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Question Number 151638 by mathdanisur last updated on 22/Aug/21

$$\underset{0}{\stackrel{2\mathit{\pi }}{\int }}{(1-\cos \mathbf{x})}^{10}\cos \left(10\mathrm{x}\right)\mathrm{dx}=?$$

Answered by Olaf_Thorendsen last updated on 22/Aug/21

$$\mathrm{I}={\int }_0^{2\pi }{(1-\cos x)}^{10}\cos \left(10x\right)dx$$$$\mathrm{I}=\mathrm{Re}{\int }_0^{2\pi }{(1-\cos x)}^{10}{e}^{10ix}dx$$$$\mathrm{I}=\mathrm{Re}{\int }_0^{2\pi }{(1-\frac{e^{ix}+e^{-ix}}{2})}^{10}{e}^{10ix}dx$$$$\mathrm{Let}u=e^{ix}$$$$\mathrm{I}=\mathrm{Re}{\int }_{\mid u\mid =1}{(1-\frac{u+\frac{1}{u}}{2})}^{10}{u}^{10}(-i\frac{du}{u})$$$$\mathrm{I}=\frac{1}{2^{10}}\mathrm{Re}{\int }_{\mid u\mid =1}-\frac{i}{u}{(u-1)}^{20}du$$$$\mathrm{I}=\frac{1}{2^{10}}\mathrm{Re}(2i\pi \left(\mathrm{Res}(-\frac{i}{u}{(u-1)}^{20},0)\right)$$$$\mathrm{I}=\frac{1}{2^{10}}\mathrm{Re}\left(2i\pi \times (-i)\right)$$$$\mathrm{I}=\frac{2\pi }{2^{10}}=\frac{\pi }{2^9}=\frac{\pi }{512}$$

Commented by mathdanisur last updated on 23/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser}\mathrm{nice}$$

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