Prove that $\displaystyle
\left(\frac12\right)!=\frac{\sqrt\pi}2.
$
$\displaystyle \Gamma(1/2)=\int_{0}^{\infty}t^{-1/2}e^{-t}dt$
using the sub $\displaystyle u^2=t \mbox{ then } 2udu=dt $
$\displaystyle \Gamma(1/2)=\int_{0}^{\infty}u^{-1}e^{-u^2}(2udu)=2\int_{0}^{\infty}e^{-u^2}du$
we could do the same sub with $\displaystyle v^2=t$ and get
$\displaystyle \Gamma(1/2)=2\int_{0}^{\infty}e^{-v^2}dv$
so
$\displaystyle [\Gamma(1/2)]^2=(2\int_{0}^{\infty}e^{-u^2}du)(2\int_{0}^{\infty}e^{-v^2}dv)=4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(u^2+v^2)dudv}$
Convert to polar coordinates the region of integration is the 1st quadrant so we get...
$\displaystyle 4 (\int_{0}^{\pi/2}d\theta)(\int_{0}^{\infty}re^{-r^2}dr)=4 \left( \frac{\pi}{2}\right) \left( \frac{1}{2}\right)=\pi$
so
$\displaystyle [\Gamma(1/2)]^2=\pi \iff \Gamma(1/2)=\sqrt{\pi}$
Yeah, that's the standard solution.
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By considering $\displaystyle \int_0^1x^n(1-x)^n\,dx$ for $\displaystyle n\ge0,$ a direct application of Beta function shows that $\displaystyle \int_0^1x^n(1-x)^n\,dx=\frac{(n!)^2}{(2n+1)!}.$
Eventually, for $\displaystyle n=\frac12$ we have $\displaystyle \int_0^1\sqrt{x(1-x)}=\bigg\{\left(\frac12\right)!\bigg\}^2\cdot\fra c12=\frac\pi8,$ this 'cause we're integrating a semicircle whose radius is $\displaystyle \frac12.$ The conclusion follows.