Part II: The Answers
550
Answers
901–1,001
From the substitution x = tanθ , you can deduce that secθ = +
1
2
x to get the solution:
1
5
1
2
3
1
1
1
5
1
2
3
1
1
2
5
2
3
2
2
5 2
2
3 2
2
+
(
) − +
(
) + + +
=
+
(
) − +
(
) + + +
x
x
x
C
x
x
x
C
Note that you can also do this problem starting with the substitution u
x
=
+
2
1
and a bit of algebra.
933.
3 4 2 1
5
2 2 1
3
5
− −
−
The given expression is
x x
dx
3
2
0
3
9
+
∫
Let x = 3 tanθ to get dx
d
= 3
2
sec θ θ. You can also find the new limits of integration by
using the original limits along with the substitution x = 3 tanθ . If x = 3, you get 3 3
= tanθ ,
or 1 = tanθ , so that π θ
4
= . Likewise, if x = 0, you get 0 3
= tanθ , or 0 = tanθ , so that 0 = θ .
Using these substitutions and the new limits of integration gives you the following:
( tan ) ( tan )
sec
( tan )
tan
se
3
3
9 3
3
9
1 3
0
4
3
2
2
0
4
3
2
θ
θ
θ θ
θ
θ
π
π
∫
∫
+ (
)
+
=
(
)
d
c c
( tan )
sec
sec
tan sec
2
0
4
3
2
2
5
0
4
3
3
3
3
3
3
θ θ
θ
θ
θ θ
θ
θ θ
π
π
(
)
=
(
)
=
∫
∫
d
d
d
Now factor out sec tan
θ
θ and use a trigonometric identity:
3
3
3
5
0
4
3
3
5
0
4
2
2
5
0
4
2
π
π
π
θ
θ θ
θ
θ
θ
θ θ
θ
∫
∫
∫
=
=
−
tan sec
tan sec sec tan
sec
d
d
1 1
2
(
) sec sec tan
θ
θ
θ θ
d
Use the substitution u = secθ so that du
d
= sec tan
θ
θ θ . You can again find the new
limits of integration; with θ π
= 4
, you get u =
=
=
sec π
4
2
2
2, and with θ = 0, you get
u = sec 0 = 1. Using the new limits and the substitution, you can get the answer as follows:
3
1
3
3
5
3
3
2
5
2
5
2
2
1
2
5
4
2
1
2
5
5
3
1
2
5
5
u
u du
u u du
u
u
−
(
)
=
−
(
)
=
−
=
−
∫
∫
3 3
5
3
1
5
1
3
3 4 2 1
5
2 2 1
3
−
−
( )
=
− −
−
550
Answers
901–1,001
From the substitution x = tanθ , you can deduce that secθ = +
1
2
x to get the solution:
1
5
1
2
3
1
1
1
5
1
2
3
1
1
2
5
2
3
2
2
5 2
2
3 2
2
+
(
) − +
(
) + + +
=
+
(
) − +
(
) + + +
x
x
x
C
x
x
x
C
Note that you can also do this problem starting with the substitution u
x
=
+
2
1
and a bit of algebra.
933.
3 4 2 1
5
2 2 1
3
5
− −
−
The given expression is
x x
dx
3
2
0
3
9
+
∫
Let x = 3 tanθ to get dx
d
= 3
2
sec θ θ. You can also find the new limits of integration by
using the original limits along with the substitution x = 3 tanθ . If x = 3, you get 3 3
= tanθ ,
or 1 = tanθ , so that π θ
4
= . Likewise, if x = 0, you get 0 3
= tanθ , or 0 = tanθ , so that 0 = θ .
Using these substitutions and the new limits of integration gives you the following:
( tan ) ( tan )
sec
( tan )
tan
se
3
3
9 3
3
9
1 3
0
4
3
2
2
0
4
3
2
θ
θ
θ θ
θ
θ
π
π
∫
∫
+ (
)
+
=
(
)
d
c c
( tan )
sec
sec
tan sec
2
0
4
3
2
2
5
0
4
3
3
3
3
3
3
θ θ
θ
θ
θ θ
θ
θ θ
π
π
(
)
=
(
)
=
∫
∫
d
d
d
Now factor out sec tan
θ
θ and use a trigonometric identity:
3
3
3
5
0
4
3
3
5
0
4
2
2
5
0
4
2
π
π
π
θ
θ θ
θ
θ
θ
θ θ
θ
∫
∫
∫
=
=
−
tan sec
tan sec sec tan
sec
d
d
1 1
2
(
) sec sec tan
θ
θ
θ θ
d
Use the substitution u = secθ so that du
d
= sec tan
θ
θ θ . You can again find the new
limits of integration; with θ π
= 4
, you get u =
=
=
sec π
4
2
2
2, and with θ = 0, you get
u = sec 0 = 1. Using the new limits and the substitution, you can get the answer as follows:
3
1
3
3
5
3
3
2
5
2
5
2
2
1
2
5
4
2
1
2
5
5
3
1
2
5
5
u
u du
u u du
u
u
−
(
)
=
−
(
)
=
−
=
−
∫
∫
3 3
5
3
1
5
1
3
3 4 2 1
5
2 2 1
3
−
−
( )
=
− −
−