Part II: The Answers
528
Answers
801–900
894.
−
+
2
7
7 2
(csc )
x
C
Start by rewriting the radical:
csc csc cot
(csc ) csc cot
5
5 2
x
x
x dx
x
x
x dx
∫
∫
=
Use the substitution u = csc x so that du = –csc x cot x
 
dx, or –du = csc x cot x
 
dx:
− ( )
= −
+
= −
+
∫ u du u C
x
C
5 2
7 2
7 2
7 2
2
7
(csc )
895.
1
3
1
5
3
5
sin
s in
x
x C
−
+
Here’s the given integral:
cos sin
3
2
x
xdx
∫
Begin by factoring out cos x and using a trigonometric identity to write cos
2
x in terms
of sin x:
cos sin cos
sin
sin cos
2
2
2
2
1
x
x
x dx
x
x
x dx
∫
∫
=
−
(
)
Now use the substitution u = sin x, which gives you du = cos x
 
dx. Putting these values
into the integral gives you the following:
1
1
1
3
1
5
1
2
2
2
2
2
4
3
5
−
(
)
=
−
(
)
=
−
(
)
=
−
+
=
∫
∫
∫
sin
sin cos
x
x
x dx
u u du
u u du
u
u C
3 3
1
5
3
5
sin
s in
x
x C
−
+
896.
27
4
The given integral is
tan sec
3
4
0
3
x
xdx
π
∫
Begin by factoring out sec
2 
x and using a trigonometric identity on one factor of sec
2 
x
to write it in terms of tan x:
tan sec
tan sec sec
tan
t an
sec
3
4
0
3
3
2
2
0
3
3
2
2
1
x
xdx
x
x
x dx
x
x
x
π
π
∫
∫
=
=
+
(
)
d dx
0
3
π
∫