Part II: The Answers
548
Answers
901–1,001
Use the substitution 2
4
x = secθ , or x = 2 secθ , so that dx
d
= 2 sec tan
θ
θ θ :
2
2
4
4
2
4
1 6
1
2
2
2
2
2
sec tan
( sec ) ( sec )
tan
sec
sec
θ
θ θ
θ
θ
θ θ
θ
θ
d
d
−
=
−
(
)
=
∫
∫
t tan
sec
t an
cos
sin
θ θ
θ
θ
θ θ
θ
d
d
C
4
1 6
1
8
1
8
2
(
)
=
=
+
∫
∫
From x = 2 secθ , or x
2
= secθ , you can deduce that sinθ =
−
x
x
2
4 . (To deduce that
sinθ =
−
x
x
2
4 , you can label a right triangle using x
2
= secθ and then find the missing
side of the triangle with the Pythagorean theorem.) Therefore, the solution becomes
1
8
1
8
4
2
sinθ + =
−





 +
C
x
x
C
931.
16
3
16 16
2
3 2
2
+
(
) −
+
+
x
x
C
Here’s the given integral:
x
x
dx
3
2
16
+
∫
Begin with the substitution x = 4 tanθ to get dx
d
= 4
2
sec θ θ . Substituting this value
into the original integral gives you the following:
( tan )
( tan )
sec
tan sec
tan
4
4
1 6
4
4
16
1
4
3
2
2
4
3
2
2
4
θ
θ
θ θ
θ
θ
θ
θ
+
=
+
(
)
=
∫
∫
d
d
t tan sec
sec
tan sec
sec
tan sec
3
2
2
4
3
2
3
3
16
4
4
4
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
(
)
=
=
∫
∫
∫
d
d
dθ θ
Now factor out tanθ and use a trigonometric identity along with a u-substitution:
4
4
1
3
2
3
2
tan tan sec
sec
s ec tan
θ
θ
θ θ
θ
θ
θ θ
∫
∫
=
−
(
)
d
d