Part II: The Answers
506
Answers
801–900
841.
1
2
1
2
tan
−
(
) +
x
C
Here’s the given integral:
tan
−
+
∫
1
2
1
x
x
dx
Let u = tan
−1 
x so that du
x
dx
= +
1
1
2
. Substituting these values into the original integral,
you get
u du u C
x
C
∫
=
+
= (
) +
−
2
1
2
2
1
2
tan
842.
ln sec x C
+
Start by rewriting the integral:
tan
sin
cos
x dx
x
x
dx
∫
∫
=
Now you can perform a u-substitution with u = cos x so that du = –sin x
 
dx, or –du =
sin x
 
dx. Substituting this expression into the original integral gives you the following:
−
= −
+
= −
+
∫
1
u
du
u C
x C
ln
ln cos
You can now use the power property of logarithms to rewrite the integral:
−
+ =
+
=
+
=
+
−
ln cos
l n (cos )
ln cos
ln sec
x C
x
C
x
C
x C
1
1
843.
e – e
1/2
The given integral is
e
xdx
x
cos sin
0
3
π
∫
Begin by using the substitution u = cos x so that du = –sin x
 
dx, or –du = sin x
 
dx. Notice
you can find the new limits of integration by noting that if x = π
3
, then u =
=
cos π
3
1
2
,
and if x = 0, then u = cos 0 = 1. With these values, you have the following:
−
=
=
= −
= −
∫
∫
e du
e du
e
e e
e e
u
u
u
1
1 2
1 2
1
1 2
1
1
12
1 2