Mathematics - Definite Integration Question with Solution | TestHub

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{1+e^{\sin x}}dx$

$\Rightarrow I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+e^{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-\frac{\mathrm{\pi }}{2}-x\right)}}dx={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+e^{\sin \left({\displaystyle -x}\right)}}dx$

$\Rightarrow I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{e^{\sin x}}{1+e^{\sin x}}dx$

Using the property, ${\int }_a^{b}f\left(x\right)dx={\int }_c^{b}f\left(a+b-x\right)dx$

$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}1\mathrm{d}x\Rightarrow I=\frac{1}{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\mathrm{d}x$

$I=\frac{1}{2}{\left[x\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\Rightarrow I=\frac{\pi }{2}$