Answers and Explanations 477
Answers
701–800
766.
π
2
72
Here’s the given problem:
sin
−
−
∫
1
2
0
1 2
1
x
x
dx
Begin by using the substitution u = sin
−1
x so that du
x
dx
=
−
1
1
2
. Find the new limits of
integration by noting that if x = 1
2
, then u =
=
−
sin
1 1
2 6
π , and if x = 0, then u = sin
−1 
0 = 0.
With these new values, you get the following answer:
sin
−
−
=
=
=
( ) − ( )






=
∫
∫
1
2
0
1 2
0
6
2
0
6
2
2
2
1
2
1
2 6
0
72
x
x
dx
u du
u
π
π
π
π
767.
π
4
Here’s the given problem:
cos
sin
x
x
dx
1
2
0
2
+
∫
π
Begin with the substitution u = sin x so that du = cos x dx. Notice that if x = π
2
, then
u =
=
sin π
2
1, and that if x = 0, then u = sin 0 = 0. With these new values, you get the
following:
cos
sin
tan
tan
tan
x
x
dx
u
du
u
1
1
1
1
0
4
0
2
0
2
2
0
1
1
0
1
1
1
+
=
+
=
=
−
= −
=
∫
∫
−
−
−
π
π
π
4 4
768.
tan ln
−
( )+
1
x C
Start with the given problem:
1
1
2
x
x
dx
+ ( )
(
)
∫
ln