Thomas Calculus 13th [Solutions]

418 Chapter 5 Integration
90.
3 /4 3 /4
csc
/4 z cot z dz [ csc z ]
3
/4 csc csc 2 2 0
4 4
91. Let u sin x du cos x dx; x 0 u 0, x u 1
2
/2 3/2
1 3/2
1
5(sin x)
cos x dx
2 5/2 5/2 1 5/2 5/2
5u du 5 u [2 u ]
0
0 5
0 2(1) 2(0) 2
0
92. Let u sin 3x du 3cos 3x dx 1 du cos3 x dx;
x u sin 3 1, x u sin 3 1
3 2
2
2 2
/2 4
15 sin 3 cos 3
/2 x x dx 1 4 1
1 15u 3
du 1 4 5 1 5 5
5
1 u du [ u ] 1 ( 1) (1) 2
2
93. Let u 1 3sin x du 6sin x cos x dx 1 du 3sin x cos x dx; x 0 u 1, x u
2
2
/2 4
3sin x cos x
4
dx 1 1
4 1/2
1/2
du 1 u du 1 u 1/2 4 1/2 1/2
[ u ]
0 2
1 3sin x 1 u 2 1 2
2 1
1 4 1 1
2
1
2
1 3sin 4
2
94. Let u 1 7 tan x
/4 2
sec x
0 (1 7 tan x)
du
2
7sec
dx
2/3
2/3
x dx
8 1 1 du
1 u 7
1 du
7
2
sec x dx; x 0
8
1 2/3
u du 1
1 7 7
1/3
1
3
u
u
8
1
1 7 tan 0 1, x u 1 7 tan 8
4
4
1/3
8
3 3 1/3 3 1/3
u (8) (1) 3
7 1 7 7 7
95.
4 4 2
4
x 1 dx 1 1 x 1 dx 1 1 x 1 16 1 15 1
1 8 2 2 1 4 2 8 ln x
x
x
1 2 8 ln 4 8 ln1 16 2
ln 4
15 ln 4 15 ln 2
16 16
g
2 1
8
96. 2 8 2
g
dx 1 2 2 12 2 3
1 3 2 3 1 12 x dx
3 ln x 12 x
x
x
1 3 ln 8 8 (ln1 12) 3 ln 8 2
12
x
2 21 2
2/3
ln 8 (ln 8) 7 ln(8 ) 7 ln 4 7
3 2 3
97.
98.
1 ( x 1) 0 u
e dx e du , where u = (x + 1), du = dx; x = 2 u = 1, x = 1 u = 0
2 1
u 0 0 1
[ e ] 1 ( e e ) e 1
0 2w
1
0 u
e dw e du , where u = 2w, du = 2 dw; w = ln 2 u ln 1 , w = 0 u = 0
ln 2 2 ln(1/4)
4
1 u 0 1 0 ln(1/4)
[ e ] 1 1 3
2 ln(1/4) [ e e ] 1
2 2 4 8
99.
ln 5 r 3/2 1
16 3/2
1 (3 r
e e 1) dr u du
3 4
,
r
r
where u 3e 1, du 3 e dr;
r = 0 u = 4, r = ln 5 u = 16
2 1/2 16 2 1/2 1/2
[ u ] 2 1 1 2 1 1
3 4 (16 4 )
3 3 4 2 3 4 6
100.
ln 9 1/2 8 1/2
e ( e 1) d u du , where u e 1, du e d ; = 0 u = 0, = ln 9 u = 8
0 0
11/2
2 3/2 8 2 3/2 3/2 2 9/2 2 32 2
[ u ]
3 0 (8 0 ) (2 0)
3 3 3 3
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