Thomas Calculus 13th [Solutions]

644 Chapter 8 Techniques of Integration
dx
1
dx
0
dx
1
dx dx
x 1 e x 1 e 1 x 1 e 0 x 1 e 1 x 1 e
68.
2 x 2 x 2 x 2 x 2 x
;
1 2 x
2
x 1
x
2
x
1
x
2 1 e
x
e
x
1
lim lim lim 1 e 2 and dx
2
x 0 x 0 x 0
0 x
diverges
1
diverges dx
0 2
x 1
e
x
diverges dx
x
2 1 e
x
69.
u x
x dx
u
;
2 2u du 2 2 2 3 2
2 2 2 2 2ln |1 |
1 x
dx u u du u u u u C
du
1 u 1 u 3
2 x
3/2
2x x 2 x 2ln 1 x C
3
70.
3 2
x 2 dx x 4x 2 dx x dx 3 dx 5 dx x 3 ln | 2 | 5 ln | 2 |
2 2
4 x
x 4
2 x 2 2 x 2 2 2 x 2
x C
71.
2
2x x dx ; x 1 sin dx cos d
For the integrand to be nonnegative x must be between 0 and 2 so x 1 is between 1 and 1 and we can take
between 2 and / 2, where cosine is nonnegative. Thus
2 2 2 2
2 1 ( 1) 1 sin cos cos .
x x x
2 2 1
2 cos cos cos 1 cos2
2
1 1 1 1
sin 2 C sin cos C
2 4 2 2
1 1 1
2
sin ( x 1) ( x 1) 2x x C
2 2
x x dx d d d
72.
dx
dx
2 2
2x x 1 ( x 1)
1
sin ( 1)
x
C
73. 2 cos x sin x
dx 2csc 2
x dx cos x dx
csc x dx 2cot x 1
2 2
sin
sin
sin x
ln | csc x cot x | C
x
x
2cot x csc x ln | csc x cot x | C
74.
2 5
sin cos d
2 5 2 2
2
d d ; u sin du cos d
sin cos sin 1 sin cos
2 2
2
2 4 6
sin 1 sin cos d sin 2sin sin cos
d
3
2 4 6 u 2 5 1 7
u 2u u du u u C
3 5 7
3
sin 2 1
sin sin
3 5 7
5 7
C
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