$\int_{0}^{\frac{\pi }{2}}\sin^{m}x \cos^{m}x=\int_{0}^{\frac{\pi }{2}}2^{-m} \cos^{m}x dx$

I tried integration by parts but got nowhere. Any help would be greatly appreciated.


$$\int_0^{\pi/2}\sin^m x \cos^m x = 2^{-m} \int_0^{\pi/2}(2 \sin x \cos x )^mdx = 2^{-m} \int_0^{\pi/2} \sin^m (2x) dx$$ Making substitution $2x = y$ gives $$2^{-(m+1)} \int_0^{\pi} \sin ^m y dy \overset{(1)}{=} 2^{-(m+1)} 2\int_0^{\pi/2}\sin^m y dy$$ Changing $y\to \pi/2-y$ gives the required result. $$2^{-m}\int_0^{\pi/2}\sin^m (\pi/2-y) dy = 2^{-m} \int_{0}^{\pi/2} \cos^m(x) dx$$ $(1)$: becuase $\sin x$ in $[0, \pi]$ is symmetric at $x=\pi/2$

  • $\begingroup$ Think there is a way to derive it from beta and gamma functions? $\endgroup$ – ClassicStyle Jun 2 '14 at 1:25
  • $\begingroup$ @TylerHG do you mean to evaluate the whole integral? $\endgroup$ – Santosh Linkha Jun 2 '14 at 3:53
  • $\begingroup$ Not really. I know that it can be evaluated with beta functions. But was wondering if you could change the integral into the new form suggested by the OP using beta and gamma funcs. $\endgroup$ – ClassicStyle Jun 2 '14 at 4:15


Set $\displaystyle2x=y\implies x=\frac\pi2\implies y=\pi$

Now $$\int_0^\pi\cdots (dy)=\int_0^\frac\pi2\cdots (dy)+\int_\frac\pi2^\pi\cdots (dy)$$

Set $\displaystyle z=y-\frac\pi2$

Finally use $\displaystyle\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$ to convert the integrand sine to cosine

  • $\begingroup$ @SantoshLinkha, May I request you to be a more patient in rectifying my answer? $\endgroup$ – lab bhattacharjee Jun 1 '14 at 6:43
  • $\begingroup$ sure, I just edited the latex ... sorry if it was bad. $\endgroup$ – Santosh Linkha Jun 1 '14 at 6:45
  • $\begingroup$ @SantoshLinkha, Actually I was going to put the answer in the current version when I found your quick edit $\endgroup$ – lab bhattacharjee Jun 1 '14 at 7:16
  • $\begingroup$ lol ... sorry again :) it would have worked anyway. $\endgroup$ – Santosh Linkha Jun 1 '14 at 7:24

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