Question

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

Answer 1

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

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

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

Adding equation (i) and (ii), we get

$\Rightarrow 2I={}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int \cos xdx=[\sin x{]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$=\sin \frac{\pi }{2}-\sin (-\frac{\pi }{2})$

$1+1$

$=2$

$\Rightarrow 2I=2$

$\Rightarrow I=1$