Answers and Explanations 547
Answers
901–1,001
sec tan
sec
sec
tan
sec tan
sec
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
π
π
π
2
2
0
3
2
0
3
0
3
1
1
−
=
=
∫
∫
∫
d
d
d
= =
=
=
−
=
∫ cos
(sin )
sin
sin
θ θ
θ
π
π
π
d
0
3
0
3
3
0
3
2
929.
x
x
x
C
4
2
2
1
−
−
( ) +
−
sin
Start by rewriting the denominator with a radical:
x
x
dx
x
x
dx
2
2
3 2
2
2
3
4
4
−
(
)
=
−
(
)
∫
∫
Use the trigonometric substitution x = 2 sinθ to get dx
d
= 2cosθ θ . These substitutions
give you the following:
( sin ) cos
sin
sin cos
cos
sin co
2
2
4 4
8
2
2
2
3
2
2
3
2
θ
θ θ
θ
θ
θ θ
θ
θ
d
d
−
(
)
=
(
)
=
∫
∫
s s
cos
tan
sec
tan
θ
θ
θ
θ θ
θ
θ
θ θ
3
2
2
1
d
d
d
C
∫
∫
∫
=
=
−
(
)
=
− +
From the substitution x = 2 sinθ , or x
2
= sinθ , you can deduce that sin
−
( ) =
1
2
x
θ and
that tanθ =
−
x
x
4
2
. (To deduce that tanθ =
−
x
x
4
2
, you can label a right triangle
using x
2
= sinθ and then find the missing side with the Pythagorean theorem.)
Therefore, the solution becomes
tan
s in
θ θ
− + =
−
−
( ) +
−
C
x
x
x
C
4
2
2
1
930.
1
8
4
2
x
x
C
−





 +
Start by rewriting the integrand:
1
4
16
1
2
4
2
2
2
2
2
x
x
dx
x
x
dx
−
=
−
∫
∫ ( )