Thomas Calculus 13th [Solutions]

Chapter 5 Practice Exercises 419
101.
e
1 1/3 1 1/3
1 g
x dx u du
x
7 1
where u = 1 + 7 ln x, du 7 dx,
x = 1
x
u = 1, x = e u = 8
3 2/3 8 3 2/3 2/3
[ u ] 3 9
14 1 (8 1 ) (4 1)
14 14 14
102.
3 2 3 ln 4
[ln( v 1)] 2 1
2
dv
1 1 1 [ln( v 1)] dv u du
v
v 1 ln 2
, where u ln( v 1), du 1 dv;
v = 1 u = ln 2, v = 3
v 1
u = ln 4;
1 3 ln 4 1 3 3 1 3 3 (ln 2) 7 3
[ u ]
3 ln 2 [(ln 4) (ln 2) ] [(2ln 2) (ln 2) ] (8 1) (ln 2)
3 3 3 3
3
g log ln8
4
103. 1
g
d 1 1
1 ln 4 1 (ln ) d u du
ln 4 0
, where u = ln , du 1 d , = 1 u = 0, = 8 u = ln 8
1 2 ln 8 2 2 (3ln 2)
[ u ] 1
9ln 2
2ln 4 0 [(ln 8) 0 ]
ln16 4ln 2 4
2
e 8(ln 3)(log 1
3 ) e 8(ln 3)(ln ) e
104. d d 8 (ln ) 1 d 8 u du , where u = ln , du 1 d ; = 1
1 1 (ln 3) 1 0
u = 0, = e u = 1
2 1 2 2
4[ u ] 0 4(1 0 ) 4
105.
106.
3/4
6
3/4 3/2
dx 3 2 dx 3 1 du , where u = 2x, du = 2 dx; x 3 u 3 , x 3
3/4 2 3/4 2 2 3/2 2 2
9 4x 3 (2 x) 3 u
4 2 4
u 3
2
1
3/2
1 1 1
3 sin u 3 sin sin 1 3 3
3 3/2
2 2 6 6 3
1/5
6
1/5 1
dx 6 5 dx 6 1 du , where u = 5x, du = 5 dx;
1/5 2 5 1/5 2 2 5 1 2 2
4 25x 2 (5 x) 2 u
x 1 u 1, x 1 u 1
5 5
1
1
6 6 1 1
sin u sin 1 sin 1 6 6 2
5 2 1 5 2 2 5 6 6 5 3 5
2
107. 3
2 3
2 3
1
2 dt 3 2
2 dt 3 2
2 3
du , where u 3 t , du 3 dt;
2 2
4 3t
2 3t
2 u
t = 2 u 2 3, t = 2 u 2 3
1
2 3
1
3 1 1
3
3 tan u
tan 3 tan 3
2 2 2 3 2 2 3 3 3
3
3
108. 1
3 1 1 1 3
dt 1 dt 1 tan t 1 tan 3 tan 1 1
3
2
3
2
3 t
2 3 3 3 3 3 4 36
3 t
3
109.
1
1
1 1
dy 2 dy 1 du , where u = 2y and du = 2 dy
1/ 3 2 1/ 3 2 1/ 3 2
y 4 y 1 (2 y) (2 y) 1 u u 1
1
1
1
1
1 1
sec u sec 2 y sec 2 sec 2/ 3
1/ 3 1/ 3
3 6 6
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