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Question Number 59647 by aliesam last updated on 12/May/19

Commented by Mr X pcx last updated on 13/May/19

$$I={\int }_0^{\pi }\frac{cosx{e}^{\mid sinx\mid }}{1+e^{tanx}}dx+{\int }_{\pi }^{2\pi }\frac{cosx{e}^{\mid sinx\mid }}{1+e^{tanx}}dx$$$$but$$$${\int }_{\pi }^{2\pi }\frac{cosx{e}^{\mid sinx\mid }}{1+e^{tanx}}dx{=}_{x=\pi +t}-{\int }_0^{\pi }\frac{cost{e}^{\mid sint\mid }}{1+e^{tanx}}dx$$$$\Rightarrow I=0.$$

Answered by MJS last updated on 12/May/19

$$f\left(x\right)=\frac{{\mathrm{e}}^{\mid \sin x\mid }\cos x}{1+{\mathrm{e}}^{\tan x}}$$$$\underset{0}{\stackrel{\pi /2}{\int }}f\left(x\right)dx=-\underset{\pi }{\stackrel{3\pi /2}{\int }}f\left(x\right)dx$$$$\underset{\pi /2}{\stackrel{\pi }{\int }}f\left(x\right)dx=-\underset{3\pi /2}{\stackrel{2\pi }{\int }}f\left(x\right)dx$$$$\Rightarrow \mathrm{answer}=0$$

Commented by MJS last updated on 13/May/19

$$f(x+\frac{\pi }{2})=-f(x+\frac{3\pi }{2})$$

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