Chapter 5: Integration - EXERCISES, Mathematics, Thomas Calculus - Engineering Mathematics PDF Download

 Page 1


374 Chapter 5: Integration
EXERCISES 5.5
Evaluating Integrals
Evaluate the indefinite integrals in Exercises 1–12 by using the given
substitutions to reduce the integrals to standard form.
1. 2.
L
x sin s2x
2
d dx, u = 2x
2
L
 sin 3x dx, u = 3x
3.
4.
L
a1 - cos 
t
2
b
2
 sin 
t
2
 dt, u = 1 - cos 
t
2
L
 sec 2t tan 2t dt, u = 2t
4100 AWL/Thomas_ch05p325-395  8/20/04  9:57 AM  Page 374
Page 2


374 Chapter 5: Integration
EXERCISES 5.5
Evaluating Integrals
Evaluate the indefinite integrals in Exercises 1–12 by using the given
substitutions to reduce the integrals to standard form.
1. 2.
L
x sin s2x
2
d dx, u = 2x
2
L
 sin 3x dx, u = 3x
3.
4.
L
a1 - cos 
t
2
b
2
 sin 
t
2
 dt, u = 1 - cos 
t
2
L
 sec 2t tan 2t dt, u = 2t
4100 AWL/Thomas_ch05p325-395  8/20/04  9:57 AM  Page 374
5.
6.
7.
8.
9.
10.
11.
a. Using b. Using 
12.
a. Using b. Using 
Evaluate the integrals in Exercises 13–48.
13. 14.
15. 16.
17. 18.
19. 20.
21. 22.
23. 24.
25. 26.
27. 28.
29. 30.
31. 32.
33.
34.
35. 36.
L
 
6 cos t
s2 + sin td
3
 dt
L
 
sin s2t + 1d
cos
2
 s2t + 1d
 dt
L
csc a
y - p
2
b cot a
y - p
2
b dy
L
 sec ay +
p
2
b tan ay +
p
2
b dy
L
x
1>3
 sin sx
4>3
- 8d dx
L
x
1>2
 sin sx
3>2
+ 1d dx
L
r
4
 a7 -
r
5
10
b
3
 dr
L
r
2
 a
r
3
18
- 1b
5
 dr
L
tan
7
 
x
2
 sec
2
 
x
2
 dx
L
sin
5
 
x
3
 cos 
x
3
 dx
L
tan
2
 x sec
2
 x dx
L
sec
2
 s3x + 2d dx
L
sin s8z - 5d dz
L
cos s3z + 4d dz
L
 
s1 +1xd
3
1x
 dx
L
 
1
1x s1 +1xd
2
 dx
L
 
4y dy
22y
2
+ 1 L
3y27 - 3y
2
 dy
L
8u2
3
u
2
- 1 du
L
u2
4
1 - u
2
 du
L
 
3 dx
s2 - xd
2
L
 
1
25s + 4
 ds
L
s2x + 1d
3
 dx
L
23 - 2s ds
u =25x + 8 u = 5x + 8
L
 
dx
25x + 8
u = csc 2u u = cot 2u
L
csc
2
 2u cot 2u du
L
 
1
x
2
 cos
2
 a
1
x
b dx, u =-
1
x
L
1x sin
2
 sx
3>2
- 1d dx, u = x
3>2
- 1
L
12s y
4
+ 4y
2
+ 1d
2
s y
3
+ 2yd dy, u = y
4
+ 4y
2
+ 1
L
 
9r
2
 dr
21 - r
3
, u = 1 - r
3
L
x
3
sx
4
- 1d
2
 dx, u = x
4
- 1
L
28s7x - 2d
-5
 dx, u = 7x - 2 37. 38.
39. 40.
41. 42.
43.
44.
45. 46.
47. 48.
Simplifying Integrals Step by Step
If you do not know what substitution to make, try reducing the integral
step by step, using a trial substitution to simplify the integral a bit and
then another to simplify it some more. Y ou will see what we mean if
you try the sequences of substitutions in Exercises 49 and 50.
49.
a. followed by then by 
b. followed by 
c.
50.
a. followed by then by 
b. followed by 
c.
Evaluate the integrals in Exercises 51 and 52.
51.
52.
Initial Value Problems
Solve the initial value problems in Exercises 53–58.
53.
54.
55.
56.
dr
du
= 3 cos
2
 a
p
4
- ub, rs0d =
p
8
ds
dt
= 8 sin
2
 at +
p
12
b, ss0d = 8
dy
dx
= 4x sx
2
+ 8d
-1>3
, ys0d = 0
ds
dt
= 12t s3t
2
- 1d
3
, ss1d = 3
L
 
sin 2u
2u cos
3
 1u
 du
L
 
s2r - 1d cos 23s2r - 1d
2
+ 6
23s2r - 1d
2
+ 6
 dr
u = 1 + sin
2
 sx - 1d
y = 1 + u
2
u = sin sx - 1d,
w = 1 + y
2
y = sin u, u = x - 1,
L
21 + sin
2
 sx - 1d sin sx - 1d cos sx - 1d dx
u = 2 + tan
3
 x
y = 2 + u u = tan
3
 x,
w = 2 + y y = u
3
, u = tan x,
L
 
