# How do I prove this derivation of a definite integral?

Q,How to prove that

$\int_{0 }^{\Pi /2}\sin ^{m}x \cos ^{n}x dx =\left [{(m-1)(m-3)(m-5)...2 or 1}\right ]\left [ \left ( n-1)\left ( n-3 \right )..2 or 1 \right ) \right ]\div \left [ \left ( m+n)(m+n-2) \right...2 or 1 ) \right ]$

I tried integrating it by parts and came up with this: $\int_{0 }^{\Pi /2}\sin ^{m}x \cos ^{n}x dx =\frac{m-1}{n+1}\int_{0 }^{\Pi /2}\sin ^{m-2}x \cos ^{n+2}x$

And if I carry on integrating by parts,

$\int_{0 }^{\Pi /2}\sin ^{m}x \cos ^{n}x dx =\frac{m-1}{n+1}\times\frac{m-3}{n+3}\int_{0 }^{\Pi /2}\sin ^{m-4}x \cos ^{n+4}x$

Finally,I will get

$\int_{0 }^{\Pi /2}\sin ^{m}x \cos ^{n}x dx =\frac{m-1}{n+1}\times\frac{m-3}{n+3}...\frac{2}{m+n-2}\int_{0 }^{\Pi /2}\sin ^{1}x \cos ^{n+m-1}x$

I am unable to figure out the proof or even claim that my attempted integration is correct.Any help or hints are will be well appreciated.

• There is a simple way i.e beta function – Behrouz Maleki Jul 16 '16 at 18:09
• @BehrouzMaleki,Sir,what is a beta function? – user312318 Jul 16 '16 at 18:09
• mathworld.wolfram.com/BetaFunction.html – Behrouz Maleki Jul 16 '16 at 18:12
• What part of the proof are you unable to figure out? BTW, what you have written is valid. – DanielWainfleet Jul 16 '16 at 18:21

set $u=\sin x$, we have $$I=\int_{0}^{1}(u^2)^{\frac{m-1}{2}}(1-u^2)^{\frac{n-1}{2}}u\,du$$