Part II: The Answers
504
Answers
801–900
837.
1
8
8
sin θ + C
Start by rewriting the given integral:
sin cos
sin
cos
7
7
θ
θ θ
θ
θ θ
∫
∫
= (
)
d
d
If you let u = sinθ , then du
d
= cosθ θ. Substitute into the original integral to find the
answer:
u du u
C
C
C
7
8
8
8
8
1
8
1
8
∫
=
+
= (
) +
=
+
sin
sin
θ
θ
838.
0
Recall that tanθ is an odd function; that is, tan
t an
−
( )= −
θ
θ . Likewise,
tan
t an
3
3
−
( )= −
θ
θ so that tan
3
θ
( ) is also an odd function. Because this function is
symmetric about the origin and you’re integrating over an interval of the form [–a, a],
you get
tan
3
3
3
0
θ θ
π
π
−
∫
=
d
.
Here’s an alternate approach: Begin by saving a factor of the tangent and use an
identity to allow you to split up the integral:
tan
t an tan
sec
tan
tan
3
3
3
2
3
3
2
3
3 1
θ θ
θ
θ θ
θ
θ θ
θ
π
π
π
π
π
π
−
−
−
∫
∫
∫
=
=
+
(
)
=
d
d
d
+ +
(
)
=
+
−
−
−
∫
∫
∫
tan sec
tan
t an sec
θ
θ θ
θ θ
θ
θ θ
π
π
π
π
π
π
2
3
3
3
3
2
3
3
d
d
d
To evaluate the first integral, rewrite the tangent:
tan
sin
cos
θ θ
θ
θ
θ
π
π
π
π
d
d
−
−
∫
∫
=
3
3
3
3
Then use the substitution u = cosθ so that du
d
= − sinθ θ , or − =
du
d
sinθ θ . You can
find the new limits of integration by noting that if θ π
= 3
or if θ
π
= − 3
, you have
cos
c os
π
π
3
3
1
2
( ) = −
( ) = :
sin
cos
θ
θ
θ
π
π
d
u
du
−
∫
∫
=
−
3
3
1
1 2
1 2
Because the upper and lower limits of integration are the same, you have the following
for the first integral:
−
=
∫
1
0
1 2
1 2
u
du