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As stated in the title, I want to prove that: $$\int_0^\frac{\pi}{2}\cos^mx \sin^mxdx=2^{-m}\int_0^\frac{\pi}{2}\cos^mxdx$$

So far, I thought that $$\int_0^\frac{\pi}{2}\cos^mx \sin^mxdx=2^{-m}\int_0^\frac{\pi}{2}\sin^m2xdx$$ But that doesn't seem to take me anywhere. I would appreciate some help! Thanks!

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    $\begingroup$ First use the substitution $u=2x$. Then use the symmetry of $\sin$ across $\pi/2$. Then use a substitution with the identity $\cos(\pi/2-x)=\sin(x)$. $\endgroup$ – anon Mar 30 '18 at 0:18
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You gave a simple answer to your question. I however would like to show you how the identity you are discussing can be generalized to almost all complex values of $m$.

This integral can be related to the Beta Function, defined as $$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}dt$$ Keeping this in mind, we begin the evaluation of our integral: $$I(m)=\int_0^{\pi/2}\sin^mt\cos^mt\ dt$$ Making the substitution $u=\sin^2t$ gives $$I(m)=\int_0^1u^{m/2}(1-u)^{m/2}\frac12u^{-1/2}(1-u)^{-1/2}du$$ $$I(m)=\frac12\int_0^1u^{\frac{m-1}2}(1-u)^{\frac{m-1}2}du$$ $$I(m)=\frac12\int_0^1u^{\frac{m+1}2-1}(1-u)^{\frac{m+1}2-1}du$$ $$I(m)=\frac12B\bigg(\frac{m+1}2,\frac{m+1}2\bigg)$$ This fact yields little significance until one recalls the identity (with $\Gamma(\cdot)$ denoting The Gamma Function) $$B(a,b)=B(b,a)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ This gives our integral: $$I(m)=\frac{\Gamma^2(\frac{m+1}2)}{2\Gamma(m+1)}=\frac{\Gamma^2(\frac{m+1}2)}{2\Gamma(2\cdot\frac{m+1}2)}$$ Using these identities (and the substitution $u=\sin^2t$), it is easily shown that $$\int_0^{\pi/2}\sin^at\ dt=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac12)}{2\Gamma(\frac{a+2}2)}=\int_0^{\pi/2}\cos^at\ dt$$ When we recall that $\Gamma(\frac12)=\sqrt{\pi}$, and the Gamma duplication formula: $$\Gamma(z)\Gamma\bigg(z+\frac12\bigg)=2^{1-2z}\sqrt{\pi}\ \Gamma(2z)$$ We can plug in $z=\frac{m+1}2$ $$\Gamma\bigg(\frac{m+1}2\bigg)\Gamma\bigg(\frac{m+1}2+\frac12\bigg)=2^{1-2\frac{m+1}2}\sqrt{\pi}\ \Gamma\bigg(2\frac{m+1}2\bigg)$$ $$\Gamma\bigg(\frac{m+1}2\bigg)\Gamma\bigg(\frac{m+2}2\bigg)=2^{-m}\Gamma\bigg(\frac12\bigg)\Gamma(m+1)$$ $$\frac{\Gamma(\frac{m+1}2)}{\Gamma(m+1)}\Gamma\bigg(\frac{m+2}2\bigg)=2^{-m}\Gamma\bigg(\frac12\bigg)$$ $$\frac{\Gamma(\frac{m+1}2)}{\Gamma(m+1)}=2^{-m}\frac{\Gamma(\frac12)}{\Gamma(\frac{m+2}2)}$$ $$\frac{\Gamma^2(\frac{m+1}2)}{\Gamma(m+1)}=\frac{\Gamma(\frac12)\Gamma(\frac{m+1}2)}{2^m\Gamma(\frac{m+2}2)}$$ $$\frac{\Gamma^2(\frac{m+1}2)}{2\Gamma(m+1)}=\frac{\Gamma(\frac12)\Gamma(\frac{m+1}2)}{2^{m+1}\Gamma(\frac{m+2}2)}$$ And recalling our previously derived value value for $I(m)$, and our identities involving the integrals of $\sin^a$ and $\cos^a$, $$I(m)=2^{-1}B\bigg(\frac{m+1}2,\frac{m+1}2\bigg)=2^{-m-1}B\bigg(\frac12,\frac{m+1}2\bigg)$$ $$I(m)=2^{-m}\int_0^{\pi/2}\sin^mt\ dt=2^{-m}\int_0^{\pi/2}\cos^mt\ dt$$ QED

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Let $2x=\pi/2-u$, then $2dx=-2du$, so:

$$2^{-m}\int_0^{\frac{\pi}{2}} \sin ^m2xdx=-2^{-m}\frac{1}{2}\int_\frac{\pi}{2}^{-\frac{\pi}{2}}\sin^m({\pi/2-u})du=-2^{-m}\int_0^\frac{\pi}{2}\cos^mudu$$ This happens because $\cos(x)$ is even, therefore $\cos^mx$ is symmetric over symmetric intervals.

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