Answers and Explanations 481
Answers
701–800
2
2 3
4
2 1
4 1 4
2 1
2 1 2
2
0
1
2
0
1
2
0
1
u
u
du
u
u
du
u
u
du
+
(
)−
−
=
+
−






=
+
− ( )
∫
∫
∫
Now you can use the substitution w u
= 2
so that 2w = u and 2
 
dw = du. You can again find
the new upper limit of integration by using the substitution w u
= 2
and noting that when
u = 1, you have w = 1
2
. Likewise, you can find the new lower limit of integration by
noting that when u = 0, you have w = 0.
2 1
2 1 2
2
2 2
1
2 1
4
1
1
2
0
1
2
0
1 2
2
0
1 2
u
u
du
w
w
dw
w
w
dw
+
− ( )
=
+
−
=
+
−
∫
∫
∫
( )
Now split the integral:
4
1
1
4
1
1
1
2
0
1 2
2
2
0
1 2
0
1 2
w
w
dw
w
w
dw
w
dw
+
−
=
−
+
−
∫
∫
∫
In the first integral, use the substitution v = 1 – w
2
so that dv = –2w
 
dw and –2
 
dv =
4w
 
dw. You can again update the limits of integration by using the substitution
v = 1 – w
2
. To find the new upper limit of integration, note that when w = 1
2
, you have
v = − ( ) =
1 1
2
3
4
2
. To find the new lower limit of integration, note that when w = 0, you
have v = 1 – 0
2
= 1. Again, use this substitution only for the first integral:
4
1
1
1
2
1
1
2
2
0
1 2
0
1 2
2
0
1 2
1
3 4
w
w
dw
w
dw
v
dv
w
dw
−
+
−
=
−
+
−
∫
∫
∫
∫
Reverse the bounds in the first integral and change the sign. Then simplify:
=
+
−
=
+
=
−
+
−
−
−
∫
∫
2
1
1
2 2
4 1 4 3
4
1 2
2
0
1 2
3 4
1
1 2
3 4
1
1
0
1 2
1
v dv
w
dw
v
w
(
)
sin
sin 1 1
2
0
4 2 3 6
1
−
= −
+
−
sin
π
Whew!