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Show that \( \int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{\tan \theta}}=\frac{\pi \sqrt{2}}{2} \) |
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Answer» \(\int\limits_0^{\pi/2}\frac{d\theta}{\sqrt{tan\theta}}\) = \(\int\limits_0^{\pi/2}\cfrac{d\theta}{\sqrt{\frac{sin\theta}{cos\theta}}}\) \(\int\limits_0^{\pi/2}\frac{\sqrt{cos\theta}}{\sqrt{sin\theta}}d\theta\) = \(\int\limits_0^{\pi/2}sin^{-1/2}\theta\,cos^{1/2}\theta d\theta\) \(=\cfrac{\Gamma(\frac{-1/2+1}2)\Gamma(\frac{1/2+1}2)}{2\Gamma(\frac{-1/2+1/2+2}2)}\) \(\Big(\)\(\because\) \(\int\limits_0^{\pi/2}\)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 |
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