# Prove that \int_0^\frac{\pi}{2}\cos^m x \sin^m xdx=2^{-m}\int_0^\frac{\pi}{2}\cos^m xdx

Prove that ${\int }_{0}^{\frac{\pi }{2}}{\mathrm{cos}}^{m}x{\mathrm{sin}}^{m}xdx={2}^{-m}{\int }_{0}^{\frac{\pi }{2}}{\mathrm{cos}}^{m}xdx$
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vanvanvan4ie
Let $2x=\frac{\pi }{2}-u$ ,then $2dx=-2du$, so:
${2}^{-m}{\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{m}2xdx=-{2}^{-m}\frac{1}{2}{\int }_{\frac{\pi }{2}}^{-\frac{\pi }{2}}{\mathrm{sin}}^{m}\left(\left\{\frac{\pi }{2}-u\right\}\right)du=-{2}^{-m}{\int }_{0}^{\frac{\pi }{2}}{\mathrm{cos}}^{\mu }du$
This happens because $\mathrm{cos}\left(x\right)$ is even, therefore ${\mathrm{cos}}^{m}x$ is symmetric over symmetric intervals.