7

Example 6 Use the substitution
evaluate.
Let
x sin
dx cos
d
1
2 dx
1 2
2 x 1 x
2
4
6
4
6
cot
( 1) (1.7321)
0.7321(4dps)
Example 7: Evaluate
Solution:
Let x
x
dx
( x 2 )
cos 2
dx 2sin 2d
1
0
2
(sin )
1 x
1 x
0
2
cos
( 1
cos ec d
4
6
4
4
sin
1
2
dx
x 2 )
dx
x
1
2
1
2
x
0
1
2
d
)
1
0 1
4
6
4
1
x
dx
x
x sin to
4
0
1 cos 2
( 2 sin 2 ) d
1 cos 2
1 cos 2
2 sin 2 d
0 1 cos 2
4
2
(
3
3
2
( x 2 ) K
Example 8: Evaluate
Solution:
Let
2 1
t tan
2
1 2
dt sec d
2 2
2dt
d
2
sec
2
2dt
2
1
tan
2
2dt
2
1
t
3
0
=
2
sin
2 sin 2 d
cos
0
4 2 sin cos
2 d
cos
2
4 sin d
0
0
4 1
4 ( 1 cos 2 ) d
2
0
2 sin 2 0
2 3
0 1 sin
d
2 d
2 tan
1 2
2
1 tan
2
2
2 sin
sin 2 d
2 cos
0
2
4
4
2
4
273