# Math Help - very basic integration problem

1. ## very basic integration problem

Hi guys. I am trying to integrate $\frac{1}{9-x^2}dx$ by trig substiution, but I'm not really sure if you can even do trig sub here since we are dealing with square roots. If I'm right, is there another method besides trig sub to evaluate this, even if it's not necessarily easier? This is what I got:

$\int\frac{1}{9-x^2}dx$

Let $x = 3sin\theta$

So $dx = 3 cos\theta d\theta$
Substitution yields $\int \frac{3cos\theta}{9-9sin^2\theta}d\theta$

Simplification yields
$
\frac{1}{3}\int \frac{cos\theta}{1-sin^2\theta}d\theta
$

or simply:
$

\frac{1}{3}\int sec\theta d\theta
$

So now we have $\frac{1}{3}ln|sec\theta+tan\theta| + C$

Using a right triangle to make substitutions, my final answer is

$
\frac{1}{3}ln\left|\frac{x}{\sqrt{x^2-9}}+\frac{3}{\sqrt{x^2-9}}\right|+C
$

So again, even if this right, is there another way to evaluate this?
Thanks for any help,
James

2. Originally Posted by james121515
Hi guys. I am trying to integrate $\frac{1}{9-x^2}dx$ by trig substiution, but I'm not really sure if you can even do trig sub here since we are dealing with square roots. If I'm right, is there another method besides trig sub to evaluate this, even if it's not necessarily easier? This is what I got:

$\int\frac{1}{9-x^2}dx$

Let $x = 3sin\theta$

So $dx = 3 cos\theta d\theta$
Substitution yields $\int \frac{3cos\theta}{9-9sin^2\theta}d\theta$

Simplification yields
$
\frac{1}{3}\int \frac{cos\theta}{1-sin^2\theta}d\theta
$

or simply:
$

\frac{1}{3}\int sec\theta d\theta
$

So now we have $\frac{1}{3}ln|sec\theta+tan\theta| + C$

Using a right triangle to make substitutions, my final answer is

$
\frac{1}{3}ln\left|\frac{x}{\sqrt{x^2-9}}+\frac{3}{\sqrt{x^2-9}}\right|+C
$

So again, even if this right, is there another way to evaluate this?
Thanks for any help,
James
If I had to do this question I would use partial fractions:

$\frac1{9-x^2}=-\frac1{x^2-9}=-\left(\frac{A}{x+3} + \frac{B}{x-3} \right)$

which will yield $A=\frac16$ and $B=-\frac16$.

3. You could use partial fractions