1 2x
16 e
2
8
16
x2
x 232dx x2
x 252 x
x 272 x 292 C
5
35
315
76.
5
0
68. Yes.
u ln x, dv x dx
v and its
antiderivatives
u and its
derivatives
x3
3x2
1
2
6x
14
cos 2x
6
18
sin 2x
0
1
16
Alternate
signs
u and its
derivatives
x2
2x
2
5
x 252
2
4
35
x 272
0
8
315
x 292
2
x 252
35x2 40x 32 C
315
66. Answers will vary.
See pages 488, 493.
4 sin d 1
Alternate
signs
1
2
1
4x3 sin 2x 6x2 cos 2x 6x sin 2x 3 cos 2x C
8
74.
v and its
antiderivatives
Alternate
signs
12 sin 2x 3x 41 cos 2x 6x 81 sin 2x 6161 cos 2x C
3
3
3
1
x3 sin 2x x2 cos 2x x sin 2x cos 2x C
2
4
4
8
64.
4
0
1
1 2x
1
1
x 3e2x dx x 3 e2x 3x2 e2x 6x e2x 6
e
C
2
4
8
16
1
e2x
4x3 6x2 6x 3 C
8
62.
x sec x dx x tan x ln cos x
0
317
Hence,
Integration by Parts
70. No. Substitution.
cos 2x
sin 2x
cos 2x
v and its
antiderivatives
x 232
72. No. Substitution.
1
4 cos 4
3 sin 12
2 cos 24
sin 24 cos C
5
x 4
25 x232 dx 1,171,875 arcsin
x5 x
2x2 25
25 x252 625x
25 x232 46,875x25 x2
128
16
64
128
14,381.0699
5
0
318
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
78. (a) dv 4 x dx ⇒
v
2
4 x12 dx 4 x32
3
⇒ du dx
ux
x
2
2
4 x dx x
4 x32 3
3
4 x32 dx
4
2
2
x
4 x32 4 x52 C 4 x32
3x 8 C
3
15
15
(b) u 4 x ⇒ x u 4 and dx du
x4 x dx u 4u12du u32 4u12 du
8
2
2
u52 u32 C u32
3u 20 C
5
3
15
2
2
4 x323
4 x 20 C 4 x32
3x 8 C
15
15
80. (a) dv 4 x dx ⇒
v
4 x12 dx
82. n 0:
2
4 x32
3
⇒ du dx
ux
2
2
x4 x dx x
4 x32 3
3
n 1:
n 2:
4 x
32 dx
4
2
x
4 x32 4 x52 C
3
15
(b) u 4 x ⇒ x 4 u and dx du
4u12 u32 du
2
8
u32 u52 C
3
5
u xn
xex dx xex ex C xex ex dx
x2ex dx x2ex 2xex 2ex C
n 3:
x3ex dx x3ex 3x2ex 6xex 6ex C
x3ex 3 x2e x dx
n 4: x 4ex dx
x 4ex 4x3ex 12x2ex 24xex 24ex C
x4 x dx 4 uu du
84. dv cos x dx ⇒
ex dx ex C
x2ex 2 xex dx
2
4 x325x 2
4 x C
15
2
4 x32
3x 8 C
15
2 32
u 20 3u C
15
2
4 x3220 3
4 x C
15
2
4 x32
3x 8 C
15
v sin x
⇒ du n x n1 dx
x n cos x dx xn sin x n xn1 sin x dx
x 4ex 4 x3ex dx
In general, x nex dx xnex nx n1ex dx.
(See Exercise 86)
1
86. dv eax dx ⇒ v eax
a
u x n ⇒ du nx n1 dx
x neax dx x neax n
a
a
x n1eax dx
Section 7.2
Integration by Parts
88. Use integration by parts twice.
1
v eax
a
(1) dv eax dx ⇒
u cos bx ⇒ du b sin bx
eax cos bx dx Therefore, 1 eax cos bx b
a
a
u sin bx ⇒ du b cos bx
eax sin bx dx eax cos bx beax sin bx b2
2
a
a2
a
b2
a2
e
ax
1
v eax
a
(2) dv eax dx ⇒
eax cos bx b eax sin bx b
a
a
a
a
eax cos bx dx
eax cos bx dx
eax
a cos bx b sin bx
a2
cos bx dx eax
a cos bx b sin bx
C.
a2 b2
eax cos bx dx 90. n 2 (Use formula in Exercise 84.)
x2 cos x dx x2 sin x 2 x sin x dx (Use formula in Exercise 83.) n 1
x2 sin x 2 x cos x cos x dx x2 sin x 2x cos x 2 sin x C
92. n 3, a 2 (Use formula in Exercise 86 three times.)
x3e2x dx x3e2x 3
2
2
1
A
9
1
9
x2e2x dx n 3, a 2
x3e2x 3 x2e2x
2
2 2
x3e2x 3x2e2x 3 xe2x 1
2
4
2 2
2
e2x 3
4x 6x2 6x 3 C
8
xe2x dx n 2, a 2
94. dv ex3 dx ⇒
ux
v 3ex3
e2x dx x3e2x 3x2e2x 3xe2x 3e2x
C n 1, a 2
2
4
4
8
96. A x3
xe
0
⇒ du dx
3
See Exercise 83.
dx
3
0
3xe 3 e
x3
3
3
x3
0
0
1 9
9ex3
9 e
dx
0
1 1
2
1 1 0.264
e
e
e
0.4
−1
4
− 0.1
−1
4
3
−1
x sin x dx x cos x sin x
0
319
320
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
98. In Example 6, we showed that the centroid of an equivalent region was
1, 8. By symmetry, the centroid of this region is 8, 1.
π
2
You can also solve this problem directly.
1
A
0
arcsin x dx x x arcsin x 1 x2
2
2
My
A
x
Mx
A
y
1
x
0
1
0
y
1
0
Example 3
x
1
2 2 0 1 1
arcsin x dx 2
8
2 arcsin x arcsin x dx 1
2
2
2
100. (a) Average 1
4
(b) Average 2
1.6t ln t 1 dt 0.8t2 ln t 0.4t2 t
1
1.6t ln t 1 dt 0.8t2 ln t 0.4t2 t
3
4
3
3.2
ln 2 0.2 2.018
12.8
ln 4 7.2
ln 3 1.8 8.035
102. c
t 30,000 500t, r 7%, t1 5
5
30,000 500te0.07t dt 500
P
0
5
60 te0.07t dt
0
Let u 60 t, dv e0.07t dt, du dt, v 60 t 1007 e P 500
0.07t
5
0
100 0.07t
e
.
7
5
100
7
e0.07t dt
0
e
$131,528.68
60 t 1007 e 10,000
49
500
104.
0.07t
5
0.07t
0
0
x 2 cos nx dx 5
x2
2
2x
sin nx 2 cos nx 3 sin nx
n
n
n
2
2
cos n 2 cos
n
n2
n
4
cos n
n2
44nn ,, ifif nn isis even
odd
1n 4
n2
2
2
106. For any integrable function, f x dx C f x dx, but this cannot be used to imply that C 0.
2 , sin x ≤ 1 ⇒ x sin x ≤ x ⇒ 108. On 0,
2
0
x sin x dx ≤
2
0
x dx.
Section 7.3
Trigonometric Integrals
321
110. fx cosx, f 0 2
(a) It cannot be solved by integration.
Section 7.3
(b) You obtain the points
4
n
xn
yn
0
0
2
1
0.05
2.05
2
0.10
2.098755
3
0.15
2.146276
80
2.8403565
4.0
0
4
0
Trigonometric Integrals
2. (a) y sec x ⇒ y sec x tan x sin x sec2 x.
(b) y cos x sec x ⇒ y sin x sec x tan x
sin x sec2 x sin x
Matches (iii)
sin x1 sec2 x
sin x tan2 x Matches i
(c) y x tan x 1 3
tan x ⇒ y 1 sec2 x tan2 x sec2 x
3
tan2 x tan2 x1 tan2 x
tan4 x Matches iv
(d) y 3x 2 sin x cos3 x 3 sin x cos x ⇒
y 3 2 cos xcos3 x 6 sin x cos2 xsin x 3 cos2 x 3 sin2 x
3 2 cos4 x 6 cos2 x1 cos2 x 3 cos2 x 31 cos2 x
8 cos4 x Matches ii
4.
cos 3 x sin4 x dx cos x1 sin2 xsin4 x dx
sin4 x sin6 xcos x dx
sin5 x sin7 x
C
5
7
6. Let u cos x, du sin x dx.
sin3 x dx 1
x
x
8. Let u sin , du cos dx.
3
3
3
cos3
x
dx 3
cos
3
x
3
1 sin 3x dx
1 sin2
2
x
3
13 cos 3x dx
3 sin
x
1
x
sin3
C
3 3
3
3 sin
x
x
sin3 C
3
3
sin x1 cos2 x dx
cos2 xsin x dx 1
cos3 x cos x C
3
sin x dx
322
10.
Chapter 7
sin5 t
dt cos t
12.
sin2 2x dx Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
sin t1 cos2 t2cos t12 dt
sin t1 2 cos2 t cos4 tcos t12 dt
4
2
cos t12 2cos t32 cos t72
sin t dt 2cos t12 cos t52 cos t92 C
5
9
1 cos 4x
1
1
dx x sin 4x C
2
2
4
14.
sin4 2 d 1
4x sin 4x C
8
1 cos 4
2
1 cos 4
d
2
1
1 2 cos 4 cos2 4 d
4
1 cos 8
d
2
1
4
1 2 cos 4 1
4
3
1
2 cos 4 cos 8 d
2
2
1 3
1
1
sin 4 sin 8 C
4 2
2
16
1
1
3
sin 8 C
sin 4 8
8
64
16. Use integration by parts twice.
dv sin2 x dx 1 cos 2x
x
sin 2x 1
2x sin 2x
⇒ v 2
2
4
4
u x2 ⇒ du 2x dx
dv sin 2x dx ⇒
1
v cos 2x
2
⇒ du dx
ux
1
1
x2 sin2 x dx x22x sin 2x 4
2
2x2 x sin 2x dx
1
1
1
1
x3 x2 sin 2x x3 2
4
3
2
x sin 2x dx
1
1
1
1
1
x3 x2 sin 2x x cos 2x 6
4
2
2
2
cos 2x dx
1
1
1
1
x3 x2 sin 2x x cos 2x sin 2x C
6
4
4
8
1
4x3 6x2 sin 2x 6x cos 2x 3 sin 2x C
24
2
18. Let u sin x, du cos x dx.
2
cos5 x dx 0
20.
