Evaluating $\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx$ How do I evaluate the definite integral $$\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx ?$$ I used trig substitution, and then a u substitution for $\sec\theta$. 
I tried doing it and got an answer of: $-\sqrt{125}+12\sqrt{5}-16$, but apparently its wrong. 
Can someone help check my error?

Following your way:
$$\begin{align}\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx &=8\int_0^{\arctan(1/2)}\tan^3\theta\sec\theta{\rm d}\theta\tag{$x=2\tan\theta$}\\&=8\int_1^{\sqrt{5}/2}(t^2-1){\rm d}t\tag{$t=\sec\theta$}\\&=8\left(\frac{t^3}3-t\right)\Bigg|_1^{\sqrt5/2}\\&=\frac83\left(\frac{5\sqrt5}8-1\right)-8\left(\frac{\sqrt5}2-1\right)\\&=\large \frac{16-7\sqrt5}3
\end{align}$$
