Hi,
This is the integral I need to take $\displaystyle \int\{x^2}/{\sqrt(9-x^2) \}$
I get to this then dont know what to do $\displaystyle \int{x^2}$dθ
Hello, highc1157!
You didn't complete the substitution . . .
$\displaystyle \displaystyle \int \frac{x^2}{\sqrt{9-x^2}}\,dx $
$\displaystyle \text{Let: }\:x = 3\sin\theta \quad\Rightarrow\quad dx = 3\cos\theta\,d\theta $
$\displaystyle \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?
Once you have an answer in terms of $\displaystyle \theta$ you will have to back-substitute $\displaystyle x = 3 \sin \theta$ to get the answer in terms of $\displaystyle x$. 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 ....