Show that $B(\alpha,\alpha)=2\int_0^{1/2}(\frac{1}{4}-(\frac{1}{2}-x)^{2})^{\alpha-1}dx$
Where $B(\alpha,\beta)=\int_0^{1}x^{\alpha-1}(1-x)^{\beta-1}dx$
I tried in many ways
$$B(\alpha,\alpha) =\frac{\Gamma(\alpha)\Gamma(\alpha)}{\Gamma(2\alpha)}$$
Trying to apply the legendre's formula: $$\Gamma(\alpha)\Gamma(\alpha+1/2)=\frac{\sqrt{\pi}}{2^{2\alpha-1}}\Gamma(2\alpha)$$
As well \begin{align*} 2\int_0^{1/2}\left(\frac{1}{4}-\left(\frac{1}{2}-x\right)^{2}\right)^{\alpha-1}dx &=2\int_0^{1/2}\left(\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}-x\right)^{2}\right)^{\alpha-1}dx\\ &=2\int_0^{1/2}(1-x)^{\alpha-1}x^{\alpha-1}dx\\ \end{align*}
but does not seem leading nowhere useful.
It follows from the fact that $x-x^2$ is symmetric about the line $x=\frac12$. Thus the integral equals twice that from $0$ to $\frac12$...