Part II: The Answers
552
Answers
901–1,001
7 6
6
7
6
7
6
9 7 9
3
16
16
2
2
2
2
2
−
−
= − −
+
= −
+
+
= −
+
+
(
) + +
= − +
+
= −
x x
x
x
x
x
x
x
x
(
)
(
)
( x x + 3
2
)
So the integral becomes
dx
x
16
3
2
3
− +
(
)
∫
(
)
Now use the substitution (
)
sin
x + =
3 4
θ , which gives you dx
d
= 4 cosθ θ . Substituting
these values into the integral gives you
4
16 4
4
16 1
4
4
2
3
2
3
2
cos
( sin )
cos
sin
cos
cos
θ θ
θ
θ θ
θ
θ θ
d
d
d
−
(
)
=
−
(
)
(
)
=
∫
∫
θ θ
θ θ
θ
θ
θ
θ θ
θ
(
)
=
=
=
=
+
∫
∫
∫
∫
3
3
3
2
2
2
4
4
1
4
1
1
16
1
16
cos
cos
cos
sec
tan
d
d
d
C
From the substitution (
)
sin
x + =
3 4
θ, or x + =
3
4
sinθ , you can deduce that
tanθ =
+
−
−
x
x x
3
7 6
2
. ( To deduce that tanθ =
+
−
−
x
x x
3
7 6
2
, you can label a right triangle
using x + =
3
4
sinθ and then find the missing side with the Pythagorean theorem.)
Therefore, the solution is
1
16
3
7 6
2
x
x x
C
+
−
−





 +
936.
2
2
2
2
4
2
2
ln x
x
x
C
+ −
+
+
+
Here’s the given integral:
dx
x
x
2
4
2
+
+
∫
This may not look like a trigonometric substitution problem at first glance because
so many of those problems involve some type of radical. However, many integrals that
contain quadratic expressions can still be solved with this method.