Part II: The Answers
532
Answers
901–1,001
sin
c os( )
sin( )
sin( )
2
1
2
1
2
1
2
1
2
2
1
2
1
4
2
x dx
x dx
x
x
C
x
x
∫
∫
=
−
(
)
=
−
(
) +
=
−
+ C C
Therefore, you get the following:
x
xdx x x
x
x
x dx
x
x
sin
s in( )
sin( )
sin(
2
2
1
2
1
4
2
1
2
1
4
2
1
2
1
4
∫
∫
=
−
(
) − −
(
)
=
−
2 2
1
4
1
8
2
1
4
1
4
2
1
8
2
2
2
x
x
x C
x
x
x
x C
)
c os( )
sin( )
cos( )
−
−
+
=
−
−
+
904.
1
4
1
6
4
6
(sin )
(sin )
x
x
C
−
+
Here’s the given integral:
sin cos
3
3
x
xdx
∫
For this problem, you can factor out either sin x or cos x and then use an identity.
For example, you can factor out cos x and then use the identity cos
2
x = 1 – sin
2 
x:
sin cos
sin cos cos
sin
s in
cos
3
3
3
2
3
2
1
x
xdx
x
x
x dx
x
x
x dx
∫
∫
∫
=
=
−
(
)
Then use the substitution u = sin x so that du = cos x
 
dx to get
u
u du
u u du
u
u
C
x
x
C
3
2
3
5
4
6
4
6
1
4
6
1
4
1
6
−
( ) = −
(
)
=
−
+
=
−
+
∫
∫
(sin )
(sin )
905.
1
3
3
sec
sec
x
x C
−
+
The given integral is
tan sec
3 x
xdx
∫
Begin by factoring out tan x and using an identity:
tan sec
tan sec tan
sec
s ec tan
3
2
2
1
x
xdx
x
x
x dx
x
x
x dx
∫
∫
∫
=
=
−
(
)