Part II: The Answers
540
Answers
901–1,001
4
4
4
16
4
4
1 6
1
2
2
2
2
sec tan
( sec ) ( sec )
sec tan
( sec )
sec
θ
θ
θ
θ
θ
θ
θ
θ
θ
−
=
−
∫
d
6 6
4
4
1 6
1
4
4
1 6
2
2
2
2
d
d
θ
θ
θ
θ
θ
θ
θ
θ
θ
∫
∫
=
−
(
)
=
sec tan
( sec )
sec
sec tan
sec
t an n
sec tan
sec ( tan )
sec
cos
2
2
2
2
4
4
4
1
4
1
1
16
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ θ
(
)
=
=
=
∫
∫
∫
d
d
d
d
∫ ∫
=
+
1
16
sinθ C
From your original substitution, you have x
4
= secθ . From this, you can deduce that
sinθ =
−
x
x
2
16 . (To deduce that sinθ =
−
x
x
2
16 , you can label a right triangle using
x
4
= secθ and then use the Pythagorean theorem to find the missing side.) Therefore,
the solution becomes
1
16
1
16
16
2
sinθ + =
−
+
C
x
x
C
919.
9
2
3
9
9
1
2
sin
−
+
−





 +
x x
x
C
Here’s the given integral:
9
2
−
∫
x dx
Begin by using the trigonometric identity x = 3 sinθ so that dx
d
= 3 cosθ θ :
9 9
3
9 1
9
2
2
2
−
(
)
=
−
=
∫
∫
∫
sin
cos
sin cos
cos
θ
θ θ
θ
θ θ
θ θ
d
d
d
Now use an identity to get
9 1
2
1
2
9
2
2
2
9
2
2
2
9
2
+
(
)
=
+
(
) +
=
+
(
) +
=
+
∫
cos( )
sin
sin cos
(
s
θ
θ
θ
θ
θ
θ
θ
θ
d
C
C
i in cos )
θ
θ + C