Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\cos \theta)^2}$?
Setting then $x = 3\cos \theta; dx = -3\sin \theta d\theta$
$$-\int \frac{(3\cos\theta)^3}{3\sin\theta}3\sin\theta d\theta$$
$$-27\int\cos^3\theta d\theta$$
$$-27\int(1-\sin ^2\theta)\cos \theta d\theta$$
Substituting again, $u=\sin \theta; du=\cos \theta d\theta$
$$-27\int(1-u^2)du $$
$$-27u + 9u^3 + C$$
$$-27\sin \theta + 9 \sin^3 \theta + C$$
$$-9\sqrt{9-x^2} + 3\sin\theta\cdot 3\sin\theta\cdot \sin \theta + C$$
$$-9\sqrt{9-x^2} + (\sqrt{9-x^2})^2 \cdot \frac{\sqrt{9-x^2}}{3} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)^\frac{3}{2} + C $$
I guess I have more doubts that I've done the algebra correctly than the substitution, but in any case I'm not getting the correct answer. Have I calculated correctly? Is the answer simplified completely?
EDIT
Answer needed to be simplified further:
$$-9\sqrt{9-x^2} + \frac{1}{3}(\sqrt{9-x^2}^2 \sqrt{9-x^2}) + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}((9-x^2)\sqrt{9-x^2}) + C$$
$$\sqrt{9-x^2} \left (-9 + \frac{1}{3}(9-x^2) \right ) + C$$
$$\sqrt{9-x^2} \left (-6 - \frac{x^2}{3} \right ) + C$$
$$ \bbox[5px,border:2px solid red]
{
- \left ( 6+ \frac{x^2}{3} \right ) \sqrt{9-x^2}
}
$$ 
This is the answer the assignment was looking for.

What you have done is absolutely correct, except where you forgot to mention that $\theta$ is in $(0, \pi)$, but you can simplify your answer further. 
The book's answer might be something like $-\frac{1}{3} \sqrt{9-x^2} (x^2+18)$, which you can get by factoring out a factor of $\sqrt{9-x^2}$:
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$
$$= \sqrt{9-x^2} \left(-9 + \frac{1}{3}(9-x^2) \right)+ C$$
and you can surely continue from here.