0
2
1 sin2 x2 cos x dx
0
2
1 2 sin2 x sin4 x cos x dx
0
sin x 8
15
sin2 x dx 2
2 3
1
sin x sin5 x
3
5
0
1
2
2
1 cos 2x dx
0
2
1
1
x sin 2x
2
2
0
4
Section 7.3
22.
sec22x 1 dx 1
tan2x 1 C
2
24.
sec6 3x dx 28.
30. Let u sec 2t, du 2 sec 2t tan 2t
32.
36.
26.
tan2 x dx sec2 x 1dx tan x x C
tan3 2t sec3 2t dt 34.
sec2
42. y sin2 x
cos2 x
2
1
1
tan 3x tan3 3x tan5 3x C
3
9
15
tan5 2x sec2 2x dx tan3 3x dx x
x
x 1
x
tan dx 2 tan
sec2
dx
2
2
2 2
2
tan2 x
sec5 x
1 2 tan2 3x tan4 3xsec2 3x dx
1
x
x
x
sec2
dx C
tan4
2
2
2
2
x
C
2
tan2
38.
1 tan2 3x2 sec2 3x dx
1
tan6 2x C
12
sec5 2t sec3 2t
C
10
6
sec4 2t sec2 2tsec 2t tan 2t dt
sec2
or
323
sec2 2t 1sec3 2t tan 2t dt
x
x
x 1
x
x
tan dx 2 sec
sec tan
dx
2
2
2 2
2
2
sec2
tan3
Trigonometric Integrals
sec2 3x 1tan 3x dx
1 3 sin 3x
1
tan 3x3 sec2 3x dx dx
3
3
cos 3x
1 2
1
tan 3x ln cos 3x C
6
3
x
C
2
cos2 d
2
2
cos5 x dx
40. s sin2 x cos3 x dx
sin2 x1 sin2 xcos x dx
sin2 x sin4 xcos x dx
sin 2
1
C
8
2
1
2 sin 2 C
16
sin3 x sin5 x
C
3
5
tan x sec4 x dx
tan12 xtan2 x 1sec2 x dx
tan52 x tan12 x sec2 x dx
2
2 72
tan x tan32 x C
7
3
sin2
1 cos 2
1
4
1 2cos d 1 cos
4
sin2 d 1
8
1 cos 2 d
2
d
324
Chapter 7
44. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
y
(b)
1
y
y
tan3 x
C
3
x
−1
1
( )
1
0, 4
−1
1
dy
sec2 x tan2 x, 0, dx
4
sec2 x tan2 x dx u tan x, du sec2 x dx
0, 41: 41 C ⇒ y 31 tan
3
46.
dy
3y tan2 x, y0 3
dx
8
48.
cos 4 cos3 d (0, 3)
−1
1
x
1
4
cos 4 cos 3 d
1
cos 7 cos d
2
sin 7 sin C
14
2
−2
50.
x
1
x
52. Let u tan , du sec2 dx.
2
2
2
sin4x cos 3x dx sin 4x cos 3x dx
1
2
1
1
cos x cos 7x C
2
7
sin x sin 7xdx
tan4
x
x
sec4 dx 2
2
csc2 3x cot 3x dx 1
3
56.
tan4
2
1
7 cos x cos 7x
C
14
54. u cot 3x, du 3 csc2 3x dx
cot3 t
dt csc t
cot 3x3 csc2 3x dx
1
cot2 3x C
6
x
x
x
tan2 1 sec2 dx
2
2
2
tan6
x
x
tan4
2
2
12 sec 2x dx
2
x
2
2 7x
tan tan5 C
7
2 5
2
cos3 t
dt sin2 t
cos t
dt sin2 t
1 sin2 tcos t
dt
sin2 t
cos t dt
1
sin t C
sin t
csc t sin t C
58.
60.
sin2 x cos2 x
dx cos x
1 sec t
dt cos t 1
1 2 cos 2 x
dx cos x
sec x 2 cos x dx lnsec x tan x 2 sin x C
cos t 1
dt
cos t 1 cos t
62.
0
3
0
sec2 ttan t dt 2
3 tan t
32
0
2
3
3
sec2 x 1 dx
0
3
tan x x
66.
4
tan2 x dx sec t dt ln sec t tan t C
64. Let u tan t, du sec2 t dt.
4
0
sin 3 cos d 1
2
3 3
sin 4 sin 2 d
1
1 1
cos 4 cos 2
2 4
2
0
Section 7.3
2
68.
sin2 x 1 dx 2
70.
sin2 x cos2 x dx 2
2
2
2
Trigonometric Integrals
325
2x
1 dx
1 cos
2
2
32 21 cos 2x dx 32 x 41 sin 2x
1
4x sin 4x
C
32
72.
2
3
2
tan31 x dx tan21 x
ln cos1 x C
2
2
2
−6
6
−3
3
−2
74.
−2
sec41 x tan1 x dx 2
sec41 x
C
4
76.
1 cos 2 d 0
2
− 3.5
3
1
3
2
4
3.5
−2
2
78.
sin6 x dx 0
2
1 5x
3
1
2 sin 2x sin 4x sin3 2x
8 2
8
6
0
5
32
80. See guidelines on page 500.
82. (a) Let u tan x, du sec2 x dx.
sec2 x tan x dx (b)
8
1 2
tan x C1
2
−4
Or let u sec x, du sec x tan x dx.
(c)
1
1
1
1
1
sec2 x C tan2 x 1 C tan2 x C tan2 x C2
2
2
2
2
2
84. Disks
y
Rx tan x
rx 0
V 2
1
1
2
4
2
tan x dx
− 12
0
2
4
−2
1
sec xsec x tan x dx sec2 x C
2
−1
4
sec x 1 dx
2
0
2 1 4
2 tan x x
0
1.348
4
π
8
π
4
x
2
2 2 sin 4 sin 2
0
326
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
2
86. (a) V cos2 x dx 0
(b) A 2
2
2
0
2
cos x dx sin x
0
1 cos 2x dx 0
1
x sin 2x
2
2
2
0
2
4
1
Let u x, dv cos x dx, du dx, v sin x.
x
2
0
y
1
2
1
4
0
2
2
sin x dx x sin x cos x
0
0
2
1
2
2
2
cos2 x dx
y
0
2
1
1 cos 2x dx
0
2
1
1
x sin 2x
4
2
x, y 2
x cos x dx x sin x
0
( π 2− 2 , π8 (
1
2
8
π
4
2 2, 8 88. dv cos x dx ⇒
π
2
x
v sin x
u cosn1 x ⇒ du n 1cosn2 x sin x dx
cosn x dx cosn1 x sin x n 1 cosn2 x sin2 x dx
cosn1 x sin x n 1 cosn2 x1 cos2 x dx
cosn1 x sin x n 1 cosn2 x dx n 1 cosn x dx
Therefore, n cosn x dx cosn1 x sin x n 1 cosn2 x dx
cosn x dx cosn1 x sin x n 1
n
n
cosn2 x dx.
90. Let u secn2 x, du n 2secn2 x tan x dx, dv sec2 x dx, v tan x.
secn x dx secn2 x tan x n 2 secn2 x tan2 x dx
secn2 x tan x n 2 secn2 xsec2 x 1 dx
secn2 x tan x n 2
secn x dx secn2 x dx
n 1 secn x dx secn2 x tan x n 2 secn2 x dx
secn x dx 92.
cos4 x dx 1
n2
secn2 x tan x n1
n1
cos3 x sin x 3
4
4
cos2 x dx secn2 x dx
cos3 x sin x 3 cos x sin x 1
4
4
2
2
dx
1
3
3
1
cos3 x sin x cos x sin x x C 2 cos3 x sin x 3 cos x sin x 3x
C
4
8
8
8
Section 7.3
94.
sin4 x cos2 x dx cos3 x sin3 x 1
6
2
cos2 x sin2 x dx
cos3 x sin x 1
cos3 x sin3 x 1
6
2
4
4
cos2 x dx
x
1
1 cos x sin x
1
C
cos3 x sin3 x cos3 x sin x 6
8
8
2
2
1
8 cos3 x sin3 x 6 cos3 x sin x 3 cos x sin x 3x
C
48
96. (a) n is odd and n ≥ 3.
2
cosn x dx 0
cosn1 x sin x
n
2
0
2
n1
n
n1
n
n 2 n1
n
n2n4
n1
n
n2n4...
cosn3 x sin x
n2
n3
n3
n3
2
0
n3
n2
cosn5 x sin x
n4
n5
cosn2 x dx
0
2
0
0
n5
2
cos x dx
0
n3
n5
n2n4...1
0
(Reverse the order)
234567. . .n n 1
234567. . .n n 1
n1
n
n3
n5
n2n4...
n1
n
n3
2
cos2 x dx
(From part (a).)
0
n5
2
n 2 n 4 . . . 2 4 sin 2x 0
n1
n
2 123456. . .n n 1
123456. . .n n 12 n3
n5
x
n2n4... 4
1
(Reverse the order)
cosn6 x dx
0
2
(b) n is even and n ≥ 2.
0
n5
n4
2
n1
n
cosn x dx cosn4 x dx
0
cosn6 x dx
2
2
2
n n 1 nn 32 nn 54 . . . sin x
1
n1
n
Trigonometric Integrals
327
328
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.4
2.
