Evaluate ${\int }_{-\pi /2}^{\pi /2} \frac{\cos x}{1+e^x}dx$

Let ${\int }_{-\pi /2}^{\pi /2} \frac{\cos x}{1+e^x}dx ...\left(1\right)$

$\left[{\int }_a^b f\left(x\right)dx={\int }_a^b f(a+b-x)dx\right]$

$={\int }_{-\pi /2}^{\pi /2} \frac{\cos \left(-\frac{\pi }{2}+\frac{\pi }{2}-x\right)}{1+e^{\left(-\frac{\pi }{2}+\frac{\pi }{2}-x\right)}}dx={\int }_{-\pi /2}^{\pi /2} \frac{\cos x}{1+e^{-x}}$

$={\int }_{-\pi /2}^{\pi /2} \frac{e^x\cos x}{1+e^x}dx ...\left(2\right)$

$\Rightarrow \left(1\right)+\left(2\right)$

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

$={\int }_{-\pi /2}^{\pi /2} \frac{\left(1+e^x\right)\cos x}{1+e^x}dx={\int }_{-\pi /2}^{\pi /2} \cos xdx$

$\left[{\int }_{-\pi /2}^{\pi /2} \cos xdx\text{ is even function }\right]$

$2I=2{\int }_0^{\pi /2} \cos xdx\Rightarrow I=[\sin x{]}_0^{\pi /2}$

$\left[{\int }_{-a}^a f\left(x\right)dx=2{\int }_0^a f\left(x\right)dx\right]$

$I=\sin \frac{\pi }{2}-\sin 0\therefore I=1$