Question

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

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

Answer: (A)

$1$

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

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

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

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

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

On adding Eqs. (i) and (ii), we get
$2I={\int }_{-\pi /2}^{\pi /2}\frac{(1+e^x)\cos x}{(1+e^x)}dx$

$={\int }_{-\pi /2}^{\pi /2}\cos xdx$

$=2{\int }_0^{\pi /2}\cos xdx$

[Since, $\cos x$ is an even function.]
$\therefore 2I=2[\sin x{]}_0^{\pi /2}=2(1-0)=2$
$\Rightarrow I=1$