Trigonometric Substitution
1
x
1
d
xx2 16 1 1 x
8 ln x2 16 x xx2 16 C 8
x2 16
2
2
dx
2
2
2
x 16 x
x 16
4.
x2
x2 16
x2 16
8 x x2 16 x2
2
2
2
x 16 x 16 x
2x 16
16 x2 x2 16
2x2 16
Indefinite integral:
2
x2
x2 16
Matches (d)
x 3 x 37 6x x2
1
d
8 arcsin
C 8
dx
4
2
1 x 342
8
16 x 3
2
4 2 x 3
1
3x
1
7 6x x2
16 x 32
x 32
2
2
216 x 3
16 x2 6x 9 16 x2 6x 9
216 x 32
216 x 32
216 x 32
16 x 32
Indefinite integral:
7 6x x2
7 6x x2 dx
Matches (c)
6. Same substitution as in Exercise 5.
10
x225
x2
dx 10
5 cos d
25
sin2
5 cos 2
2
225 x2
csc2 d cot C C
5
5
5x
8. Same substitution as in Exercise 5
x2
dx 25 x2
25 sin2 25
5 cos d 5 cos 2
1
25
25
sin 2 C sin cos C
2
2
2
25
x
x
arcsin
2
5
5
25 x2
5
10. Same substitution as in Exercise 9
x2 4
x
1 cos 2 d
dx 2 tan 2 sec tan d 2
2 sec 2tan C 2
x2 4
2
C 21 25 arcsin
5x x
25 x2 C
tan2 d 2 sec2 1 d
arcsec
1
7 6x x2
2
2x C x
2
4 2 arcsec
2x C
Section 7.4
12. Same substitution as in Exercise 9
x3
dx 4
x2
8 sec3 2 sec tan d 8 sec4 d
2 tan 8 1 tan2 sec2 d 8 tan 8 x2 4
3
2
3 x 4 4 C 31
2
14. Same substitution as in Exercise 13.
Trigonometric Substitution
tan3 8
C tan 3 tan2 C
3
3
1
3
x2 4 12 x2 4 C x2 4 x2 8 C
9x3
tan3 sec3 dx 9
sec2 d 9 sec2 1sec tan d 9
sec C
2
1 x
sec 3
3 sec sec2 3 C 31 x2 1 x2 3 C 31 x2x2 2 C
16. Same substitution as in Exercise 13
x2
dx 1 x22
x2
dx 1 x2 4
1
2
tan2 sec2 d
sec4 1 cos 2 d sin2 d
1
sin 2
1
sin cos C
2
2
2
x
1
arctan x 2
1 x2
1
1 x2
C 21 arctan x 1 x x C
2
18. Let u x, a 1, and du dx.
1 x2 dx 20.
1
x1 x2 ln x 1 x2 C
2
x
1
dx 9 x2122x dx
2
9 x
2
9 x212 C
22.
1
25 x2
dx arcsin
x
C
5
(Power Rule)
24. Let u 16 4x2, du 8x dx.
x16 4x2 dx 1
8
16 4x2128x dx 1
2
16 4x232 C 4 x232 C
12
3
26. Let u 1 t 2, du 2t dt.
t
1
dt 1 t 232
2
28. Let 2x 3 tan , dx 4x2 9
x4
1 t 2322t dt C
3
sec2 d, 4x2 9 3 sec .
2
dx 8
9
8
C
27 sin3 3 sec 32 sec2 d
324 tan4 1
1 t 2
cos d
sin4 8
csc3 C
27
4x2 932
C
27x3
4x 2 + 9
2x
θ
3
329
330
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
30. Let 2x 4 tan , dx 2 sec2 d, 4x2 16 4 sec .
1
dx x4x2 16
1
4
2 sec2 d
2 tan 4 sec sec 1
d tan 4
x2 + 4
x
csc d
θ
2
1
1 x2 4 2
ln csc cot C ln
C
4
4
x
32. Let x 3 tan , dx 3 sec2 d, x2 3 3 sec2 .
x2
1
dx 332
1
3
3 sec2 d
33 sec3 x2 + 3
x
cos d
θ
3
1
sin C
3
x
C
3x2 3
34. Let u x2 2x 2, du 2x 2 dx.
x 1x2 2x 2 dx 1
2
1
x2 2x 2122x 2 dx x2 2x 232 C
3
36. Let x sin , x sin2 , dx 2 sin cos d, 1 x cos .
1 x
x
dx cos 2 sin cos d
sin 1
2 cos2 d
x
θ
1− x
1 cos 2 d
sin cos C
arcsinx x1 x C
38. Let x tan , dx sec2 d, x2 1 sec2 .
1
x3 x 1
dx x 4 2x2 1
4
x4
4x3 4x
dx 2x2 1
1
lnx 4 2x2 1 4
1
1
lnx2 1 2
2
1
dx
x2 12
d
sec4 θ
1
1 cos 2 d
1
1
lnx2 1 sin cos C
2
2
1
x
lnx2 1 arctan x 2
C
2
x 1
x2 + 1
x
sec2 Section 7.4
40. u arcsin x, ⇒ du x arcsin x dx 1
x2
dx, dv x dx ⇒ v 2
2
1 x
x2
1
x2
arcsin x dx
2
2 1 x2
x sin , dx cos d, 1 x2 cos x arcsin x dx Trigonometric Substitution
x2
1
arcsin x 2
2
x2
1
sin2 cos d arcsin x cos 2
4
1 cos 2 d
1
1
1
x2
x2
arcsin x sin 2 C arcsin x sin cos C
2
4
2
2
4
1
1
x2
arcsin x arcsin x x1 x2 C 2x2 1 arcsin x x1 x2 C
2
4
4
42. Let x 1 sin , dx cos d, 1 x 12 2x x2 cos .
x2
2x x2
dx x2
1 x 12
dx
1
1 sin 2cos d
cos 1 2 sin sin2 x−1
θ
1 − (x − 1)2
d
3
1
2 sin cos 2 d
2
2
1
3
2 cos sin 2 C
2
4
1
3
2 cos sin cos C
2
2
1
3
arcsinx 1 22x x2 x 12x x2 C
2
2
3
1
arcsinx 1 2x x2x 3 C
2
2
44. Let x 3 2 sec , dx 2 sec tan d, x 32 4 2 tan .
x
dx x 6x 5
2
x
dx x 32 4
2 sec 3
2 sec tan d
2 tan 2 sec2 3 sec d
2 tan 3 ln sec tan C1
2
x 33 4
2
3 lnx 2 3 x 32 4
2
C1
x2 6x 5 3 ln x 3 x2 6x 5 C
331
332
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
46. Same substitution as in Exercise 45
(a)
1
dt 1 t252
32
Thus,
0
cos d
cos5 sec4 d tan2 1 sec2 d
1 3
1
t
tan tan C 3
3 1 t 2
t
1
t3
2 52 dt 1 t 31 t 232 1 t 2
3
t
1 t 2
C
32
0
32
338
3 3 23 3.464.
31432 14
(b) When t 0, 0. When t 32, 3. Thus,
32
0
1
1 3
dt tan tan 1 t252
3
48. (a) Let 5x 3 sin , dx 1
3 3 3 23 3.464.
3
3
cos d, 9 25x2 3 cos .
5
9 1 cos 2
d
5
2
0
3
3 cos cos d
5
9 25x2 dx 9
1
sin 2 C
10
2
9
sin cos C
10
5x 5x
9
arcsin
10
3
3
35
Thus,
3
9 25x2 dx 0
9 25x2
3
C
9
5x 5x9 25x2
arcsin
10
3
9
35
0
9 9
.
10 2
20
3
(b) When x 0, 0. When x , .
5
2
35
Thus,
9 25x2 dx 2
109 sin cos 0
0
9 9
.
10 2
20
50. (a) Let x 3 sec , dx 3 sec tan d, x2 9 3 tan .
x2 9
x2
dx 3 tan 3 sec tan d
9sec2 tan2 d
sec sin2 d
cos 1 cos2 d
cos sec cos d
ln sec tan sin C
ln
—CONTINUED—
x
3
x2 9
9
C
3
x
x2
Section 7.4
Trigonometric Substitution
333
50. —CONTINUED—
6
Hence,
x2 9
x2
3
dx ln
.
3
(b) When x 3, 0; when x 6, 6
Hence,
x2 9
x2
3
52.
54.
x2 9
x2 9
x
3
3
x
dx ln sec tan sin 3
0
6
3
ln 2 3 3
ln 2 3 2
3
2
.
.
1
75
x2 2x 1132 dx x 1x2 2x 26x2 2x 11 ln x2 2x 11 x 1 C
4
2
x2x2 4 dx 1 3 2
1
x x 4 xx2 4 2 ln x x2 4 C
4
2
56. (a) Substitution: u x2 1, du 2x dx
(b) Trigonometric substitution: x sec 58. (a) x2 y k2 25
1
(b) Area square circle
4
y
Radius of circle 5
k 2 52 52 50
1
25 52 25 1 4
4
(0, k)
k 52
5
1
(c) Area r 2 r 2 r 2 1 4
4
5
5
x
60. (a) Place the center of the circle at 0, 1; x2 y 12 1. The depth d satisfies 0 ≤ d ≤ 2. The volume is
V32
6
d
1 y 12 dy
0
1
arcsiny 1 y 11 y 12
2
d
(Theorem 7.2 (1))
0
3arcsind 1 d 11 d 12 arcsin1
(b)
3
3 arcsind 1 3d 12d d 2.
2
d
10
V6
(d)
1 y 12 dy
0
dV dV
dt
dd
0
dd
61 d 12
dt
1
dt 4
2
0
(c) The full tank holds 3 9.4248 cubic meters. The
horizontal lines
y
⇒ dt (e)
1
241 d 12
0.3
9
3
3
, y
, y
4
2
4
intersect the curve at d 0.596, 1. 0, 1.404. The dipstick would have these markings on it.
0
2
0
The minimum occurs at d 1, which is the widest
part of the tank.
334
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
62. Let x h r sin , dx r cos d, r 2 x h2 r cos .
y
Shell Method:
hr
V 4
xr 2 x h2 dx
x
hr
4
2
2
h2 2
4 r 2
2
1 cos 2 d r
2
2 r 2h
h r sin r cos r cos d 4 r 2
1
sin 2
2
2
2
h−r
2
4 r 3
h r sin cos2 d
sin cos2 d
3
h+r
2
2
cos3 h
2
r
2 2 r 2h
2
x−h
θ
r 2 − (x − h)2
1
64. y x2, y x, 1 y2 1 x2
2
4
s
1 x2 dx 0
12xx
2
4
1 ln x x2 1
(Theorem 7.2)
0
1
417 ln4 17 9.2936
2
66. (a) Along line: d1 a2 a4 a1 a2
y
Along parabola: y x2, y 2x
(a, a 2)
a
d2 1 4x2 dx
0
y = x2
a
1
2x4x2 1 ln 2x 4x2 1
4
(Theorem 7.2)
0
x
(0, 0)
1
2a4a2 1 ln2a 4a2 1 4
(b) For a 1, d1 2 and d2 5
2
1
ln2 5 1.4789.
4
For a 10, d1 10101 100.4988
d2 101.0473.
(c) As a increases, d2 d1 → 0.
