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Differentiation Questions

Question Number 98244 by ~blr237~ last updated on 12/Jun/20

$$\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:{xe}^{{cosx}} {cos}\left({sinx}\right){dx}=\:\mathrm{2}\pi^{\mathrm{2}} \\$$

Answered by maths mind last updated on 15/Jun/20

$${e}^{{cos}\left({x}\right)+{isin}\left({x}\right)} ={e}^{{cos}\left({x}\right)} \left({cos}\left({sin}\left({x}\right)\right)+{isin}\left({sin}\left({x}\right)\right)\right) \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} {xe}^{{cos}\left({x}\right)+{isin}\left({x}\right)} {dx} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {xe}^{{e}^{{ix}} } {dx} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \underset{{k}\geqslant\mathrm{0}} {\sum}{x}.\frac{{e}^{{ikx}} }{{k}!} \\$$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{k}!}\int_{\mathrm{0}} ^{\mathrm{2}\pi} {xe}^{{ikx}} {dx} \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} {xe}^{{ikx}} {dx}=\left[{x}\frac{{e}^{{ikx}} }{{ik}}\right]−\int\frac{{e}^{{ik}} }{{ik}}{dx} \\$$$$=\frac{\mathrm{2}\pi}{{ik}}\:,{k}\neq\mathrm{0} \\$$$${if}\:{k}=\mathrm{0} \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} {xdx}=\mathrm{2}\pi^{\mathrm{2}} \\$$$${Re}\left\{\int_{\mathrm{0}} ^{\mathrm{2}\pi} {xe}^{{e}^{{ix}} } {dx}\right\}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {xe}^{{cos}\left({x}\right)} {cos}\left({sin}\left({x}\right)\right){dx} \\$$$$={Re}\left[\mathrm{2}\pi^{\mathrm{2}} +\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2}\pi}{{ik}.{k}!}\right] \\$$$$=\mathrm{2}\pi^{\mathrm{2}} \\$$$$\Sigma\frac{\mathrm{1}}{{kk}!}=\int\frac{{e}^{{x}} }{{x}}{dx}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{k}} }{{kk}!} \\$$$$\Rightarrow\int{xe}^{{cos}\left({x}\right)} {sin}\left({sin}\left({x}\right)\right){dx}=−\mathrm{2}\pi{E}_{{i}} \left(\mathrm{1}\right) \\$$

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