# Tricky trig sub integration

• Oct 1st 2010, 12:52 PM
highc1157
Tricky trig sub integration
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
• Oct 1st 2010, 01:03 PM
pickslides
What substitution did you make?

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

Also remember $\displaystyle \displaystyle \sin^2 \theta = \frac{1-\cos \theta}{2}$
• Oct 1st 2010, 01:07 PM
Soroban
Hello, highc1157!

You didn't complete the substitution . . .

Quote:

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

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

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

Got it?
• Oct 1st 2010, 02:30 PM
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 :(
• Oct 2nd 2010, 03:57 PM
mr fantastic
Quote:

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 $\displaystyle \theta$ 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 ....