18 tan
2
 x sec
2
 x
s2 + tan
3
 xd
2
 dx
L
3x
5
2x
3
+ 1 dx
L
x
3
2x
2
+ 1 dx
L
A
x - 1
x
5
 dx
L
t
3
s1 + t
4
d
3
 dt
L
su
4
- 2u
2
+ 8u - 2dsu
3
- u + 2d du
L
ss
3
+ 2s
2
- 5s + 5ds3s
2
+ 4s - 5d ds
L
 
cos 2u
2u sin
2
 2u
 du
L
 
1
u
2
 sin 
1
u
 cos 
1
u
 du
L
 
1
1t
 coss1t + 3d dt
L
 
1
t
2
 cos a
1
t
- 1b dt
L
 
sec z tan z
2sec z
 dz
L
2cot y csc
2
 y dy
5.5 Indefinite Integrals and the Substitution Rule 375
4100 AWL/Thomas_ch05p325-395  8/20/04  9:57 AM  Page 375
Page 3


374 Chapter 5: Integration
EXERCISES 5.5
Evaluating Integrals
Evaluate the indefinite integrals in Exercises 1–12 by using the given
substitutions to reduce the integrals to standard form.
1. 2.
L
x sin s2x
2
d dx, u = 2x
2
L
 sin 3x dx, u = 3x
3.
4.
L
a1 - cos 
t
2
b
2
 sin 
t
2
 dt, u = 1 - cos 
t
2
L
 sec 2t tan 2t dt, u = 2t
4100 AWL/Thomas_ch05p325-395  8/20/04  9:57 AM  Page 374
5.
6.
7.
8.
9.
10.
11.
a. Using b. Using 
12.
a. Using b. Using 
Evaluate the integrals in Exercises 13–48.
13. 14.
15. 16.
17. 18.
19. 20.
21. 22.
23. 24.
25. 26.
27. 28.
29. 30.
31. 32.
33.
34.
35. 36.
L
 
6 cos t
s2 + sin td
3
 dt
L
 
sin s2t + 1d
cos
2
 s2t + 1d
 dt
L
csc a
y - p
2
b cot a
y - p
2
b dy
L
 sec ay +
p
2
b tan ay +
p
2
b dy
L
x
1>3
 sin sx
4>3
- 8d dx
L
x
1>2
 sin sx
3>2
+ 1d dx
L
r
4
 a7 -
r
5
10
b
3
 dr
L
r
2
 a
r
3
18
- 1b
5
 dr
L
tan
7
 
x
2
 sec
2
 
x
2
 dx
L
sin
5
 
x
3
 cos 
x
3
 dx
L
tan
2
 x sec
2
 x dx
L
sec
2
 s3x + 2d dx
L
sin s8z - 5d dz
L
cos s3z + 4d dz
L
 
s1 +1xd
3
1x
 dx
L
 
1
1x s1 +1xd
2
 dx
L
 
4y dy
22y
2
+ 1 L
3y27 - 3y
2
 dy
L
8u2
3
u
2
- 1 du
L
u2
4
1 - u
2
 du
L
 
3 dx
s2 - xd
2
L
 
1
25s + 4
 ds
L
s2x + 1d
3
 dx
L
23 - 2s ds
u =25x + 8 u = 5x + 8
L
 
dx
25x + 8
u = csc 2u u = cot 2u
L
csc
2
 2u cot 2u du
L
 
1
x
2
 cos
2
 a
1
x
b dx, u =-
1
x
L
1x sin
2
 sx
3>2
- 1d dx, u = x
3>2
- 1
L
12s y
4
+ 4y
2
+ 1d
2
s y
3
+ 2yd dy, u = y
4
+ 4y
2
+ 1
L
 
