Algebra Questions

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

$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}\boldsymbol{\pi}} {\int}}\left(\mathrm{1}\:-\:\mathrm{cos}\boldsymbol{\mathrm{x}}\right)^{\mathrm{10}} \:\mathrm{cos}\left(\mathrm{10x}\right)\:\mathrm{dx}\:=\:? \\$$

Answered by Olaf_Thorendsen last updated on 22/Aug/21

$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{1}β\mathrm{cos}{x}\right)^{\mathrm{10}} \mathrm{cos}\left(\mathrm{10}{x}\right)\:{dx} \\$$$$\mathrm{I}\:=\:\mathrm{Re}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{1}β\mathrm{cos}{x}\right)^{\mathrm{10}} {e}^{\mathrm{10}{ix}} \:{dx} \\$$$$\mathrm{I}\:=\:\mathrm{Re}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{1}β\frac{{e}^{{ix}} +{e}^{β{ix}} }{\mathrm{2}}\right)^{\mathrm{10}} {e}^{\mathrm{10}{ix}} \:{dx} \\$$$$\mathrm{Let}\:{u}\:=\:{e}^{{ix}} \\$$$$\mathrm{I}\:=\:\mathrm{Re}\int_{\mid{u}\mid=\mathrm{1}} \left(\mathrm{1}β\frac{{u}+\frac{\mathrm{1}}{{u}}}{\mathrm{2}}\right)^{\mathrm{10}} {u}^{\mathrm{10}} \:\left(β{i}\frac{{du}}{{u}}\right) \\$$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{10}} }\mathrm{Re}\int_{\mid{u}\mid=\mathrm{1}} β\frac{{i}}{{u}}\left({u}β\mathrm{1}\right)^{\mathrm{20}} \:{du} \\$$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{10}} }\mathrm{Re}\left(\mathrm{2}{i}\pi\left(\mathrm{Res}\left(β\frac{{i}}{{u}}\left({u}β\mathrm{1}\right)^{\mathrm{20}} ,\mathrm{0}\right)\right)\right. \\$$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{10}} }\mathrm{Re}\left(\mathrm{2}{i}\piΓ\left(β{i}\right)\right) \\$$$$\mathrm{I}\:=\:\frac{\mathrm{2}\pi}{\mathrm{2}^{\mathrm{10}} }\:=\:\frac{\pi}{\mathrm{2}^{\mathrm{9}} }\:=\:\frac{\pi}{\mathrm{512}} \\$$

Commented by mathdanisur last updated on 23/Aug/21

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