\(\int_{-\pi}^{\pi}\frac{\cos x}{1+a^x}dx=\)

0

The correct answer is Option (3) → 0

\(I=\int\limits_{-\pi}^{\pi}\frac{\cos x}{1+a^x}dx\)  ...(1)

$=\int\limits_{-\pi}^{\pi}\frac{\cos(π-π-x)}{1+a^{π-π-x}}dx$

$=\int\limits_{-\pi}^{\pi}\frac{\cos (-x)}{1+a^{-x}}dx$

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

Add eq. (1) + eq. (2)

$2I=\int\limits_{-\pi}^{\pi}\cos x dx$

$=\left[\sin x\right]_{-\pi}^{\pi}$

$=[\sin \pi+\sin \pi]$

$=0$