68. (a)
(b) y 0 for x 72
25
− 10
80
−5
(c) y x x2
x
x
, y 1 , 1 y 2 1 1 72
36
36
72
s
1 1
0
36
2
x
36
2
72
dx 36
18 x
36
0
1 1
x
36
2
x
36
1 1
2 ln 1 2 1
2
ln 1 361 dx
2
x
36
1 1
36
2 ln 1 2
x
36
2 18 ln
2
72
0
2 1
2 1
82.641
Section 7.4
Trigonometric Substitution
y
70. First find where the curves intersect.
y 2 16 x 42 6
1 4
x
16
(4, 4)
4
2
16 2 16x 4 2 x4
x
−2
−2
16 2 16x 2 128x 162 x4
x4
16x2
2
4
0
−6
128x 0
1
1
1 2
x dx 42 x3
4
4
12
4
My x
0
x4
16
1 2
x dx 4
4
0
4 y=
16
4
3
8
x16 x 42 dx
4
8
4
0
8
x 416 x 42 dx 4
16 416 x 42 dx
4
13 16 x 4 8
2 32
4
2 16 arcsin
x4
x 416 x 42
4
1
16 1632 2 16
3
2
16
16 643 16 112
3
4
Mx 0
dx 8
2
1 1 2
x
2 4
4
1
16 x 42 dx
2
321 x5 8x x 6 4 64
416
32
64 32 5
6
15
5 4
3 8
0
4
x
My 1123 16 112 48 28 12
4.89
A
163 4
16 12
4 3
y
Mx
41615
104
1.55
A
163 4 54 3
x, y 4.89, 1.55
72. Let r L tan , dr L sec2 d, r 2 L2 L2 sec2 .
1
R
R
0
b
2mL
2mL
dr r 2 L232
R
2m
RL
a
L sec2 d
L3 sec3 b
cos d
RL sin 2m
RL
b
a
r
r 2 L2
2m
LR 2 L2
r 2 + L2
r
θ
L
a
2m
6
10
−4
xx 4x2 4x 32 ⇒ x 0, 4
A
4
R
0
8
4
16 − (x − 4)
2
335
336
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
0.8
74. (a) Finside 48
1
0.8 y21 y 2 dy
0.8
96 0.8
1
0.8
1 y2 dy 1
y1 y 2 dy
2 arcsin y y1 y 31 y 0.8
96
0.8
1
2
961.263 121.3 lbs
2 32
1
0.4
(b) Foutside 64
1
0.4 y21 y 2 dy
0.4
128 0.4
128
1
0.4
1 y 2 dy y1 y 2 dy
1
0.4
arcsin y y1 y 2 31 1 y 232
2
76. S 1520.4 111.2t 15.8t2
(a)
0.4
1
92.98
78. False
60
x2 1
x
dx 0
7
tan sec tan d
sec tan2 d
0
1
(b) St 1520.4 111.2t 15.8t212111.2 31.6t
2
S5 2.71
(c) Average value 80. True
1
1
St dt 68.24
x21 x2 dx 2
1
2
12
10
1
x21 x2 dx 2
0
Section 7.5
sin2 cos cos d 2
0
2
sin2 cos2 d
0
Partial Fractions
2.
4x 2 3
A
B
C
x 53 x 5 x 52 x 53
6.
2x 1
A Bx C
Dx E
2
2
xx 2 12
x
x 1
x 12
8.
1
1
A
B
4x 2 9 2x 32x 3 2x 3 2x 3
4.
1 A2x 3 B2x 3
3
2,
1
1
dx 4x2 9
6
2x 1 3 dx 2x 1 3 dx
1
ln 2x 3 ln 2x 3 C
12
1
2x 3
ln
C
12 2x 3
10.
x2
x2
A
B
x 2 4x 3 x 1x 3 x 1 x 3
x1
dx x2 4x 3
1
6.
When x 1 6A, A When x 32 , 1 6B, B 16 .
2
x 1
dx
x 1x 3
1
dx ln x 3 C
x3
Section 7.5
12.
Partial Fractions
337
5x2 12x 12 A
B
C
xx 2x 2
x
x2 x2
5x2 12x 12 Ax2 4 Bxx 2 Cxx 2
When x 0, 12 4A ⇒ A 3. When x 2, 16 8B ⇒ B 2. When x 2, 32 8C ⇒ C 4.
14.
5x2 12x 12
dx x3 4x
3
dx x
2
dx x2
4
dx 3 ln x 2 ln x 2 4 ln x 2 C
x2
x3 x 3
2x 1
A
B
x1
x1
x2 x 2
x 2x 1
x2 x1
2x 1 Ax 1 Bx 2
When x 2, 3 3A, A 1. When x 1, 3 3B, B 1.
x3 x 3
dx x2 x 2
16.
x1
1
1
dx
x2 x1
x2
x2
x ln x 2 ln x 1 C x ln x 2 x 2 C
2
2
x2
A
B
xx 4 x 4
x
18.
2x 3
A
B
x 12 x 1 x 12
x 2 Ax Bx 4
2x 3 Ax 1 B
When x 4, 6 4A, A 32 .
When x 0, 2 4B, B 12 .
x2
dx x 2 4x
When x 1, B 1. When x 0, A 2.
32
12
dx
x4
x
2x 3
dx x 12
2
1
dx
x 1 x 12
2 ln x 1 3
1
ln x 4 ln x C
2
2
20.
1
C
x1
4x2
4x2
4x2
A
B
C
x3 x2 x 1 x2x 1 x 1 x2 1x 1 x 1 x 1 x 12
4x2 Ax 12 Bx 1x 1 Cx 1
When x 1, 4 2C ⇒ C 2. When x 1, 4 4A ⇒ A 1. When x 0, 0 1 B 2 ⇒ B 3.
x3
4x2
dx x1
x2
1
dx x1
3
dx x1
ln x 1 3 ln x 1 22.
2
dx
x 12
2
C
x 1
6x
A
6x
Bx C
x3 8 x 2x2 2x 4 x 2 x2 2x 4
6x Ax2 2x 4 Bx Cx 2
When x 2, 12 12A ⇒ A 1. When x 0, 0 4 2C ⇒ C 2. When x 1, 6 7 B 21 ⇒ B 1.
6x
dx x3 8
1
dx x2
x 2
dx x2 2x 4
1
dx x2
3 arctan
x 1
dx x2 2x 4
1
3
x1
arctan
C
ln x2 2x 4 3
2
3
1
ln x2 2x 4 2
ln x 2 ln x 2 3x 1
3
C
3
dx
x2 2x 1 3
338
24.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x2 x 9 Ax B
Cx D
2
2
x2 92
x 9
x 92
x2 x 9 Ax Bx2 9 Cx D
Ax3 Bx2 9A Cx 9B D
By equating coefficients of like terms, we have A 0, B 1, D 0, and C 1.
x2 x 9
x2 92
26.
1
dx x2 9
x
dx
x2 92
x
1
1
arctan
C
3
3
2x2 9
x 2 4x 7
A
Bx C
x 1x 2 2x 3 x 1 x 2 2x 3
x 2 4x 7 Ax 2 2x 3 Bx Cx 1
When x 1, 12 6A. When x 0, 7 3A C. When x 1, 4 2A 2B 2C. Solving these equations we have
A 2, B 1, C 1.
x 2 4x 7
dx 2
x3 x 2 x 3
1
dx x1
2 ln x 1 28.
x 1
dx
x 2 2x 3
1
ln x 2 2x 3 C
2
x 2 x 3 Ax B
Cx D
2
2
x 2 32
x 3
x 32
x 2 x 3 Ax Bx 2 3 Cx D
Ax 3 Bx 2 3A Cx 3B D
By equating coefficients of like terms, we have A 0, B 1, 3A C 1, 3B D 3. Solving these equations we have
A 0, B 1, C 1, D 0.
x2 x 3
dx x4 6x 2 9
30.
1
x
dx
x 2 3 x 2 32
1
3
arctan
x
3
1
C
2x 2 3
x1
A
B
C
2
x 2x 1
x
x
x1
x 1 Axx 1 Bx 1 Cx 2
When x 0, B 1. When x 1, C 2. When x 1, 0 2A 2B C. Solving these equations we have A 2,
B 1, C 2.
5
1
x1
dx 2
x 1
x2
5
1
1
dx x
5
1
5
1
dx 2
x2
1
1
1
2 ln x 2 ln x 1
x
2 ln
2 ln
x
1
x1
x
5 4
3 5
5
1
1
dx
x1
5
Section 7.5
x2
x1
2
7
6x 2 1
1
dx 3 ln
C
x 2x 13
x
x
x 1 2x 12
1
32.
0
34.
x2 x
dx x1
1
0
0
x2
2x 1
dx x ln x 2 x 1
x1
0 1 ln 3
1
2, 1: 3 ln
1
dx Partial Fractions
4
1
1 2 7
1
C 1⇒ C 2 3 ln
2
2 1 2
2
(2, 1)
− 4.7
4.7
−4
36.
x2
3, 4:
x3
1
2
C
dx ln x 2 4 2
4 2
2
x 4
8
2
22 1
1
ln 5 C 4 ⇒ C ln 5
2
5
5
2
(3, 4)
−8
8
−2
38.
x2x 9
1
5
dx 2 ln x 2 C
x 3 6x 2 12x 8
x 2 x 22
10
3, 2: 0 1 5 C 2 ⇒ C 4
(3, 2)
− 10
10
−3
40.
x2 x 2
dx arctan x ln x 1 C
x3 x2 x 1
10
2, 6: arctan 2 0 C 6 ⇒ C 6 arctan 2
(2, 6)
−2
5
−2
42. Let u cos x, du sin x dx.
44.
1
A
B
uu 1
u
u1
1 Au 1 Bu
1 Au 1 Bu
When u 0, A 1. When u 1, B 1, u cos x.
du sin dx.
sin x
1
dx du
cos x cos2 x
uu 1
1
A
B
, u tan x, du sec2 x dx
uu 1
u
u1
1
du
u
ln u 1 ln u C
ln
u1
C
u
ln
cos x 1
C
cos x
sec2 x dx
tan xtan x 1
1
du u1
When u 0, A 1.
When u 1, 1 B ⇒ B 1.
ln 1 sec x C
1
du
uu 1
1
1
du
u u1
ln u ln u 1 C
ln
u
C
u1
ln
tan x
C
tan x 1
339
340
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
46. Let u ex, du ex dx.
1
A
Bu C
2
u2 1u 1 u 1
u 1
1 Au2 1 Bu Cu 1
When u 1, A 12 .
