The improper integral $ \int_{0}^{1}\frac{\sqrt{x}}{\sqrt{1-x^6}} dx $ I've ran into this improper integral:
$$
\int_{0}^{1}\frac{\sqrt{x}}{\sqrt{1-x^6}} dx
$$
I've tried the substitution $t=\sqrt x$ without success.
Any help will be greatly appreciated.
Thank you,
Yaron.
 A: Substitute $u = x^6$, $x=u^{1/6}$, $dx = (1/6) u^{-5/6} du$.  Then look up what a beta function is.  http://en.wikipedia.org/wiki/Beta_function
I get after substitution
$$\frac16 \int_0^1 du \, u^{-5/6} u^{1/12}(1-u)^{-1/2}$$
which takes the value
$$\frac16 \frac{\Gamma\left( \frac14 \right ) \Gamma \left ( \frac12 \right)}{\Gamma\left ( \frac{3}{4} \right)} = \frac{\Gamma^2\left(\frac14\right)}{6 \sqrt{2\pi}}$$
A: Substitute $t=x^6$ and you get
$$
\begin{align}
\int_0^1\frac{\sqrt{x}}{\sqrt{1-x^6}}\,\mathrm{d}x
&=\frac16\int_0^1t^{-3/4}(1-t)^{-1/2}\,\mathrm{d}t\\
&=\frac16\mathrm{B}\left(\frac14,\frac12\right)\tag{$\ast$}\\
&=\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac12\right)}{6\,\Gamma\left(\frac34\right)}\\
&=\frac{\sqrt\pi}{6}\frac{\Gamma\left(\frac14\right)}{\Gamma\left(\frac34\right)}\\
&=\frac{\Gamma\left(\frac14\right)^2}{6\sqrt{2\pi}}
\end{align}
$$
The last equation is due to the reflection formula: $\displaystyle\Gamma(x)\,\Gamma(1-x)=\frac\pi{\sin(\pi x)}$
$(\ast)$ follows from $\displaystyle\mathrm{B}(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\,\mathrm{d}t=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$
