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.

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


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


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