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...
$$
\frac {16}3 \int \cos^2\theta \,d\theta\
$$

Notice, this standard formula $$\int\sqrt{a^2-x^2}dx=\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\sin^{-1}\left(\frac{x}{a}\right)\right)$$ Hence, we have 
$$\int\sqrt{16-9x^2}dx=\int\sqrt{(4)^2-(3x)^2}dx$$ let $3x=u\implies 3dx=du$$$=\frac{1}{3}\int\sqrt{(4)^2-(u)^2}du $$ $$=\frac{1}{3}\cdot\frac{1}{2}\left(u\sqrt{(4)^2-u^2}+(4)^2\sin^{-1}\left(\frac{u}{4}\right)\right)$$ $$=\frac{1}{6}\left(3x\sqrt{(4)^2-(3x)^2}+16\sin^{-1}\left(\frac{3x}{4}\right)\right)$$ $$=\color{blue}{\frac{1}{2}x\sqrt{16-9x^2}+\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)}$$ 