When u 0, 1 A C.
1
1
1
When u 1, 1 2A 2B 2C. Solving these equations we have A 2 , B 2 , C 2 , u e x, du e x dx.
48.
e
2x
ex
dx 1e x 1
1
2
1
1
ln u 1 ln u2 1 arctan u C
2
2
1
2 ln e x 1 ln e 2x 1 2 arctan e x C
4
1
du
u2 1u 1
u 1 1 du uu 11 du
2
1
A
B
a2 x2 a x a x
1
A
C
B
2
x2a bx
x
x
a bx
50.
1 Axa bx Ba bx Cx 2
1 Aa x Ba x
When x 0, 1 Ba ⇒ B 1a.
When x ab, 1 Ca2b2 ⇒ C b2a2.
When x 1, 1 a bA a bB C ⇒
When x a, A 12a.
When x a, B 12a.
1
1
dx a2 x2
2a
1
1
dx
ax ax
1
ln a x ln a x C
2a
A ba 2.
1
dx x a bx
2
1
ax
ln
C
2a a x
52.
dy
4
, y0 5
dx x2 2x 3
54. (a)
8
(b)
−1
56. A 2
0
y
3
2
3
dx 14
0
2x 6
0
5
2
2
3
2
1
dx
16 x2
14 4 x
ln
8
4x
7
ln 7 2.595
4
3
0
x
(From Exercise 46)
−3 −2 −1
1
b
b
ln x ln a bx C
a2
ax a2
a bx
b
1
ln
C
ax a2
x
1
x
b
ln
C
ax a2 a bx
Nx
A1 B1x
An Bn x
. . .
Dx ax2 bx c
ax2 bx cn
−2
7
1
dx
16 x2
ba2 1a
b2a2
2 dx
x
x
a bx
Nx
A1
A2
Am
. . .
Dx px q px q2
px qm
3
3
1
2
3
Section 7.5
58. (a)
Partial Fractions
341
(e) k 1, L 3
y
6
i y0 5: y 5
4
15
5 2e3t
1
32
3
ii y0 : y 2
12 52e3t 1 5e3t
3
2
1
5
t
1
2
3
4
5
6
(b) The slope is negative because the function is
decreasing.
(c) For y > 0, lim yt 3.
0
B
A
dy
yL y
y
Ly
(d)
dy
kyL y
dt
(f )
dy
dy
d 2y
k y
L y
0
dt 2
dt
dt
1
1
1 AL y By ⇒ A , B L
L
dy
yL y
1
L
1
dy y
5
0
t →
⇒ y
k dt
1
dy Ly
⇒
L
2
From the first derivative test, this is a maximum.
1
ln y ln L y kt C1
L
ln
y
k L t LC1
Ly
C2ekLt When t 0,
y
Ly
y
y0
y0
C2 ⇒
ekLt.
L y0
L y L y0
y0L
.
y0 L y0ekLt
Solving for y, you obtain y 3
60. (a) V 0
2x 2
dx 4
2
x 1
3
0
3
4
0
x2
x2
dx
12
1
1
dx
x2 1 x2 12
(partial fractions)
1
x
arctan x 2
2
x 1
x
x2 1
4 arctan x 2 arctan x —CONTINUED—
dy
dy
L y
dt
dt
y
k dt
3
0
3
(trigonometric substitution)
0
2 arctan 3 3
5.963
10
342
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
60. —CONTINUED—
3
(b) A 0
x
1
A
x2
3
0
2x2
1
dx 2
x 1
ln 10
2
ln 10
3
0
3
0
0
ln 10
3
2
0
3
1
2x 2 arctan x
ln 10
1 1
y
A 2
3
2x
dx lnx2 1
1
0
y
2
2
dx
x2 1
(1.521, 0.412)
1
2
3 arctan 3 1.521
ln 10
2
2x 2
dx x2 1
ln 10
(partial fractions)
1
x
2
arctan x arctan x 2
ln 10
2
x 1
2 1
x
arctan x ln 10 2
2x2 1
3
0
3
x2
dx
x2 12
1
1
dx
x2 1 x2 12
2
−1
3
0
x
1
3
(trigonometric substitution)
0
1
x
arctan x 2
ln 10
x 1
3
0
1
3
arctan 3 0.412
ln 10
10
x, y 1.521, 0.412
B
1
1
A
1
,A
,B
y0 xz0 x y0 x z0 x
z0 y0
z0 y0
62. (a)
1
z0 y0
(Assume y0 z0)
1
1
dx kt C
y0 x z0 x
1
z x
ln 0
kt C, when t 0, x 0
z0 y0 y0 x
C
z
1
ln 0
z0 y0 y0
kt
1
z x
z
ln 0
ln 0
z0 y0
y0 x
y0
y0z0 x
z0 y0k t
0 0 x
z y
ln
y0z0 x
ez0 y0kt
z0 y0 x
x
(b) (1) If y0 < z0, lim x y0.
t →
(2) If y0 > z0, lim x z0.
t →
y0z0ez0 y0kt 1
z0ez0 y0kt y0
(c) If y0 z0, then the original equation is
1
dx y0 x 2
k dt
y0 x1 kt C1
x 0 when t 0 ⇒
1
C1
y0
kt y0 1
1
1
kt y0 x
y0
y0
y0 x y0
kt y0 1
x y0 y0
kt y0 1
As t → , x → y0 x0.
Section 7.6
Section 7.6
Integration by Tables and Other Integration Techniques
Integration by Tables and Other Integration Techniques
2. By Formula 13: b 2, a 5
2
2 1
4
1
5 4x
x
dx ln
3 x22x 52
3 25 x5 2x 5 2x 5
2 4x 5
x
8
ln
C
375 2x 5
75 x2x 5
4. By Formula 29: a 3
C
6. By Formula 41:
1
1 x 9
x
dx x2 9 arcsec
C
3
x
3
3
2
8. By Formulas 51 and 47:
cos3 x
dx 2
x
u x, du 10. By Formula 71:
1
1
dx 1 tan 5x
5
2
x
9 x 4
2
1 x dx
x sin x
3
1
dx
2
x
1 1
u ln cos u sin u C
5 2
1
5x ln cos 5x sin 5x C
10
u 5x, du 5 dx
12. By Formula 85:
a 21, b 2
ex2 sin 2x dx ex2
1
sin 2x 2 cos 2x C
14 4 2
4 x2 1
e
sin 2x 2 cos 2x C
17
2
14. By Formulas 90 and 91:
ln x3 dx xln x3 3 ln x2 dx
xln x3 3x2 2 ln x ln x2 C
xln x3 3ln x2 6 ln x 6 C
16. (a) By Formula 89:
x 4 ln x dx x5
x5 1 5
x ln x C
2 1 4 1 ln x C 5
25
5
(b) Integration by parts: u ln x, du x 4 ln x dx x5
ln x 5
x5
1
dx, dv x 4 dx, v x
5
x5 1
x5
x5
dx ln x C
5 x
5
25
1
2x
dx
2 32 x 22
1
x2
arcsin C
2
3
2
1
2
cos x
dx sin
x cos 2 x 2 C
3
3
2
x
1
5 dx
1 tan 5x
dx cos3 x
cos
2
343
344
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
18. (a) By Formula 24: a 75, x u, and
(b) Partial fractions:
1
1
x 75
dx ln
C
x 2 75
2
75 x 75
3
30
ln
A
1
B
x 2 75 x 75 x 75
1 A x 75 B x 75 x 75
C
x 75
x 75: 1 2A
75 ⇒ A 3
1
1
30
2
75 10
3
x 75: 1 2B
75 ⇒ B 1
dx x 2 75
20. By Formula 21:
22. By Formula 79:
x
1 x
30
330
330
dx
x 75 x 75
3
30
3
ln
x 75
C
x 75
2
dx 2 x
1 x C
3
arcsec 2x dx 1
2x arcsec 2x ln 2x 4x 2 1 C
2
u 2x, du 2 dx
24. By Formula 52:
26. By Formula 7:
28. By Formula 14:
x sin x dx sin x x cos x C
1 sin
3
1
2
2x 2
dx arctan
C arctanx 1 C
x 2 2x 2
2
4
30. By Formula 56:
2
C
x2
1
25
10 ln 3x 5
dx 3x 3x 52
27
3x 5
d 32. By Formula 71:
1
1
3 2 d
3 1 sin 3
1
dt t 1 ln t2
u ln t, du u e x, du e x dx
1
tan 3 sec 3 C
3
34. By Formula 23:
ex
1
dx e x ln cos e x sin e x C
1 tan e x
2
1
1
dt arctanln t C
1 ln t2 t
1
dt
t
36. By Formula 26:
1
x
x 2 3 3 ln x x 2 3 C
2
38. By Formula 27:
1
3x2
2 2 3x2 3 dx
27
3 x 2 dx x 2
2 3x2 dx 1
3x18x 2 2
2 9x 2 4 ln 3x 2 9x 2 C
827
Section 7.6
40. By Formula 77:
2
3
x arctanx32 dx 42. By Formula 45:
arctanx32
Integration by Tables and Other Integration Techniques
32
x dx
2 32
x arctanx32 ln 1 x3 C
3
ex
ex
dx C
2x
32
1 e 1 e 2x
u e x, du e x dx
44. By Formula 27:
2x 32
2x 32 4 dx 1
2x 32
2x 3)2 42 dx
2
1
2x 32x 32 2
2x 32 4 ln 2x 3 2x 32 4 C
8
u 2x 3, du 2 dx
46. By Formula 31:
cos x
sin2 x 1
dx ln sin x sin2 x 1 C
u sin x, du cos x dx
48.
3x
dx 3x
3x
9 x 2
3
1
dx 9 x 2
3 arcsin
50. By Formula 67:
dx
un
a bu
du x
9 x 2 C
3
du
a bu
2
, v a bu
b
a bu
2n n1
2u n
a bu u a bu du
b
b
a bu
2n au n1 bu n
2u n
a bu du
b
b
a bu
2u n
2na
u n1
un
a bu du 2n
du
b
b a bu
a bu
54.
2u n
2n n1
a bu u a bu du
b
b
Therefore, 2n 1
tan3 d x
dx
9 x 2
52. Integration by parts: w u n, dw nu n1 du, dv un
a bu
un
a bu
du 2 n
un1
u a bu na
du and
b
a bu
2
un1
un
a bu na
du .
