The value of $\int\limits_{-\pi}^{\pi}\frac{cos^2x}{1+a^x}dx$, a > 0, is

$\pi/2$

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

$=\int\limits_{-\pi}^{\pi}\frac{cos^2t\,a^t}{1+a^t}dx=\int\limits_{-\pi}^{\pi}\frac{cos^2x\,a^x}{1+a^x}dx$

$⇒2I=\int\limits_{-\pi}^{\pi}cos^2x\,dx=2\int\limits_{0}^{\pi}cos^2x\,dx=4\int\limits_{-0}^{\pi/2}\frac{1+cos2x}{2}dx=\pi$

$⇒I=\frac{\pi}{2}$

Hence (C) is the correct answer.