Thomas Calculus 13th [Solutions]

386 Chapter 5 Integration
72.
csc x dx csc x(1) dx csc x csc x cot x
csc x cot x
dx . Let
2
u csc x cot x du (csc x csc x cot x) dx.
csc x dx du
u
ln u C ln csc x cot x C
2
73. Let u 3t 1 du 6t dt 2du 12t dt
2 3 3 1 4 1 4 1 2 4
s 12 t(3t 1) dt u (2 du) 2 u C u C (3t 1) C;
4 2 2
4 2 4
s 3 when t 1 3 1 (3 1) C 3 8 C C 5 s 1 (3t
1) 5
2 2
2
74. Let u x 8 du 2x dx 2 du 4x dx
2 1/3 1/3 3 2/3 2/3 2 2/3
y 4 x( x 8) dx u (2 du) 2 u C 3u C 3( x 8) C;
2
2/3 2 2/3
y 0 when x 0 0 3(8) C C 12 y 3( x 8) 12
75. Let u t du dt
12
2 2
s 8sin t dt 8sin u du 8 u 1 sin 2u C 4 t 2sin 2 t C;
12 2 4 12 6
s 8 when t 0 8 4 2sin C C 8 1 9
12 6 3 3
s 4 t 2sin 2t 9 4t 2sin 2t
9
12 6 3 6
76. Let u du d
4
2 2
r 3cos d 3cos u du 3 u 1 sin 2u C 3 3 sin 2 C;
4 2 4 2 4 4 2
r when 0 3 3 sin C C 3 r 3 3 sin 2 3
8
8 8 4 2 2 4 2 4 4 2 2 4
r 3 3 sin 2 3 r 3 3 cos 2 3
2 4 2 8 4 2 4 8 4
77. Let u 2t du 2 dt 2 du 4 dt
2
ds 4sin 2 t dt (sin u)( 2 du) 2cosu C
2 1 2cos 2 t C
2 1;
dt
at t 0 and ds 100 we have
dt
100 2cos C
2 1 C1
100 ds
dt
2cos 2 t
2
100
s 2cos 2t 100 dt (cosu 50) du sin u 50u C
2 2 sin 2t 50 2 t C
2 2 2;
at t 0 and s 0 we have 0 sin 50 C
2 2 2 C2
1 25
s sin 2t 100t 25 (1 25 ) s sin 2t 100t
1
2 2
2 2
78. Let u tan 2x du 2sec 2x dx 2du 4sec 2 x dx;
v 2x dv 2dx 1 dv dx
2
dy 2 2 2
4sec 2x tan 2 x dx u(2 du) u C1 tan 2 x C1;
dx
dy
2 2 2
at x 0 and 4 we have 4 0 C
dx
1 C1
4 dy tan 2x 4 (sec 2x 1) 4 sec 2x
3
dx
2 2
y (sec 2x 3) dx (sec v 3) 1 dv 1 tan v 3 v C 1
2 2 2 2 tan 2x 3 x C
2
2;
at x 0 and y 1 we have 1 1 (0) 0 C 1
2 2 C2
1 y tan 2x 3x
1
2
79. Let u 2t du 2 dt 3du 6dt
s 6 sin 2 t dt (sin u)(3 du) 3 cos u C 3 cos 2 t C;
at t 0 and s 0 we have 0 3cos 0 C C 3 s 3 3cos 2t s 3 3cos( ) 6 m
2
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