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Question Number 143506 by tugu last updated on 15/Jun/21

$$\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{e^{-x}+e^x}dx=?$$$$$$

Answered by Dwaipayan Shikari last updated on 15/Jun/21

$${\int }_1^{\mathrm{\infty }}\frac{1}{e^x+e^{-x}}dx{e}^x=t$$$$={\int }_e^{\mathrm{\infty }}\frac{dt}{t^2+1}={\left[{tan}^{-1}\left(t\right)\right]}_e^{\mathrm{\infty }}=\frac{\pi }{2}-{tan}^{-1}\left(e\right)={tan}^{-1}\left(\frac{1}{e}\right)=0.3525$$

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