2n 1b
a bu
uncos u du un sin u n un1sin u du
w un, dv cos u du, dw nun1 du, v sin u
tan2 2
tan d
tan2 ln cos x C
2
345
346
56.
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
ln un du uln un nln un1
1uu du uln u
w ln un, dv du, dw nln un1
58.
x
x 2 2x dx 0, 0:
n
n ln un1 du
1u du, v u
1
2x 2 2x32 3x 1
x 2 2x 3 ln x 1 x 2 2x C
6
15
1
3 ln 1 C 0 ⇒ C 0
6
−6
6
(0, 0)
−15
60.
2 2x x 2
x1
0, 2 :
dx 2 2x x2
3 ln
3 2 2x x 2
x1
3
(0, 2 )
C
2 3 ln 3 2 C 2 ⇒ C 3 ln 3 2
−2.5
1
−6
62.
0, 1: C 1 ⇒ y 64.
sin 1 sin 1 sin d ln
cos 1 sin 2 1 sin cos 1 sin 1 sin ln
2 1 sin cos C
10
1
sin sin d d
1 cos2 1 cos 2
−8
8
(0, 1)
−2
2
66.
0
1
d 3 2 cos arctancos C
1
0
1
2
0
u tan
68.
cos d 1 cos cos 1 cos d
1 cos 1 cos 70.
2
1
d sec tan 2u
1 u2
21 u2
3
1 u2
1
du
5u2 1
25 arctan 5 u
1
0
2
5
arctan 5
1
d
1cos sin cos cos cos2 d
sin2 csc cot cot2 d
ln 1 sin C
csc cot csc2 1 d
csc cot C
cos d
1 sin u 1 sin , du cos d
Section 7.6
2
72. A 0
1
2
0
1
2
2x dx
2
1 ex
1
4
2
1 2
2
x ln1 e x 2
1
1
4 ln1 e4 ln 2
2
2
347
y
x
2 dx
1 ex
2
Integration by Tables and Other Integration Techniques
x
1
2
0
0.337 square units
74. Log Rule:
1
du, u ex 1
u
78. Formula 16 with u e2x
76. Integration by parts
5
82. W 80. A reduction formula reduces an integral to the sum of a
function and a simpler integral. For example, see Formula
50, 54.
0
500x
26 x 2
dx
5
250
26 x 2122x dx
0
500
26 x 2
500 26 1
2049.51 ft lbs
84.
2
1
20
0
2500
5000
dt 1 e4.81.9t
1.9
2
0
1.9 dt
1 e4.81.9t
2500
4.8 1.9t ln1 e4.81.9t
1.9
2500
1 ln1 e 4.8 ln1 e4.8
1.9
2500
1e
3.8 ln
1.9
1 e4.8
86. (a)
401.4
20
k
6x 2ex2
dx 50
0
By trial and error, k 5.51897.
0
5.51897
(b)
0
88. True
6x 2ex2 dx
2
−1
5.52
0
5
0
348
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
Section 7.7
Indeterminate Forms and L’Hôpital’s Rule
1 ex
1
x →0
x
2. lim
1
−1
0.1
0.01
0.001
0.001
0.01
0.1
0.9516
0.9950
0.9995
1.00005
1.005
1.0517
x
f x
6x
4. lim
x →
3x2 2x
3.4641
exact: 63
5
1
10
102
103
104
105
f x
6
3.5857
3.4757
3.4653
3.4642
3.4641
x →1
(b) lim
x →1
8. (a) lim
x →0
x →
(b) lim
x →
0
100
0
2x2 x 3
2x 3x 1
lim
lim 2x 3 5
x →1
x →1
x1
x1
2x2 x 3
ddx2x2 x 3
4x 1
lim
5
lim
x
→1
x →1
x1
ddxx 1
1
sin 4x
sin 4x
lim 2
21 2
x →0
2x
4x
10. (a) lim
−2
x
6. (a) lim
(b) lim
x →0
sin 4x
4 cos 4x
ddxsin 4x
lim
lim
2
x →0
x →0
2x
ddx2x
2
2x 1
2x 1x2 0
lim
0
2
x
→
4x x
4 1x
4
2x 1
ddx2x 1
2
lim
lim
0
4x2 x x → ddx4x2 x
x → 8x 1
x2 x 2
2x 1
lim
3
x →1
x →1
x1
1
12. lim
14. lim
x →2
4 x2
x2
lim
x →2
lim
x →2
16. lim
x →0
ex 1 x
ex 1
lim
3
x →0
x
3x2
lim
x →0
ex
6x sin ax
a cos ax a
lim
x→0 sin bx
x→0 b cos bx
b
20. lim
x →
x1
1
lim
0
x2 2x 3 x → 2x 2
x2
2x
2
lim x lim x 0
x→ ex
x→ e
x→ e
28. lim
x4 x2
1
x
4 x2
2
18. lim ln x lim 2 ln x
x→1 x2 1
x→1 x2 1
lim
2x
2x
lim
1
1
x2
x→1
x→1
24. lim
1
arctan x 4
11 x2 1
lim
x →1
x →1
x1
1
2
22. lim
26. lim
x→ x3
3x2
lim
x→
x1
1
Section 7.7
30. lim
x→ x2
x
lim
x→
1 1x2
1
Indeterminate Forms and L’Hôpital’s Rule
32. lim
x2
x→ sin x
0
x
Note: Use the Squeeze Theorem for x > .
e x2
12e x2
lim
x→ x
x→ 1
ln x 4
4 ln x
4x
lim
lim
x→ x3
x→ x→ 3x2
x3
36. lim
34. lim
lim
x→ 4
0
3x3
38. (a) lim x3 cot x 0
(b) lim x3 cot x lim
(c)
x→0
40. (a) lim x tan
x→ x→0
x→0
1
sin x
1
≤
≤
x
x
x
x3
tan x
lim
x→0
3x2
sec2 x
0
(b) lim x tan
x→ 1
0
x
1
tan1x
lim
x x→ 1x
1
lim
x→ 0
1x2 sec21x
1x2
3
lim sec2
x→ −1
(c)
1x 1
2
1
10
−1
42. (a) lim e x x2x 1
x→0
(b) Let y lim e x x2x.
x→0
ln y lim
x→0
2 ln
ex
x
1
1
(b) Let y lim 1 .
x
x→ 1
x
x
x
x→ x
ln y lim
2e x 1e x x
lim
4
x→0
1
x→ x ln1 1x lim ln1 1x1x
x→
1 1x
1x2
Thus, ln y 4 ⇒ y e 4 54.598.
(c)
44. (a) lim 1 lim
x→ 60
1x2
lim
x→ 1
1
1 1x
Thus, ln y 1 ⇒ y e1 e. Therefore,
lim 1 x→ 0
1
x
x
e.
2
0
(c)
5
0
−1
10
349
350
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
48. (a) lim 3x 4
x4 00
46. (a) lim 1 x1x 0
x →
x→4
(b) Let y lim 3x 4
x4.
(b) Let y lim 1 x
1x.
x→ ln y lim
x→ lim
x→ x→4
ln y lim x 4ln3x 4
ln1 x
x
x→4
111 x 0
ln3x 4
1x 4
lim
1x 4
1x 42
x→4
x→4
Thus, ln y 0 ⇒ y e0 1.
Therefore, lim 1 x1x 1.
lim x 4
0
x →
(c)
lim
x→4
Hence, lim 3x 4
x4 1.
5
x→4
(c)
2
10
0
−1
4
7
0
2 x
x
50. (a) lim cos
x→0
00
2 x .
x
(b) Let y lim cos
x→0
x 1 4 x x41 1 x1
1
x1
lim
lim x 4
x 4
x 4
52. (a) lim
(b)
x →2
2
x →2
2
2
2
x →2
2 x
ln y lim x ln cos
x→0
lim
x →2
000
2 x
x
Hence, lim cos
x→0
(c)
lim
x →2
1.
(c)
2
1 2x 1 2x
1
4xx 1
0.25
2
4
2
− 0.25
0
3
0
54. (a) lim
x→0
(b) lim
x→0
10x x3 (c)
2
10x x3 lim 10xx 3 2
x→0
10
0
5
2
−20
56. (a)
(b) Let y sin xx, then ln y x lnsin x.
2
−
lim
x →0
2
−1
lnsin x
cos xsin x
x2
2x
lim
lim
lim
0
2
x →0
x →0 tan x
x →0 sec2 x
1x
1x
Therefore, since ln y 0, y 1 and lim sin xx 1.
x →0
1
8
Section 7.7
58. (a)
Indeterminate Forms and L’Hôpital’s Rule
x3
3x2
6x
6
lim 2x lim 2x lim 2x 0
2x
x→ e
x→ 2e
x→ 4e
x→ 8e
(b) lim
1
−1
5
−3
62. Let f x x 25 and gx x.
60. See Theorem 7.4.
64. lim
x3
3x2
6x
6
lim
lim
0
lim
e2x x→ 2e2x x→ 4e2x x→ 8e2x
66. lim
ln x2
2 ln xx
lim
x→ x3
3x2
x→ x→ x→ xm
mxm1
lim
enx x→ nenx
mm 1xm2
n2enx
lim
2 ln x
3x3
lim
lim
2x
2
lim 3 0
x→ 9x
9x2
. . . lim
x→ x→ 70.
68. lim
x→ x→ m!
0
nmenx
x
1
5
10
20
30
40
50
100
ex
x5
2.718
0.047
0.220
151.614
4.40 105
2.30 109
1.66 1013
2.69 1033
ln x
x
Horizontal asymptote: y 0 (See Exercise 29)
74. y 72. y x x, x > 0
lim x x and lim x x 1
x→ x→0
dy x1x ln x1 1 ln x
0
dx
x2
x2
No horizontal asymptotes
ln y x ln x
Critical number:
1
1 dy
x
ln x
y dx
x
0, e
Sign of dydx:
y f x: Increasing
1
Relative maximum: e,
e
dy
x x1 ln x 0
dx
e1, 0
Increasing
1 1
1
Relative minimum: e1, e1e ,
e e
e, Decreasing
x e1
Critical number:
xe
Intervals:
0, e1
Sign of dydx:
y f x: Decreasing
Intervals:
1
−1
4
(e, 1e (
1e
−4
4
(
−1
((
1
1, 1 e
e e
(
4
−1
sin x 1
0
x→ x
76. lim
(Numerator is bounded)
Limit is not of the form 00 or .
