Thomas Calculus 13th [Solutions]

416 Chapter 5 Integration
68.
dx
du ,
2 2
( x 3) ( x 3) 25 u u 25
1 1 1
sec u C 1 sec x 3
5 5 5 5
where u = x + 3 and du = dx
C
69.
sin
1
x
e
2
x x
2
u
dx e du , where
u sin x
e C e C
1
u
sin
1
x and
du
2
dx
x x
2
70.
1
sin x 1/2
dx u du , where u
2
1 x
2 3/2 2 1 3/2
3 u C 3 (sin x)
C
sin
1
x and
du
1
dx
x
2
71.
1
1 y
2
1
1/2
dy dy u du , where
1 2 1
tan y(1 y ) tan y
1/2 1
2u C 2 tan y C
1
dy
u tan y and du
2
1 y
72.
1 2
(tan x) 2
dx u du , where
2
1 x
1 3 1 1 3
3 u C 3 (tan x)
C
1
u tan x and du dx
2
1 x
73.
1 2 3 2 1 3 2 3 2
(3x 4x 7) dx [ x 2x 7 x ]
1
1 [1 2(1) 7(1)] [( 1) 2( 1) 7( 1)] 6 ( 10) 16
74.
1 3 2 4 3 1 4 3
(8
0 s 12 s 5) ds [2 s 4 s 5 s ] 0 [2(1) 4(1) 5(1)] 0 3
2
75. 4
2 2 1 2
dv 4 4
1
2
1 4 v dv [ 4 v ] 1 2 1
2
v
76.
27 4/3 1/3 27 1/3 1/3
x dx [ 3 x ] 1
1
1 3(27) ( 3(1) ) 3 3(1) 2
3
4
dt
4
dt
4 3/2 1/2 4 2 ( 2)
t dt [ 2 t ]
1 1 1
1
1
t t t
4 1
77.
3/2
1/2
78. Let x 1 u dx 1 u du 2 dx du ; u 1 x 2, u 4 x 3
2
u
4 1/2
(1 u ) 3 1/2 3/2
3
du x (2 dx) 2 2 x 4 3/2 3/2
(3 ) 4 (2 ) 4 3 8 2 4 (3 3 2 2)
1 u
2 3 2 3 3 3 3
79. Let u 2x 1 du 2 dx 18 du 36 dx; x 0 u 1, x 1 u 3
1 2 3 3
36 dx 3 3
18u
du 18u
9 9 9 8
0
3
1
2
2 2 2
(2x
1) 1 u 1 3 1
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