The value of $\int\limits_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x, a>0$, is
Answer & explanation
Correct answer: option 3
We know that
$\int\limits_0^a f(x) d x=\int\limits_0^a\{f(x)+f(-x)\} d x$
∴ $I=\int\limits_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x$
$\Rightarrow I=\int_\limits0^\pi\left\{\frac{\cos ^2 x}{1+a^x}+\frac{\cos ^2(-x)}{1+a^{-x}}\right\} d x=\int\limits_0^\pi \cos ^2 x d x$
$\Rightarrow I=\frac{1}{2} \int\limits_0^\pi(1+\cos 2 x) d x=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]_0^\pi=\frac{\pi}{2}$