Practicing Success
The value of $\int\limits_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x, a>0$, is |
$\pi$ $a~\pi$ $\pi/2$ $2~\pi$ |
$\pi/2$ |
We know that $\int\limits_0^a f(x) d x=\int\limits_0^a\{f(x)+f(-x)\} d x$ ∴ $I=\int\limits_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x$ $\Rightarrow I=\int_\limits0^\pi\left\{\frac{\cos ^2 x}{1+a^x}+\frac{\cos ^2(-x)}{1+a^{-x}}\right\} d x=\int\limits_0^\pi \cos ^2 x d x$ $\Rightarrow I=\frac{1}{2} \int\limits_0^\pi(1+\cos 2 x) d x=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]_0^\pi=\frac{\pi}{2}$ |