Question Number 137707 by mohammad17 last updated on 05/Apr/21
Answered by Dwaipayan Shikari last updated on 05/Apr/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{tan}\theta}\:{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\frac{\mathrm{1}}{\mathrm{2}}} \theta\:\:{cos}^{−\frac{\mathrm{1}}{\mathrm{2}}} \theta\:{d}\theta \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{2}}=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$