Differentiation Questions

Question Number 129490 by 676597498 last updated on 16/Jan/21

$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \mathrm{x}\sqrt{\mathrm{tanx}}\mathrm{dx} \\$$

Answered by mnjuly1970 last updated on 16/Jan/21

$$\:\Omega=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {x}.{sin}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right).{cos}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right){dx}\: \\$$$$\:\:\:\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right).{cos}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right)−\Omega \\$$$$\mathrm{2}\Omega=\frac{\pi}{\mathrm{4}}\left[\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right).{cos}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right){dx}\right] \\$$$$\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\:\left(\beta\left(\frac{\mathrm{3}}{\mathrm{4}}\:,\:\frac{\mathrm{3}}{\mathrm{4}}\right)\right)=\frac{\pi}{\mathrm{4}}\ast\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)} \\$$$$\:\:\:\therefore\:\:\Omega=\frac{\pi}{\mathrm{8}}\ast\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\pi}{\mathrm{4}\sqrt{\pi}\:}\ast\Gamma\:^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\$$