Answers and Explanations 567
Answers
901–1,001
Integrating the second term of the integrand is a bit more difficult. Begin by using a
trigonometric substitution:
−
+
(
)
∫
17
3
2
2
x
dx with x = 3 tanθ so that dx
d
= 3
2
sec θ θ .
Using these substitutions gives you the following:
−
+
(
)
=
−
(
) +
(
)
=
−
+
(
∫
∫
17
3
17 3
3
3
17 3
3
1
2
2
2
2
2
2
2
x
dx
d
sec
tan
sec
tan
θ
θ
θ
θ
θ ) )
(
)
= −
(
)
= −
= −
∫
∫
∫
2
2
2
2
2
2
17 3
9
17 3
9
1
17 3
9
d
d
d
d
θ
θ
θ
θ
θ
θ
θ θ
sec
sec
sec
cos
∫ ∫
∫
∫
= −
+
(
)
= −
+
(
)
= − −
+
17 3
9
1
2
1
2
17 3
18
1
2
17 3
18
1
2
cos( )
cos( )
s
θ
θ
θ
θ
θ
d
d
i in( )
sin cos
sin cos
2
17 3
18
1
2
2
17 3
18
θ
θ
θ
θ
θ
θ
θ
(
) +
= −
+
(
) +
= −
+
(
) +
C
C
C
Using the substitution x = 3 tanθ gives you
x
3
= tanθ , from which you can deduce
that tan
− 




 =
1
3
x
θ , sinθ =
+
x
x
2
3
, and cosθ =
+
3
3
2
x
.
So the antiderivative of the second term becomes
−
+
+
= −





 +
+






+

−
17 3
18
17 3
18
3
3
3
3
1
2
2
(
sin cos )
tan
θ
θ
θ C
x
x
x
x
 






 


 
+
= −





 +
+





 +
−
C
x
x
x
C
17 3
18
3
3
3
1
2
tan
Combining the two solutions gives you the answer: