integral evaluation | 9math

integral evaluation

Posted December 9th, 2007 by nico

in

Analysis questions

Show that:

$\int_0^{\frac{\pi}{2}}\sin^{2n}x\:dx=\frac{\pi}{2}\frac{1}{2}\frac{3}{4}\frac{5}{6}...\frac{2n-1}{2n}$

$\int_0^{\frac{\pi}{2}}\sin^{2n+1}x\:dx=\frac{2}{3}\frac{4}{5}\frac{6}{7}...\frac{2n}{2n+1}$