Question Number 132925 by metamorfose last updated on 17/Feb/21
$$\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}\left({x}\right)}{dx}\right)^{\mathrm{2}} +\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}\left({x}\right)}{dx}\right)^{\mathrm{2}} <\mathrm{8}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$
Answered by Dwaipayan Shikari last updated on 17/Feb/21
$$\Phi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}\left({x}\right)}\:{dx}=\mathrm{2}\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}=\mathrm{2}\sqrt{\mathrm{2}}\pi\frac{\mathrm{1}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$\Psi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}\left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}\left({x}\right)}{dx}=\mathrm{2}\sqrt{\mathrm{2}}\pi.\frac{\mathrm{1}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$\Phi^{\mathrm{2}} +\Psi^{\mathrm{2}} =\frac{\mathrm{16}\pi^{\mathrm{2}} }{\Gamma^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$
Commented by Ñï= last updated on 17/Feb/21
$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}….? \\ $$$$\Gamma\left(\mathrm{1}−{x}\right)\Gamma\left({x}\right)=\frac{\pi}{{sin}\pi{x}}…? \\ $$$${Confused}\:{why}\:{you}\:{got}\:{that}\:{result}. \\ $$