Terrence Moore
2022-04-28
Answered

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

Answered 2022-04-29
Author has **12** answers

Let $2x=\frac{\pi}{2}-u$ ,then $2dx=-2du$ , so:

${2}^{-m}{\int}_{0}^{\frac{\pi}{2}}{\mathrm{sin}}^{m}2xdx=-{2}^{-m}\frac{1}{2}{\int}_{\frac{\pi}{2}}^{-\frac{\pi}{2}}{\mathrm{sin}}^{m}\left(\{\frac{\pi}{2}-u\}\right)du=-{2}^{-m}{\int}_{0}^{\frac{\pi}{2}}{\mathrm{cos}}^{\mu}du$

This happens because$\mathrm{cos}\left(x\right)$ is even, therefore ${\mathrm{cos}}^{m}x$ is symmetric over symmetric intervals.

This happens because

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