Int sqrt over 9-x/x dx
Answers
Explanation:
We want
∫
√
9
−
x
2
x
d
x
.
In general, when encountering an integral involving
√
a
2
−
x
2
,
make the trigonometric substitution
x
=
a
sin
θ
.
In this case,
a
2
=
9
,
a
=
3
,
and so our substitution is
x
=
3
sin
θ
.
Solving for
d
x
yields
d
x
=
3
cos
θ
d
θ
.
Apply the substitution:
∫
3
cos
θ
√
9
−
9
sin
2
θ
3
sin
θ
d
θ
Simplify:
∫
cos
θ
√
9
(
1
−
sin
2
θ
)
sin
θ
d
θ
Recall the identity
sin
2
θ
+
cos
2
θ
=
1
This identity also tells us that
1
−
sin
2
θ
=
cos
2
θ
Rewrite with the identity applied:
3
∫
cos
θ
√
cos
2
θ
sin
θ
d
θ
=
3
∫
(
cos
2
θ
sin
θ
)
d
θ
Now, to solve the resultant integral, we're best off rewriting again with the identity reversed:
3
∫
1
−
sin
2
θ
sin
θ
d
θ
=
3
(
∫
csc
θ
d
θ
−
∫
sin
θ
d
θ
)
∫
csc
θ
d
θ
=
ln
|
csc
θ
−
cot
θ
|
. Memorize this, it's a fairly common integral.
∫
sin
θ
d
θ
=
−
cos
θ
3
ln
|
csc
θ
−
cot
θ
|
+
3
cos
θ
+
C
(Don't forget the constant of integration).
We now need to rewrite in terms of
x
.
We said
x
=
3
sin
θ
,
so
sin
θ
=
x
3
,
csc
θ
=
1
x
3
=
3
x
.
To solve for cosine, recall
sin
2
θ
+
cos
2
θ
=
1
,
x
2
9
+
cos
2
θ
=
9
9
,
cos
2
θ
=
9
−
x
2
9
,
cos
θ
=
√
9
−
x
2
3
.
Finally, we end up with
∫
√
9
−
x
2
x
d
x
=
ln
∣
∣
∣
3
−
√
9
−
x
2
3
∣
∣
∣
+
√
9
−
x
2
+