Question Number 196928 by SANOGO last updated on 03/Sep/23
$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} {x}\:{dx} \\ $$
Commented by Frix last updated on 03/Sep/23
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}^{{n}} \:{x}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{B}\:\left(\frac{\mathrm{1}}{\mathrm{2}},\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)\:= \\ $$$$=\frac{\sqrt{\pi}\:\Gamma\:\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\:\Gamma\:\left(\frac{{n}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$