$\int\frac{x^3}{\sqrt{4+x^2}}$ I was trying to calculate 
$$\int\frac{x^3}{\sqrt{4+x^2}}$$
Doing $x = 2\tan(\theta)$, $dx = 2\sec^2(\theta)~d\theta$, $-\pi/2 < 0 < \pi/2$ I have:
$$\int\frac{\left(2\tan(\theta)\right)^3\cdot2\cdot\sec^2(\theta)~d\theta}{2\sec(\theta)}$$
which is 
$$8\int\tan(\theta)\cdot\tan^2(\theta)\cdot\sec(\theta)~d\theta$$
now I got stuck ... any clues what's the next substitution to do?
I'm sorry for the formatting. Could someone please help me with the formatting?

Let $u = x^2 + 4$, $du = 2x\,dx$:
\begin{align*}
I &= \frac{1}{2} \int \frac{u - 4}{\sqrt{u}}du
\end{align*}
Should be easy to take it from there.
