Thomas Calculus 13th [Solutions]

Chapter 4 Practice Exercises 331
104. Our trial solution based on the chain rule is
d
d
2
7 C . Thus
2
7
7
2
d
2
7 C . Differentiate the solution to check:
2
7 C.
105. Our trial solution based on the chain rule is
d
dx
1 4 3/4 3 4 1/4
(1 x ) C x (1 x ) . Thus
3
1 4 3/4
3 (1 x ) C . Differentiate the solution to check:
3 4 1/4 1 4 3/4
x (1 x ) dx (1 x ) C.
3
8/5
106. Our trial solution based on the chain rule is 5
8 (2 x)
C . Differentiate the solution to check:
d 5 8/5 3/5
3/5 8/5
(2 x) C (2 x ) . Thus (2 x) dx 5 (2 x) C.
dx 8 8
107. Our trial solution based on the chain rule is 10 tan s C . Differentiate the solution to check:
10
d 2
10 tan s sec s
2
C . Thus sec s ds 10 tan s C.
ds 10 10
10 10
108. Our trial solution based on the chain rule is 1 cot s C . Differentiate the solution to check:
d
ds
1 2
cot s C csc s . Thus
2
csc s ds 1 cot s C.
109. Our trial solution based on the chain rule is 12 csc 2 C . Differentiate the solution to check:
d
d
1
2
csc 2 C csc 2 cot 2 . Thus csc 2 cot 2 d 1 csc 2 C.
2
110. Our trial solution based on the chain rule is
d
d
3sec C sec tan . Thus
3 3 3
3sec C . Differentiate the solution to check:
3
sec tan cot 2 d 3sec C.
3 3 3
111. Our trial solution based on the chain rule is x
2
sin x
2
C . Differentiate the solution to check:
d x 1 1
2
sin x cos x sin x
2
C . Thus sin x dx x sin x C.
dx 2 2 2 2 2 4
4 2 2
112. Our trial solution based on the chain rule is x 1
2 2 sin x C . Differentiate the solution to check:
d x 1 1 1
2
sin cos cos x
2
x C x . Thus cos x dx x 1 sin x C.
dx 2 2 2 2 2
2 2 2
113.
3
x
x dx 3ln x x C
2
2
114. 5 2 dx 5x 2 2 dx 5x 1 2 tan
1 x C
2 2 2
x x 1 x 1
115. 1 t t 1 t 1 t t
e e dt e e C e e C
2 2 1 2
t
116.
s
6
s 5
(5 s ) ds 5 s C 117.
ln 5 6
1
( )d C
2
2
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