Answers and Explanations 549
Answers
901–1,001
Let u = secθ to get du
d
= sec tan
θ
θ θ . Using these values in the integral, you get
4
1
4
3
4 1
3
3
2
3
3
3
3
u
du
u u
C
C
−
(
) =
−





 +
=
−
(
) +
∫
(sec ) sec
θ
θ
From the first substitution, you get
x
4
= tanθ , from which you can deduce that
16
4
2
+
=
x
secθ . ( You can find the value 16
4
2
+
=
x
secθ by using a trigonometric
identity for tangent and secant.) Inserting this value into the preceding equation and
simplifying a bit gives you the answer:
4
3
16
4
4
16
4
16
3
16 16
16
3
2
3
3
2
2
3
2
+





 −
+





 +
=
+
(
) − + +
=
+
x
x
C
x
x
C
x
2 2
3 2
2
3
16 16
(
) −
+
+
x
C
932.
1
5
1
2
3
1
1
2
5 2
2
3 2
2
+
(
) − +
(
) + + +
x
x
x
C
Here’s the given integral:
x
x
dx
5
2
1
+
∫
Begin with the substitution x = tanθ so that dx
d
= sec
2
θ θ :
(tan ) sec
tan
tan sec
tan sec tan
θ
θ θ
θ
θ
θ θ
θ
θ
θ θ
5
2
2
5
4
1
d
d
d
+
=
=
∫
∫
∫
Next, use an identity to get
tan
sec tan
sec
s ec tan
2
2
2
2
1
θ
θ
θ θ
θ
θ
θ θ
(
)
=
−
(
)
∫
∫
d
d
Now use the substitution u = secθ so that du
d
= sec tan
θ
θ θ :
u
du
u
u
du
u
u
u C
2
2
4
2
5
3
5
3
1
2
1
5
2
3
1
5
2
3
−
(
)
=
−
+
(
)
=
−
+ +
=
−
+
∫
∫
(sec )
(sec )
θ
θ
s secθ + C