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!