Evaluating the indefinite integral $\int\sqrt{16-9x^2}\,dx$ I need to solve the integral below, but I just can't figure how.

$$\int \sqrt{16-9x^2}\,dx$$

I have tried to replace $9x^2$ with $16\sin^2\theta$. I get to a point where I have the function below. Please let me know whether I'm on the right track, and please explain to me how to finish it...
$$
\frac {16}3 \int \cos^2\theta \,d\theta\
$$

Hint: $$\cos^2\theta=\frac{1}{2}(1+\cos(2\theta))$$

Edit: Also, after some calculations, I am pretty sure you made an error (can anyone double check this?)
You have got $\displaystyle\frac{4}{3}\int\cos^2\theta d\theta$, but the substitution  $9x^2=16\sin^2\theta$ gives us the following:
$$\sqrt{16-9x^2}=\sqrt{16-16\sin^2\theta}=\sqrt{16(1-\sin^2\theta)}=\sqrt{16\cos^2\theta}=4\cos\theta$$
and since $x=\frac{4}{3}\sin\theta$, we have got $dx=\frac{4}{3}\cos\theta d\theta$, so
$$\displaystyle\int\sqrt{16-9x^2}dx=\int4\cos\theta\cdot\left(\frac{4}{3}\cos\theta  d\theta\right)=\frac{16}{3}\int\cos^2\theta d\theta$$
and here is the way to do the rest 

$\displaystyle\frac{16}{3}\int\cos^2\theta d\theta=\frac{16}{3}\int\left(\frac{1}{2}\left(1+\cos 2\theta)\right)\right)=\frac{8}{3}\int\left(1+\cos 2\theta\right)=\frac{8}{3}\left(\theta +\frac{1}{2}\sin2\theta+C\right)$.
 Since $\sin\theta =\frac{3x}{4}$, we have  $\theta=\sin^{-1}\left(\frac{3x}{4}\right)$ and $\sin2\theta = \sqrt{16-9x^2}$ and so $\displaystyle\int\sqrt{16-9x^2}dx$ evaluates to $\displaystyle\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)+\frac{1}{2}x\sqrt{16-9x^2}+C$

Edit: (To address OP's questions, as this is comment is not properly showing up in the comments section). After multiplying $\frac{8}{3}$ with $\frac{1}{2}\sin\theta$, we have $\displaystyle\frac{4}{3}\sin2\theta=\frac{4}{3}(2\sin\theta\cos\theta)=\frac{8}{3}\sin\theta\cos\theta=\frac{8}{3}\cdot\left(\frac{3x}{4}\right)\cdot\left(\frac{\sqrt{16-9x^2}}{4}\right)$, then cancel you get it.
