integrate: cosx/1+e^x. lower limit -pi/2 and upper limit pi/2

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$$\begin{gathered} \mathrm{Given} \\ \mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\frac{\mathrm{cosx}}{1+{\mathrm{e}}^{\mathrm{x}}}\right)d\mathrm{x} ---\left(1\right) \\ \mathrm{since} \mathrm{we} \mathrm{know}, \mathrm{if} \mathrm{we} \mathrm{replace} \mathrm{x} \mathrm{by} \frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{2}-\mathrm{x}=-\mathrm{x}, \mathrm{we} \mathrm{will} \mathrm{get} \mathrm{same} \mathrm{integral} \mathrm{I} \\ \mathbf{The}\mathbf{ }\mathbf{above}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{property}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{definite}\mathbf{ }\mathbf{integral} \\ \mathbf{hence}\mathbf{:} \\ \mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\frac{\cos \left(-\mathrm{x}\right)}{1+{\mathrm{e}}^{-\mathrm{x}}}\right)d\mathrm{x} \\ \mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\frac{\cos \left(\mathrm{x}\right)}{1+{\displaystyle \frac{1}{{\mathrm{e}}^{\mathrm{x}}}}}\right)d\mathrm{x} \\ \mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\frac{{\mathrm{e}}^{\mathrm{x}}.\cos \left(\mathrm{x}\right)}{{\mathrm{e}}^{\mathrm{x}}+{\displaystyle 1}}\right)d\mathrm{x} ---\left(2\right) \\ \mathrm{Adding} 1 \mathrm{and} 2 \\ 2\mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\left(\frac{{\mathrm{e}}^{\mathrm{x}}.\cos \left(\mathrm{x}\right)}{{\mathrm{e}}^{\mathrm{x}}+{\displaystyle 1}}\right)+\left(\frac{\mathrm{cosx}}{1+{\mathrm{e}}^{\mathrm{x}}}\right)\right)d\mathrm{x} \\ 2\mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{\left({\mathrm{e}}^{\mathrm{x}}+1\right).\cos \left(\mathrm{x}\right)}{{\mathrm{e}}^{\mathrm{x}}+{\displaystyle 1}}\mathrm{dx} \\ 2\mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\mathrm{cosx} \mathrm{dx} \\ 2\mathrm{I}={}_{-\frac{\mathrm{\pi }}{2}}{}^{\frac{\mathrm{\pi }}{2}}\left[\mathrm{sinx}\right] =\sin \frac{\mathrm{\pi }}{2}-\sin \left(-\frac{\mathrm{\pi }}{2}\right)=1+1=2 \\ \mathbf{hence} \\ \mathbf{2}\mathbf{I}\mathbf{=}\mathbf{2} \\ \mathbf{I}\mathbf{=}\mathbf{1} \end{gathered}$$

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