Question

$\because {\int }_0^{\frac{\pi }{2}}{\sin }^m{xcos}^{n}xdx=\frac{n-1}{m+n}\frac{n-3}{m+n-2}....\frac{1}{m+2}\frac{m-1}{m}\frac{m-3}{m-2}.....\frac{1}{2}\frac{\pi }{2}$; if m is even, n is even

Answer 1

${\int }_0^{\pi /2}{\sin }^{m}xdx=\int {\cos }^{n-1}x.{\sin }^{m}x\cos xdx$

Put $\sin x=t$

$\cos xdx=dt$

$\therefore I=\int \left({\sqrt{1-t^2}}^{n-1}\right)t^{m}dt$

Applying Integration By Path

$=\int \left({\sqrt{1-t^2}}^{n-1}\right)\frac{t^{m+1}}{m+1}-{\int }_0^1-\left({\sqrt{1-t^2}}^{n-3}\right)t^{m+2}\frac{n-1}{m+1}dt$

Resultituting $t=\sin x$

$\therefore I=0+{\int }_0^{\pi /2}\cos x^{n-2}\sin x^{m+2}dx\left(\frac{n-1}{m+1}\right)$

$\therefore I=\frac{n-1}{m+1}{\int }_0^{\pi /2}\sin x^m(1-{\cos }^{2}x){\cos }^{n-2}xdx$

$\therefore I=\frac{n-1}{m+1}[\int {\sin }^{m}x{\cos }^{n-2}xdx-\int {\sin }^{n}x{\cos }^{n}xdx]$

$\therefore \frac{n-1}{m+1}\int {\sin }^{m}x{\cos }^{n-2}xdx-\left(\frac{n-1}{m+1}\right)I$

$\therefore I\left(\frac{m+n}{m+1}\right)=\frac{n-1}{m+1}\int {\sin }^{m}x{\cos }^{n-2}xdx$

$\therefore I=\frac{n-1}{m+n}\underset{\downarrow }{\underset{}{\int {\sin }^{m}x{\cos }^{n-3}x\cos xdx}}$

Solving in a similar manner yirld the required answer.

$\therefore I=\frac{n-1}{m+n}.\frac{n-3}{m+n-2}$