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?
 A: A way to compute this is as follows:
\begin{align*}
\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\mathrm d x &=\int_0^1\frac{4x+x^3-4x}{\sqrt{4+x^2}}\mathrm d x\\
&=\int_0^1x\sqrt{4+x^2}\mathrm d x -2\int_0^1\frac{2x}{\sqrt{4+x^2}}\mathrm d x\\
&=\left.\frac{1}{3}(4+x^2)^{3/2}\right|_0^1-4\left.\left(4+x^2\right)^{1/2}\right|_0^1\\
&=\frac{5\sqrt{5}-8}{3}-4\left(\sqrt{5}-2\right)\\
&=\boxed{\color{blue}{\dfrac{16-7\sqrt{5}}{3}}}
\end{align*}
A: By the change of variable $x^2+4=t$, we have $2x\,dx=dt$:
$$\int_0^1\frac {x^3}{\sqrt {4+x^2}}\,dx=\frac12\int_4^5\frac{t-4}{\sqrt t}\,dt=\frac13t\sqrt t-4\sqrt t\Big|_4^5=\frac{16-7\sqrt5}3.$$
A: Following your way:
$$\begin{align}\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx &=8\int_0^{\arctan(1/2)}\tan^3\theta\sec\theta{\rm d}\theta\tag{$x=2\tan\theta$}\\&=8\int_1^{\sqrt{5}/2}(t^2-1){\rm d}t\tag{$t=\sec\theta$}\\&=8\left(\frac{t^3}3-t\right)\Bigg|_1^{\sqrt5/2}\\&=\frac83\left(\frac{5\sqrt5}8-1\right)-8\left(\frac{\sqrt5}2-1\right)\\&=\large \frac{16-7\sqrt5}3
\end{align}$$
A: Hint:
$$\int_0^1\frac{x^3}{\sqrt{4+x^2}}dx=\frac{1}{2}\int_0^1\frac{x^22x}{\sqrt{4+x^2}}dx=\int_0^1\frac{(x^2+4-4)(4+x^2)'}{\sqrt{4+x^2}}dx$$
A: Alternately, let $x=2\sinh t$, and use the fact that $\cosh^2t-\sinh^2t=1,~\sinh't=\cosh t$, and $\cosh't=\sinh t$. You will finally arrive at $\displaystyle\int\sinh^3t~dt=\int(\cosh^2t-1)~d(\cosh t)=$ $=\displaystyle\int(u^2-1)~du$, which is trivial to evaluate.