L’Hôpital’s Rule does not apply.
ex
0
0
x → 1 ex
10
78. lim
Limit is not of the form 00 or .
L’Hôpital’s Rule does not apply.
351
352
Chapter 7
80. A P 1 r
n
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
nt
r
ln A ln P nt ln 1 ln P n
lim
n →
ln 1 r
n
lim
r
n
1
nt
1
r
n2 1 rn
1
2
nt
n →
1
nt
ln 1 lim
n →
rt
1
1
r
n
rt
Since lim ln A ln P rt, we have lim A eln Prt eln Pert Pert. Alternatively,
n→ r
lim A lim P 1 n →
n →
n
n→ nt
1 n lim P
n →
r
nr rt
Pert.
82. Let N be a fixed value for n. Then
N 1x N2
N 1N 2x N3
N 1!
x N1
lim
lim
. . . lim
0.
x
x
x→ e
x→ x→ x→ e
ex
ex
lim
84. f x xk 1
k
k 1,
k=1
f x x 1
1
10x 0.1 1
0.1
x 0.1
f x k 0.01,
x 0.01 1
100x 0.01 1
f x 0.01
k →0
(See Exercise 68)
6
k 0.1,
lim
k = 0.1
k = 0.01
−2
10
−2
xk 1
xk ln x
lim
ln x
k →0
k
1
1
86. f x , gx x2 4, 1, 2
x
fc
f 2 f 1
g2 g1
gc
88. f x ln x, gx x3, 1, 4
f 4 f 1
fc
g4 g1
gc
12 1c2
3
2c
ln 4 1c
1
2 3
63
3c
3c
3c3 ln 4 63
1
1
3
6
2c
c3 2c3 6
3
3
c 90. False. If y exx2, then
y x2ex 2xex xex x 2 ex x 2
.
x4
x4
x3
c
21
ln 4
ln214 2.474
3
92. False. Let f x x and gx x 1. Then
lim
x →
x
1, but lim x x 1
1.
x →
x1
Section 7.8
e1x ,
0,
94. gx 2
Improper Integrals
x0
y
x0
1.5
gx g0
e1x
lim
g0 lim
x →0
x →0
x0
x
2
e1x
e1x
, then ln y ln
Let y x
x
2
2
0.5
1
1 x2 ln x
2 ln x . Since
x
x2
x
−2
−1
1
2
− 0.5
ln x
1x
x2
lim
lim 0
lim x2 ln x lim
2
3
x →0
x →0 1x
x →0 2x
x →0
2
1 x x
2
we have lim
x →0
ln x
2
. Thus, lim y e
x →0
0 ⇒ g0 0.
Note: The graph appears to support this conclusion—the tangent line is horizontal at 0, 0.
98. lim x ln 21ln x
96. lim f xgx
x →0
x →a
Let y x ln 21ln x, then:
y f xgx
ln y gx ln f x
ln y lim gx ln f x ln 2
1 ln x
x →a
As x → a, ln y ⇒
, and hence y . Thus,
lim ln y lim
x →0
lim f xgx .
x →0
ln 2ln x
ln 2x
lim
x →0
1 ln x
1x
x→0
Thus, lim y eln 2 2.
x →
Improper Integrals
2. Infinite discontinuity at x 3.
4
3
1
dx lim
b →3
x 332
x 332 dx
2
b 3
4
b
lim 2x 312
b →3
lim 2 b →3
4
b
Diverges
4. Infinite discontinuity at x 1.
2
0
1
dx x 123
1
0
1
dx x 123
b
lim
b→1
0
b→1
2
1
1
dx
x 123
1
dx lim
c→1
x 123
3 x 1
lim 3
Converges
ln 2ln x
1 ln x
lim ln 2 ln 2
x →a
Section 7.8
ln x b
0
2
c
1
dx
x 123
3 x 1
lim 3
c→1
2
c
0 3 3 0 6
353
354
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
6. Infinite limit of integration.
0
8.
0
e2x dx lim
b→
ex dx 0. You need to evaluate the limit.
0
lim
b→ 0
1 2x
e
2
lim
b→
b
b
e2x dx
b
e ex dx lim
b→
0
1
1
0
2
2
lim
b→ b
x
0
e
1 1
b
Converges
10.
5
1
b
dx lim
b→ x3
1
b→
xex2 dx lim
x 1ex dx lim
eax sin bx dx lim
b→
0
b→ 18.
e
x2
c→
0
x 1ex dx lim xex
b→
0
ax
e
a sin bx b cos bx
a2 b2
x2
0
x3
dx lim
b→ 12
b
0
x2
b→
x
dx lim
b→ 1
1 lnx
2
lim
2
b
0
1 1
lim
b
0
b→ 0 0 by L’Hôpital’s Rule.
b
e
b
20.
1
ln x
dx lim
b→ x
b→ b
0
1
2 1
x2
x
x2 1 2
dx
Diverges
b→ lim
34
0
b
b
0 2
a b2 a 2 b2
22.
163x lim eb22b 4 4 4
b
c
4x14 dx
1
2x 4
b
0
b→ b→ xex2 dx lim
b
lim
b
b→ dx lim
5
2
1
0
16.
b
4
4
x
1
2
14.
12.
25x lim
5
dx
x3
24.
0
b
1
ln x
dx
x
ln x2
2
b
1
Diverges
b
ex
dx lim ln1 ex
b→ 1 ex
ln 2
0
b
Diverges
0
1
2
Diverges
26.
6
30.
x
x b
dx lim 2 cos
b→
2
2
0
0
x
Diverges since cos does not approach a limit as x → .
2
sin
0
4
dx lim
b→6
6 x
0
4
b
8
dx lim 8 ln x
b→0
x
4
b
32.
0
b
e
ln x2 dx lim
b→0
2 ln x dx
0
lim 2x ln x 2x
b→0
0
e
b
80 86
lim 2e 2e 2b ln b 2b
86
0
sec d lim
Diverges
e
46 x12 dx
8
dx lim
b→0
x
Diverges
0
2
34.
0
b
lim 86 x12
b→6
4
28.
b→ 2
b→0
lnsec tan b
0
2
,
36.
0
1
x
dx lim arcsin
b→2
2
4 x2
b
0
2
Section 7.8
2
38.
0
1
dx lim
b→2
4 x2
b
0
1 1
1
1 2x
dx lim
ln
b→2
4 2x 2x
4 2x
Diverges
3
40.
1
2
2
dx x 283
b
0
Improper Integrals
355
0
3
2x 283 dx 1
2x 283 dx
2
b
lim
b→2
1
2x 283 dx
3
2x 283 dx lim
c→2
6
lim x 253
b→2
5
b
c
6
lim x 253
c→2
5
1
3
c
Diverges
42.
Thus,
1
1
dx x ln x
0
1
dx x
ln
x
1
lim lnln x
b→1
44. If p 1,
e
e
lim
1
1
1
dx ln ln x C
x ln x
c→ 1
dx lim ln x
a→0
x
a
lim ln a .
a→0
Diverges. If p 1,
e
1
1
dx
x ln x
0
lnln x
.
1
x1p
dx lim
a→0
xp
1p
1
a
lim
a→0
1 1 p 1a p
.
1p
1
if 1 p > 0 or p < 1.
1p
This converges to
e
Diverges
1
46. (a) Assume
gx dx L (converges).
a
Since 0 ≤ f x ≤ gx on a, , 0 ≤
f x dx ≤
a
gx dx diverges, because otherwise, by part (a), if
a
1
48.
0
3 x
56.
f x dx converges.
a
gx dx converges, then so does
dx f x dx.
1
3
converges.
1 13 2
1
1
≥ on 2, and
x
x 1
a
50.
x 4ex dx converges.
0
See Exercise 44, p 13 .
(See Exercise 45.)
1
2
x
1
1
≤ 32 on 1, and
x
x1 x
54. Since
gx dx L and
a
1
52. Since
a
(b)
dx diverges by Exercise 43,
2
1
1
x32
1
x 1
dx diverges.
dx converges by Exercise 43,
1
1
≥ since x ln x < x on 2, . Since
x
x ln x
1
1
2
x
1
x1 x
dx diverges by Exercise 43,
2
dx converges.
1
x ln x
dx diverges.
58. See the definitions, pages 540, 543.
60. Answers will vary.
(a)
1e
Converges
62. f t t
ex
2x
dx
(b)
x dx
(Example 4)
Diverges
Fs st 1e s
test dt lim
b→
0
1
2
1
,s > 0
s2
st
b
0
356
Chapter 7
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
64. f t eat
Fs 66. f t sin at
eat est dt 0
etas dt
0
1 e
as
b
t as
0
1
1
,s > a
as sa
68. f t sinh at
est sinh at dt 0
est sin at dt
b→
2
est
s sin at a cos at
a2
a
a
,s > 0
s2 a2 s2 a2
eat
1
dt 2
2
at
s
0
et sa et sa dt
0
b
1
1
1
et sa et sa
b→ 2 s a
s a
lim
0
e
est
0
lim
0
lim
b→
Fs Fs 0
0
1
1
1
2 s a s a
1
1
a
1
2
,s > a
2 s a s a
s a2
70. (a) A 1
1
1
dx x2
x
1
1
(b) Disk:
V
1
(c) Shell:
V 2
1
x
x1 dx lim 2
ln x
b
2
b→ 1
1
dx lim 3
b→ x4
3x
b
1
Diverges
72. x 22 y2 1
2x 2 2yy 0
y x 2
y
1 y 2 1 3
S 4
1
x
dx 4
y
x 2 2y 2 3
1
1
(Assume y > 0.
y
x
dx 4
1 x 22
3
1
74. (a) Fx W
(b)
2
1 x 22
dx
4
0 2 arcsin1 2 arcsin1 8
lim 4
1 x 22 2 arcsinx 2
a→1
b→3
x2
1 x 22
b
a
K
K
,5
, K 80,000,000
x2
40002
80,000,000
80,000,000
dx lim
b→ x2
x
4000
80,000,000
W
10,000 2
x
b
4000
80,000,000
10,000
b
b 8000
Therefore, 4000 miles above the earth’s surface.
b
4000
20,000 mi-ton
80,000,000
20,000
b
2
3
b
0
Section 7.8
76. (a)
2 2t5
e
dt 5
4
(b)
0
0
b
2 2t5
e
dt lim e2t5 1
b→ 5
0
4
2 2t5
e
dt e2t5
5
0
e85 1
0.7981 79.81%
(c)
Improper Integrals
25 e dt lim te
2t5
t
0
2t5
b→
5
e2t5
2
b
0
5
2
5
78. (a) C 650,000 25,0001 0.08te0.06t dt
0
650,000 25,000 $778,512.58
$905,718.14
1 0.06t
t 0.06t
1
e
0.08
e
e0.06t
0.06
0.06
0.062
5
0
10
(b) C 650,000 25,0001 0.08te0.06t dt
0
650,000 25,000 (c) C 650,000 1 0.06t
t 0.06t
1
e
e
0.08
e0.06t
0.06
0.06
0.062
10
0
25,0001 0.08te0.06t dt
0
t
e
0.06
650,000 25,000 lim
0.06t
b→ 80. (a)
1
1
xn
1
1
e0.06t
0.062
b
0
$1,622,222.22
(b) It would appear to converge.
y
1
1.00
dx will converge if n > 1 and will diverge if n ≤ 1.
