Answers and Explanations 551
Answers
901–1,001
934.
1
6
1
1
3
3
6
sin ( )
−
+
−
(
) +
x
x
x
C
Start by rewriting the integrand:
x
x dx
x
x dx
2
6
2
32
1
1
−
=
−
∫
∫
( )
Use a substitution, letting w = x
3
, to get dw = 3x
2 
dx, or 1
3
2
dw x dx
=
. Putting these
values into the integral gives you
x
x dx
w dw
2
3
2
2
1
1
3
1
−
=
−
∫
∫
( )
Now use the trigonometric substitution w = 1sinθ , which gives you dw
d
= cosθ θ .
Using these values in the integral gives you the following:
1
3
1
1
3
1
3
2
2
2
−
=
=
∫
∫
∫
sin cos
cos cos
cos
θ
θ θ
θ
θ θ
θ θ
d
d
d
Now you can use the identity cos
c os( )
2
1
2
1
2
θ
θ
=
+
(
) to rewrite the integral:
1
3
1
2
1
2
1
6
1
2
1
6
1
2
2
1
6
+
(
)
=
+
(
)
=
+
(
) +
=
+
∫
∫
cos( )
cos( )
sin( )
θ
θ
θ
θ
θ
θ
θ
d
d
C
1 1
2
2
1
6
sin cos
(
sin cos )
θ
θ
θ
θ
θ
(
) +
=
+
+
C
C
From w = sinθ , you can deduce that sin
−
=
1 w θ and that cosθ = −
1
2
w . (Note that
you can find the value for cosθ by using an identity for sine and cosine.) Substituting
these values into the integral gives you
1
6
1
1
2
sin
−
+
−
(
) +
w w
w
C
Replacing w with x
3
gives you the solution:
1
6
1
1
3
3
6
sin ( )
−
+
−
(
) +
x
x
x
C
935.
1
16
3
7 6
2
x
x x
C
+
−
−





 +
Rewrite the denominator as a radical:
dx
x x
dx
x x
7 6
7 6
2
3 2
2
3
−
−
(
)
=
−
−
(
)
∫
∫
Now you complete the square on the quadratic expression: