$\int\frac{x^3}{\sqrt{4+x^2}}$ 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?

You have not chosen an efficient way to proceed. However, let us continue along that path.
Note that $\tan^2\theta=\sec^2\theta-1$. So you want
$$\int 8(\sec^2\theta -1)\sec\theta\tan\theta\,d\theta.$$
Let $u=\sec\theta$.
Remark: My favourite substitution for this problem and close relatives is a variant of the one used by Ayman Hourieh. Let $x^2+4=u^2$. Then $2x\,dx=2u\,du$, and $x^2=u^2-4$. So
$$\int \frac{x^3}{\sqrt{x^2+4}}\,dx=\int \frac{(u^2-4)u}{u}\,du=\int (u^2-4)\,du.$$
