Question Number 59647 by aliesam last updated on 12/May/19
Commented by Mr X pcx last updated on 13/May/19
$${I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{cosx}\:{e}^{\mid{sinx}\mid} }{\mathrm{1}+{e}^{{tanx}} }\:{dx}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\frac{{cosx}\:{e}^{\mid{sinx}\mid} }{\mathrm{1}+{e}^{{tanx}} }{dx} \\ $$$${but} \\ $$$$\int_{\pi} ^{\mathrm{2}\pi} \:\:\frac{{cosx}\:{e}^{\mid{sinx}\mid} }{\mathrm{1}+{e}^{{tanx}} }\:{dx}\:=_{{x}=\pi\:+{t}} \:\:\:−\int_{\mathrm{0}} ^{\pi} \:\frac{{cost}\:{e}^{\mid{sint}\mid} }{\mathrm{1}+{e}^{{tanx}} }{dx} \\ $$$$\Rightarrow\:{I}\:=\mathrm{0}\:. \\ $$
Answered by MJS last updated on 12/May/19
$${f}\left({x}\right)=\frac{\mathrm{e}^{\mid\mathrm{sin}\:{x}\mid} \mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{e}^{\mathrm{tan}\:{x}} } \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}{f}\left({x}\right){dx}=−\underset{\pi} {\overset{\mathrm{3}\pi/\mathrm{2}} {\int}}{f}\left({x}\right){dx} \\ $$$$\underset{\pi/\mathrm{2}} {\overset{\pi} {\int}}{f}\left({x}\right){dx}=−\underset{\mathrm{3}\pi/\mathrm{2}} {\overset{\mathrm{2}\pi} {\int}}{f}\left({x}\right){dx} \\ $$$$\Rightarrow\:\mathrm{answer}\:=\:\mathrm{0} \\ $$
Commented by MJS last updated on 13/May/19
$${f}\left({x}+\frac{\pi}{\mathrm{2}}\right)=−{f}\left({x}+\frac{\mathrm{3}\pi}{\mathrm{2}}\right) \\ $$