Integral, Trig Substitution - Stuck

[phyzmatix]

Homework Statement

Use trigonometric substitution to evaluate

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

The Attempt at a Solution

Let x=3\sin\theta
then dx=3\cos\theta d\theta

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

=\int{\frac{9\sin^2\theta}{3\sqrt{1-\sin^2\theta}}}\ 3\cos\theta \ d\theta

=\int{\frac{9\sin^2\theta}{3\sqrt{\cos^2\theta}}}\ 3\cos\theta \ d\theta

=\int{9\sin^2\theta \ d\theta}

=\frac{9}{2} \int{(1-\cos2\theta)}\ d\theta

let w=2\theta
then dw=2\ d\theta

=\frac{9}{4} \int{(1-\cos w)}\ dw

=\frac{9}{4}[w-\sin w] + c \ \mbox{(substituting everything back in)}

=\frac{9}{4}[2\sin^{-1}(\frac{x}{3})-\sin(2\sin^{-1}(\frac{x}{3})]+c

Is this correct so far? And if so, now what?

I'm stumped

Thanks!
phyz

 

[linearfish]

Your integration seems fine. I would clean up your substitution at the end. Use:

sin2 \theta = 2 sin \theta cos \theta

 

[snipez90]

Well at this step:

=\frac{9}{2} \int{(1-\cos2\theta)}\ d\theta

you could just find the antiderivative rather easily without another substitution. Then you can just relate theta to x instead of relating w to theta to x.

After you get the antiderivative in terms of x, that's pretty much it. Take the derivative to check.

 

[phyzmatix]

Your integration seems fine. I would clean up your substitution at the end. Use:

sin2 \theta = 2 sin \theta cos \theta

That's exactly what I needed. Thank you very much!