9r
2
 dr
21 - r
3
, u = 1 - r
3
L
x
3
sx
4
- 1d
2
 dx, u = x
4
- 1
L
28s7x - 2d
-5
 dx, u = 7x - 2 37. 38.
39. 40.
41. 42.
43.
44.
45. 46.
47. 48.
Simplifying Integrals Step by Step
If you do not know what substitution to make, try reducing the integral
step by step, using a trial substitution to simplify the integral a bit and
then another to simplify it some more. Y ou will see what we mean if
you try the sequences of substitutions in Exercises 49 and 50.
49.
a. followed by then by 
b. followed by 
c.
50.
a. followed by then by 
b. followed by 
c.
Evaluate the integrals in Exercises 51 and 52.
51.
52.
Initial Value Problems
Solve the initial value problems in Exercises 53–58.
53.
54.
55.
56.
dr
du
= 3 cos
2
 a
p
4
- ub, rs0d =
p
8
ds
dt
= 8 sin
2
 at +
p
12
b, ss0d = 8
dy
dx
= 4x sx
2
+ 8d
-1>3
, ys0d = 0
ds
dt
= 12t s3t
2
- 1d
3
, ss1d = 3
L
 
sin 2u
2u cos
3
 1u
 du
L
 
s2r - 1d cos 23s2r - 1d
2
+ 6
23s2r - 1d
2
+ 6
 dr
u = 1 + sin
2
 sx - 1d
y = 1 + u
2
u = sin sx - 1d,
w = 1 + y
2
y = sin u, u = x - 1,
L
21 + sin
2
 sx - 1d sin sx - 1d cos sx - 1d dx
u = 2 + tan
3
 x
y = 2 + u u = tan
3
 x,
w = 2 + y y = u
3
, u = tan x,
L
 
18 tan
2
 x sec
2
 x
s2 + tan
3
 xd
2
 dx
L
3x
5
2x
3
+ 1 dx
L
x
3
2x
2
+ 1 dx
L
A
x - 1
x
5
 dx
L
t
3
s1 + t
4
d
3
 dt
L
su
4
- 2u
2
+ 8u - 2dsu
3
- u + 2d du
L
ss
3
+ 2s
2
- 5s + 5ds3s
2
+ 4s - 5d ds
L
 
cos 2u
2u sin
2
 2u
 du
L
 
1
u
2
 sin 
1
u
 cos 
1
u
 du
L
 
1
1t
 coss1t + 3d dt
L
 
1
t
2
 cos a
1
t
- 1b dt
L
 
sec z tan z
2sec z
 dz
L
2cot y csc
2
 y dy
5.5 Indefinite Integrals and the Substitution Rule 375
4100 AWL/Thomas_ch05p325-395  8/20/04  9:57 AM  Page 375
57.
58.
59. The velocity of a particle moving back and forth on a line is
for all t. If when find
the value of s when 
60. The acceleration of a particle moving back and forth on a line is
for all t. If and 
when find s when 
Theory and Examples
61. It looks as if we can integrate 2 sin x cos x with respect to x in
three different ways:
a.
b.
c.
 =-
cos 2x
2
+ C
3
.
2 sin x cos x = sin 2x 
L
2 sin x cos x dx =
L
 sin 2x dx
 =-u
2
+ C
2
=-cos
2
 x + C
2
u = cos x, 
L
2 sin x cos x dx =
L
-2u du
 = u
2
+ C
1
= sin
2
 x + C
1
u = sin x, 
L
2 sin x cos x dx =
L
2u du
t = 1 sec. t = 0, 8 m/sec
y = s = 0 a = d
2
s>dt
2
= p
2
 cos pt m>sec
2
t = p>2 sec.
t = 0, s = 0 y = ds>dt = 6 sin 2t m>sec
d
2
y
dx
2
= 4 sec
2
 2x tan 2x, y¿s0d = 4, ys0d =-1
d
2
s
dt
2
=-4 sin a2t -
p
2
b, s¿s0d = 100, ss0d = 0
Can all three integrations be correct? Give reasons for your an-
swer.
62. The substitution gives
The substitution gives
Can both integrations be correct? Give reasons for your answer.
63. (Continuation of Example 9.)
a. Show by evaluating the integral in the expression
that the average value of over a full cycle
is zero.
b. The circuit that runs your electric stove is rated 240 volts rms.
What is the peak value of the allowable voltage?
c. Show that
L
1>60
0
sV
max
d
2
 sin
2
 120 pt dt =
sV
max
d
2
120
.
V = V
max
 sin 120 pt
1
s1>60d - 0
 
L
1>60
0
V
max
 sin 120 pt dt
L
sec
2
 x tan x dx =
L
u du =
u
2
2
+ C =
sec
2
 x
2
+ C.
u = sec x
L
sec
2
 x tan x dx =
L
u du =
u
2
2
+ C =
tan
2
 x
2
+ C.
u = tan x
376 Chapter 5: Integration
4100 AWL/Thomas_ch05p325-395  8/20/04  9:57 AM  Page 376