# Thread: Tricky trig sub integration

1. ## Tricky trig sub integration

Hi,

This is the integral I need to take $\int\{x^2}/{\sqrt(9-x^2) \}$

I get to this then dont know what to do $\int{x^2}$

2. What substitution did you make?

Did you try $x = 3\sin\theta$

Also remember $\displaystyle \sin^2 \theta = \frac{1-\cos \theta}{2}$

3. Hello, highc1157!

You didn't complete the substitution . . .

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

$\text{Let: }\:x = 3\sin\theta \quad\Rightarrow\quad dx = 3\cos\theta\,d\theta$

$\displaystyle \text{Substitute: }\;\int \frac{\overbrace{9\sin^2\!\theta}^{x^2}}{\underbra ce{3\cos\theta}_{\sqrt{9-x^2}}}\,\underbrace{(3\cos\theta\,d\theta)}_{dx} \;=\;9\!\1\int \sin^2\theta\,d\theta$

Got it?

4. yea i get that thanks a lot Sorobon,

but i'm still having trouble i used the identity for the integral of sine squared and my answer looks nothing like the answer in my book

5. Originally Posted by highc1157
yea i get that thanks a lot Sorobon,

but i'm still having trouble i used the identity for the integral of sine squared and my answer looks nothing like the answer in my book
Once you have an answer in terms of $\theta$ you will have to back-substitute $x = 3 \sin \theta$ to get the answer in terms of $x$. If you need more help, please show all the details of your working (not just one or two steps). I'm sure your classnotes and textbook will have examples to follow ....