0.50
0.25
u
1
0.06t
0.75
(c) Let dv sin x dx ⇒
b
t
e
0.06
1
1
1
dx lim b→ x2
x
1
1
b
1
dx lim ln x
b→ x
0.08
1
x
v cos x
⇒ du sin x
cos x
dx lim b→0
x
x
cos 1 1
b
1
x
−5
− 0.25
1
dx
x2
1
15
20
cos x
dx
x2
cos x
dx
x2
Converges
82. (a) Yes, the integral is not defined at x 2.
(b)
5
(c) As n → , the integral approaches 4
4 .
(d) In 2
0
4
dx
1 tan xn
I2 3.14159
I4 3.14159
I8 3.14159
I12 3.14159
0
−2
2
357
358
Chapter 7
84. (a) f x Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
1
2
ex70 18
32
(b) P72 ≤ x <
0.2525
(c) 0.5 P70 ≤ x ≤ 72 0.5 0.2475 0.2525
90
f x dx 1.0
These are the same answers because by symmetry,
50
P70 ≤ x <
0.4
0.5
and
50
0.5 P70 ≤ x <
90
P70 ≤ x ≤ 72 P72 ≤ x <
− 0.2
86. False. This is equivalent to Exercise 85.
.
88. True
Review Exercises for Chapter 7
2.
2
xe x1 dx 2
1 x1
e
2x dx
2
4.
x
1
dx 1 x2122x dx
2
1 x
2
1 2
e x1 C
2
1 1 x212
C
2
12
1 x2 C
6.
u4 3u2 du 2x2x 3 dx 10.
8.
x
x 4 2x2 x 1
1 2
x 4 2x2 1
x 12
22x 332
x 1 C
5
u 2x 3, x u5
u3 C
5
x4 2x2 x 1
dx x2 12
u2 3
, dx u du
2
dx x
1
2
2x
dx
x2 12
1
C
2x2 1
x2 1ex dx x2 1ex 2 xex dx x2 1ex 2xex 2 ex dx exx2 2x 1 1
(1) dv ex dx
⇒
v ex
u x2 1 ⇒ du 2x dx
(2) dv ex dx ⇒
ux
⇒ du dx
12. u arctan 2x, du v ex
2
dx, dv dx,v x
1 4x2
arctan 2x dx x arctan 2x x arctan 2x 14.
lnx2 1 dx 2x
dx
1 4x2
1
ln1 4x2 C
4
1
2
lnx2 1 dx
1
x ln x2 1 2
1
x ln x2 1 2
x2
x2
dx
1
dx 1
dx
x2 1
1 x1
1
C
x ln x2 1 x ln
2
2 x1
dv dx
⇒
vx
u lnx2 1 ⇒ du 2x
dx
x2 1
Review Exercises for Chapter 7
16.
ex arctanex dx ex arctanex ex arctanex ⇒
dv ex dx
20.
x
dx 2
sin2
ex
dx
1 e2x
or
tan sec4 d 22.
tan3 tan sec2 d sec3 sec tan d cos 2sin cos 2 d x2 9
x
dx 24.
1
ln1 e2x C
2
1
1
1
1
1 cos x dx x sin x C x sin x C
2
2
2
tan sec4 d e2x
dx
1 e2x
v ex
u arctan ex ⇒ du 18.
359
1 4
1
tan tan2 C1
4
2
1
sec4 C2
4
cos 2 sin2 sin cos 2 d
1
sin cos 3cos sin d sin cos4 C
4
3 tan 3 sec tan d
3 sec x
3 tan2 d
x2 − 9
θ
3
3 sec2 1 d
3tan C
x2 9 3 arcsec
3x C
x 3 sec , dx 3 sec tan d, x2 9 3 tan 26.
9 4x2 dx 1
9 2x2 2 dx
2
2 9 arcsin
1
2
9
2x
x
arcsin
9 4x2 C
4
3
2
1
28.
2x
2x9 4x2 C
3
sin 1
d 1 2 cos2 2
1
2
1
2 sin d
1 2 cos2 arctan 2 cos C
u 2 cos , du 2 sin d
360
Chapter 7
30. (a)
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
x4 x dx 64 tan3 sec3 d
(b)
x4 x dx 2 u4 4u2 du
64 sec4 sec2 sec tan d
64 sec3 3 sec3 5 C
15
24 x32
3x 8 C
15
2u3 2
3u 20 C
15
24 x32
3x 8 C
15
u2 4 x, dx 2u du
x 4 tan2 , dx 8 tan sec2 d,
4 x 2 sec (c)
x4 x dx u32 4u12 du
(d)
x4 x dx 4 x32 dx
2u32
3u 20 C
15
4
2x
4 x32 4 x52 C
3
15
24 x32
3x 8 C
15
24 x32
3x 8 C
15
dv 4 x dx ⇒
2
v 4 x32
3
⇒ du dx
ux
2x3 5x2 4x 4
4
3
2x 3 x2 x
x
x1
32.
34.
u 4 x, du dx
2x
2
4 x32 3
3
2x3 5x2 4x 4
dx x2 x
2x 3 3
4
dx x2 3x 4 ln x 3 ln x 1 C
x
x1
4x 2
A
B
3x 12 x 1 x 12
4x 2 3Ax 1 3B
Let x 1: 2 3B ⇒ B 2
3
Let x 2: 6 3A 3B ⇒ A 36.
4x 2
4
dx 3x 12
3
sec2 d tan tan 1
4
3
2
1
dx x1
3
1
du uu 1
1
du u1
ln u 1 ln u C ln
u tan , du sec2 d
1
A
B
uu 1
u
u1
1 Au 1 Bu
Let u 0: 1 A ⇒ A 1
Let u 1: 1 B
4
1
2
2
1
dx lnx 1 C 2 lnx 1 C
x 12
3
3x 1
3
x1
1
du
u
tan 1
C ln 1 cot C
tan Review Exercises for Chapter 7
38.
x
2 3x
dx 24 3x
2 3x C (Formula 21)
27
40.
x
1
1
dx
2 dx 1 ex
2 1 eu
6x 8
2 3x C
27
u x2
1
u ln1 eu C
2
(Formula 84)
1
2
x2 ln1 e x C
2
42.
3
3
1
dx 3 dx
2 3x3x2 1
2x9x2 1
44.
(Formula 33)
tann x dx 50.
u 3x
1
1
1
dx
dx 1 tan x
1 tan x
46.
3
arcsec 3x C
2
u x
11
x lncos x sin x C (Formula 71)
2
tann2xsec2 x 1 dx
tann2 x sec2 x dx 1
tann1 x n1
1 x dx 48.
4u4 4u2 du 4u5 4u3
4
C 1 x 323x 2 C
5
3
15
u 1 x, x u4 2u2 1, dx 4u3 4u du
52.
3x3 4x
Cx D
Ax B
2
2
x2 12
x 1
x 12
3x3 4x Ax Bx2 1 Cx D
Ax3 Bx2 A Cx B D
A 3, B 0, A C 4 ⇒ C 1,
BD0 ⇒ D0
3x3 4x
x
dx dx 3 2
x2 12
x 1
54.
x
dx
x2 12
3
1
lnx2 1 C
2
2x2 1
sin cos 2 d sin2 2 sin cos cos2 d
1 sin 2 d 1
dx
u 2x, du 2x
tann2 x dx
u4u3 4u du csc2x
1
dx 2 csc2x
dx
x
2x
2 ln csc2x cot2x C
tann2 x dx
1
1
cos 2 C 2 cos 2 C
2
2
361
362
Chapter 7
56. y 4 x2
2x
Integration Techniques, L’Hôpital’s Rule, and Improper Integrals
dx 2 cos 2 cos d
4 sin csc sin d
ln csc cos cos C
ln
4 x2
2 4 x2
C
x
2
x 2 sin , dx 2 cos d, 4 x 2 2 cos 1 cos d 58. y sin 1 cos d 1 cos 12sin d 21 cos C
u 1 cos, du sin d
1
60.
x
x 2x 4
0
2
0
dx 2 ln x 4 ln x 2
1
62.
xe3x dx 0
e9 3x 1
2
3x
0
1
5e6 1 224.238
9
2 ln 3 2 ln 4 ln 2
ln
3
64.
0
9
0.118
8
x
22 x
1 x
dx 3
1 x
3
0
4 4 8
3 3 3
1
dx
25 x2
1
x5
ln
10 x 5
4
66. A 0
68. By symmetry, y 0.
A 4 5
1 44
x
4
17
3.4
5
70. s y
4
0
1
1
1
ln ln 9 0.220
10 9 10
1 sin2 2x dx 3.82
0
3
2
1
(3.4, 0)
x
−1
1
3
4
5
−2
−3
x, y 3.4, 0
72. lim
x →0
76.
sin x
cos x
1
lim
sin 2 x x →0 2 cos 2 x 2 2
74. lim xex lim
2
x →
x →
y lim x 1ln x
x →1
ln y lim ln x lnx 1
x →1
lim
x →1
lnx 1
lim
x →1
1
ln x
lim 2xln x 0
x →1
Since ln y 0, y 1.
1
1
2
ln x
x1
ln2 x
x
lim
lim
x →1
x →1
1 1
x1
1
x ln2 x
x
x2
x
1
2 lim
2 0
x → 2xe x
ex