Using trig to integrate $x^2/\sqrt{16-x^2}$ I'm trying to integrate:   $\int\frac{x^2}{\sqrt{16-x^2}}\ dx$
My try was to convert $x$ to $4\sin(u)$ and $dx$ to $4\cos(u)du$, but I'm not sure. Thanks.

Notice, let $x=4\sin u\implies dx=4\cos u$ 
$$\begin{align}
\int \frac{x^2}{\sqrt{16-x^2}}\ dx&=\int \frac{16\sin^2u}{\sqrt{16-16\sin^2u}}(4\cos u\ du)\\\\
&=\int \frac{16\sin^2u}{\cos u}(4\cos u\ du)\\\\
&=\int \sin^2 u\ du\\\\
&=\int \frac{1-\cos (2u)}{2}\ du\\\\
&=\frac{1}{2}\int \ du-\frac{1}{2}\int \cos (2u)\ du\\\\
&=\frac{1}{2}\int \ du-\frac{1}{4}\int \cos (2u)\ d(2u)
\end{align}$$
