Part II: The Answers
524
Answers
801–900
883.
x
x
x
x C
2
1
1
2
1
2
1
2
tan
t an
−
−
−
+
+
Here’s the given integral:
x
xdx
tan
−
∫
1
Use integration by parts with u = tan
−1 
x so that du
x
dx
= +
1
1
2
and let dv = x
 
dx
so that v x
=
2
2
:
x
xdx x
x
x
x
dx
x
x
x
x
dx
tan
t an
tan
−
−
−
∫
∫
∫
=
−
+
=
−
+ −
+
1
2
1
2
2
2
1
2
2
2
1
2 1
2
1
2
1 1
1
= =
−
− +






=
−
+
+
−
−
−
∫
x
x
x
dx
x
x
x
x C
2
1
2
2
1
1
2
1
2
1
1
1
2
1
2
1
2
tan
tan
t an
Note: The numerator and denominator are almost the same for the expression x
x
2
2
1 +
,
allowing you to add 1 and subtract 1 (in the second line) and then break it up into two
fractions (in the third line); this approach lets you avoid long division. Another way to
simplify x
x
2
2
1 +
is to use polynomial long division to arrive at x
x
x
2
2
2
1
1
1
1
+
= − +
.
884.
3
4
π
Here’s the given integral:
sin ( )
2
0
3 2
2θ θ
π
d
∫
Recall that sin ( )
c os( )
2
1
2
1
2
θ
θ
=
−
(
) so that sin ( )
c os( )
2 2
1
2
1
4
θ
θ
=
−
(
) . Using this identity
in the integral gives you
1
2
1
4
1
2
1
4
4
1
2
3
2
1
4
6
0
0
3 2
0
3 2
−
(
)
=
−
(
)
=
−
−
∫
cos( )
sin( )
sin( )
θ
θ
θ
θ
π
π
π
π
d
− −
(
)






=
1
4
0
3
4
sin( )
π
885.
π
8
1
4
+
The given integral is
cos ( )
2
0
4
θ θ
π
d
∫
Recall that cos ( )
c os( )
2
1
2
1
2
θ
θ
=
+
(
) . Using this identity gives you the following: