Originally Posted by

**james121515** Hi guys. I am trying to integrate $\displaystyle \frac{1}{9-x^2}dx$ 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:

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

Let $\displaystyle x = 3sin\theta$

So $\displaystyle dx = 3 cos\theta d\theta$

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

Simplification yields

$\displaystyle

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

$

or simply:

$\displaystyle

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

$

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

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

$\displaystyle

\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