Answers and Explanations 543
Answers
901–1,001
1
1
5
1
5
1
5
1
2
1
2
1
2 5
2
2
−






=
=
+
(
)
=
+
∫
∫
∫
sin
c os
cos
cos( )
θ
θ θ
θ θ
θ
θ
θ
d
d
d
s sin
sin cos
(
sin cos )
2
2
1
2 5
2
2
1
2 5
θ
θ
θ
θ
θ
θ
θ
(
) +
=
+
(
) +
=
+
+
C
C
C
From the substitution x 5 = sinθ , you can deduce that sin
−
( ) =
1
5
x
θ and that
cosθ = −
1 5
2
x . (To deduce that cosθ = −
1 5
2
x , you can label a right triangle using
x 5
1
= sinθ and find the missing side of the triangle with the Pythagorean theorem.)
Therefore, the solution becomes
1
2 5
5
5 1 5
1
2 5
5
5 25
2
1
2
1
sin
sin
−
−
( ) +
−
)
(
+
=
( ) + −
)
(
+
x
x
x
C
x
x
x
C
923.
1
3
3
9
2
ln x
x
x
C
−
−
+
The given integral is
dx
x
x
9
2
−
∫
Begin by using the trigonometric substitution x = 3 sinθ so that dx
d
= 3 cosθ θ :
3
3
9 9
9 1
3
2
2
cos
sin
s in
cos
sin
s in
cos
sin c
θ
θ
θ
θ
θ θ
θ
θ
θ
θ θ
θ
−
=
−
(
)
=
∫
∫
d
d
d
d
o os
csc
ln csc
cot
2
1
3
1
3
θ
θ
θ θ
θ
θ
d
d
C
∫
∫
=
=
−
+
From the substitution x = 3 sinθ , you can deduce that cscθ = 3
x
and that
cotθ =
−
9
2
x
x
. (To deduce that cotθ =
−
9
2
x
x
, you can label a right triangle using
x
3
= sinθ and find the missing side of the triangle with the Pythagorean theorem.)
Therefore, the solution becomes
1
3
3
9
2
ln x
x
x
C
−
−
+