<s>[INST] 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...
$$
\frac {16}3 \int \cos^2\theta \,d\theta\
$$
 [/INST] $\frac{16}{3}\int(cos^2\theta)$ 
= $\frac{16}{3}\int\frac{1-cos2\theta}{2}$
= $\frac{16}{6}\int(1-cos2\theta)$
= $\frac{16}{6}(\int(1) - \int(cos2\theta))$
= $\frac{16}{6}\theta$ - $\frac{16}{6}\int(cos2\theta)$
= $\frac{8}{3}\theta$ - $\frac{8}{3}\int(cos2\theta)$
= $\frac{8}{3}\theta$ - $\frac{8}{3}(\frac{1}{2})\int(cos(v))$, substitute $v = 2\theta$
= $\frac{8}{3}\theta$ + $\frac{4}{3}sin(v)$
= $\frac{8}{3}\theta$ + $\frac{4}{3}sin(2\theta)$
= $\frac{8}{3}\theta$ + $\frac{4}{3}(2) sin(\theta) cos
(\theta)$

= $\frac{8}{3}arcsin(\frac{3}{4}x)$ + $\frac{8}{3}(\frac{3x}{4})$$\frac{sqrt (16-9x^2)}{4}$
= $\frac{8}{3}arcsin(\frac{3}{4}x)$ + $\frac{1}{2}x[sqrt (16-9x^2)]$ + C
</s>