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

0

The correct answer is Option (3) → 0

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

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

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

eq (1) + eq (2)

so $2I=\int_{-\pi}^{\pi}\frac{(1+a^x)}{(1+a^x)}\cos x dx$

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

$⇒I=0$