\(\int\limits_0^{\pi/2}\frac{d\theta}{\sqrt{tan\theta}}\) = \(\int\limits_0^{\pi/2}\cfrac{d\theta}{\sqrt{\frac{sin\theta}{cos\theta}}}\)
\(\int\limits_0^{\pi/2}\frac{\sqrt{cos\theta}}{\sqrt{sin\theta}}d\theta\) = \(\int\limits_0^{\pi/2}sin^{-1/2}\theta\,cos^{1/2}\theta d\theta\)
\(=\cfrac{\Gamma(\frac{-1/2+1}2)\Gamma(\frac{1/2+1}2)}{2\Gamma(\frac{-1/2+1/2+2}2)}\)
\(\Big(\)\(\because\) \(\int\limits_0^{\pi/2}\)sinm θ cosn θ dθ = \(\cfrac{\Gamma(\frac{m+1}2)\Gamma(\frac{n+1}2)}{2\Gamma(\frac{m+n+2}2)}\)\(\Big)\)
\(=\cfrac{\Gamma(1/4)\Gamma(3/4)}{2\Gamma(1)}\)
\(=\frac{\pi}{2sin(\pi/4)}\) (\(\because\) \(\Gamma(n)\Gamma(1-n)=\frac{\pi}{sin\,n\pi}\) and here n = 1/4 and \(\Gamma\)(1) = 0! = 1)
\(=\frac{\pi}{2\times1/\sqrt2}\) (\(\because\) sin \(\pi/4\) = 1/√2)
\(=\frac{\pi}{\sqrt2}\) or \(\frac{\pi/2}2\)
Hence proved
