The value of $\int_{-\pi}^\pi \frac{\cos ^2 x}{1+\alpha^x} \mathrm{~d} x, \alpha0$ is

Question

The value of $\int_{-\pi}^\pi \frac{\cos ^2 x}{1+\alpha^x} \mathrm{~d} x, \alpha>0$ is

Aicharya · Accepted Answer

from (i)+(ii) $\begin{aligned} & 2 I=\int_{-\pi}^\pi \frac{\left(1+\alpha^xight) \cos ^2 x}{\left(1+\alpha^xight)} \mathrm{d} x=\int_{-\pi}^\pi \cos ^2 x \mathrm{~d} x \ & \Rightarrow 2 I=2 \int_0^\pi \cos ^2 x \mathrm{~d} x \quad\left[\because \cos ^2(\pi-x)=\cos ^2 xight] \ & \Rightarrow I=\int_0^\pi \frac{1+\cos 2 x}{2} \mathrm{~d} x\left[\frac{x}{2}+\frac{\sin 2 x}{4}ight]_0^\pi=\frac{\pi}{2}\end{aligned}$

The value of $\int_{-\pi}^\pi \frac{\cos ^2 x}{1+\alpha^x} \mathrm{~d} x, \alpha>0$ is

Solution