Thomas Calculus 13th [Solutions]

620 Chapter 8 Techniques of Integration
28.
1 4r dr
b
1 2 1 2 1
lim 2sin r lim 2sin b 2sin 0 2 0
1 r
2
b 1 b 1
0 4
0
29.
2 1
2
ds
1 1
lim sec s lim sec 2 sec b 0
s s
3 3
b b
1 2
1 1 b 1
30.
4 1
4
dt 1 1 1 4 1 1
lim sec t lim sec sec b 1 1 0
t t
2 2 2 2 2 2 2 3 2 6
b b
2 2
4 2 b 2
31.
32.
4
dx
b
4
4
lim dx lim dx
b
lim 2 x lim 2 x
1 x b 0 1 x c 0 c x b 0 1 c 0 c
lim 2 b 2 ( 1) lim 2 4 2 c 0 2 2 2 0 6
b 0 c 0
2
dx
1
dx
2
dx
2
0 1 0 1 1
lim 2 1 b
x lim 2 x 1
x x x 1 b 1 0 c 1
c
lim 2 1 b 2 1 0 lim 2 2 1 2 c 1 0 2 2 0 4
b 1 c 1
33. d
2 2 1 2 1
1 lim ln b
2
5 6 3 lim ln b
1
b 3 ln 1 3 0 ln 2
ln 2
b
b
b
34. dx
1 1 2 1 1 1 1 1 1
lim ln x 1 ln x 1 tan x lim ln x tan x
0 2
( x 1) x 1 2 4 2 2 2
b 0 b x 1
2
b
0
lim 1 ln b 1 1 tan 1 1 1 1 1
b ln tan 0 1 ln1 1 1 ln1 1 0
b
b 1
1
2 2 2 2 2 2 2 2 2 2 4
35.
36.
/2
b
tan d lim ln cos lim ln |cos b| ln1 lim ln |cos b | , the integral
0
0
b b b
2 2 2
diverges
/2 /2
cot
0 lim ln sin lim ln1 ln |sin | lim
0
b
b b 0 b 0
ln |sin | , 37.
0
1
ln x
2 dx
x
1
1/3
ln x
2 dx is bounded, so convergence is determined by
x
0
1/3
ln x
2 dx .
x
On (0, 1/ 3], ln x 1 and
hence
0
1
ln x
2 dx diverges.
x
ln x 1
2 2 .
x x
Since
0
1/3
1
2 dx
x
diverges to
, so does
0
1/3
ln x
2 dx
x
and
Copyright
2014 Pearson Education, Inc.