Part II: The Answers
534
Answers
901–1,001
908.
−
−
+
1
3
1
5
3
5
(cot )
(cot )
x
x
C
Begin by factoring out csc
2 x and using an identity to get the following:
cot csc
cot csc csc
cot
c ot
csc
2
4
2
2
2
2
2
2
1
x
xdx
x
x
x dx
x
x
x dx
∫
∫
∫
=
=
+
(
)
Now use the substitution u = cot x so that du = –csc
2 
x
 
dx, or –du = csc
2 
x
 
dx:
−
+
( ) = − −
= −
−
+
= −
−
+
∫
∫
u
u du
u u du
u
u
C
x
x
C
2
2
2
4
3
5
3
5
1
3
5
1
3
1
5
(cot )
(cot )
909.
cos(cos )
cos (cos )
cos (cos )
θ
θ
θ
−
+
+
2
3
1
5
5
3
C
Here’s the given integral:
sin sin (cos )
θ
θ θ
5
d
∫
Begin by performing the substitution u = cosθ so that du
d
= − sinθ θ , or − =
du
d
sinθ θ .
Using these values in the integral gives you − ∫ sin
5 u du. Now factor out sin u and use
some algebra and a trigonometric identity to write sin
2 
u in terms of cos u:
−
=−
= − (
)
= −
−
(
)
∫
∫
∫
sin
s in sin
sin
sin
cos
sin
5
4
2
2
2
2
1
u du
u
u du
u
udu
u
udu
∫ ∫
Again use a substitution to integrate: Let w = cos u so that dw = –sin u
 
du,
or –dw = sin u
 
du. Using these values yields the integral 1
2
2
−
(
)
∫ w dw. Now
simply expand the integrand and integrate:
1
1 2
2
3
1
5
2
2
2
4
3
5
−
(
) = − +
(
)
= −
+
+
∫
∫
w dw
w w dw
w
w
w C
Then undo all the substitutions:
w
w
w C
u
u
u C
−
+
+
=
−
+
+
=
−
2
3
1
5
2
3
1
5
2
3
3
5
3
5
3
cos
cos
cos
cos(cos )
cos (cos
θ
θ ) )
cos (cos )
+
+
1
5
5
θ C