Thomas Calculus 13th [Solutions]

Chapter 5 Practice Exercises 415
57. 4 2t
9 1
2
dt du , where u t 25, du = 2t dt; t = 0 u = 25, t = 4 u = 9
0
2
t 25 25 u
9
ln u ln 9 ln 25 ln 9 ln 25 ln 9
25
25
58.
tan(ln v)
dv tan u du sin u du , where u = ln v and du 1 dv
v
cosu
v
ln cosu C ln cos(ln v)
C
59.
60.
3
(ln x) 3
u du , where u = ln x and du 1 dx
x
x
2
u 1 2
C
2 2 (ln x ) C
1 2 2
csc (1 ln r) dr csc u du , where u = 1 + ln r and du 1 dr
r
r
cot u C cot(1 ln r)
C
61.
2
x
3 1 u
x dx 3 du , where
2
1 u 1 x
(3 ) C 3
2ln 3 2ln 3
2
u
C
2
x and du = 2x dx
62.
tan x 2 u
2 sec x dx 2 du , where u = tan x and
du
2
sec
x dx
tan x
1 u
(2 ) C 2
ln 2 ln 2
C
63.
64.
3dr
3 du
1 4( r 1)
2
1 u
2 2
, where u = 2(r 1) and du = 2 dr
3 1 3 1
sin u C sin 2( r 1) C
2 2
6 dr
6 du , where u = r + 1 and du = dr
2 2
4 ( r 1) 4 u
1 1
6sin u C 6sin r 1 C
2 2
65. dx du , where u = x 1 and du = dx
2 2
2 ( x 1) 2 u
1 1 1
tan u C 1 tan x 1 C
2 2 2 2
66. dx 1 du , where u = 3x + 1 and du = 3 dx
2 3 2
1 (3x
1) 1 u
1 1 1 1
tan u C tan (3x 1) C
3 3
67.
dx 1 du , where u = 2x 1 and du = 2 dx
2 2 2
(2x 1) (2x 1) 4 u u 4
1 1 1 1
sec u C 1 sec 2x
1 C
2 2 2 4 2
Copyright
2014 Pearson Education, Inc.