Answers and Explanations 555
Answers
901–1,001
938.
8
3
4
3 7 6
2
1
2
sin
(
)
−
−
( ) +
−
+
−
+
x
x
x x
C
Here’s the given integral:
7 6
2
+
−
∫
x x dx
First complete the square on the quadratic expression underneath the square root:
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
)
Therefore, you have the integral
7 6
16
3
2
2
+
−
=
− −
∫
∫
x x dx
x
dx
(
)
Now you can use the substitution (
)
sin
x − =
3 4
θ , which gives you dx
d
= 4 cosθ θ .
Substituting these values into the integral gives you
16
3
16 4
4
16 1
4
16
2
2
2
− −
=
−
=
−
(
)
=
∫
∫
∫
(
)
( sin ) cos
sin
cos
cos
x
dx
d
d
θ
θ θ
θ
θ θ
2 2
2
4
16
θ
θ θ
θ θ
∫
∫
=
cos
cos
d
d
Then use the identity cos
c os( )
2
1
2
1
2
θ
θ
=
+
(
) to simplify the integral:
16
16 1
2
1
2
8 1
2
8
1
2
2
2
cos
c os( )
cos( )
sin( )
θ θ
θ
θ
θ
θ
θ
θ
d
d
d
∫
∫
∫
=
+
(
)
=
+
(
)
=
+
( (
) +
=
+
(
) +
=
+
+
C
C
C
8
1
2
2
8
θ
θ
θ
θ
θ
θ
sin cos
(
sin cos )
From the substitution (
)
sin
x − =
3 4
θ , you get x − =
3
4
sinθ and sin
−
−
( ) =
1
3
4
x
θ . You
can deduce that cosθ =
+
−
7 6
4
2
x x by labeling a right triangle using x − =
3
4
sinθ and
then finding the missing side of the triangle with the Pythagorean theorem. Putting
these values into the antiderivative gives you the solution:
8
3
4
3
4
7 6
4
8
3
4
1
2
1
sin
sin
−
−
−
( ) + −
( )
+
−
















+
=
−
( )
x
x
x x
C
x
+ +
−
+
−
+
(
)
x
x x
C
3 7 6
2
2