\(\int_{-\pi}^{\pi}\frac{\cos x}{1+a^x}dx=\)
Answer & explanation
Correct answer: option 3
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)
$2I=\int\limits_{-\pi}^{\pi}\cos x dx$
$=\left[\sin x\right]_{-\pi}^{\pi}$
$=[\sin \pi+\sin \pi]$
$=0$