Question

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

• A. $\pi$
• B. $a\pi$
• C. $\frac{\pi }{2}$
• D. $2\pi$

Answer 1

The correct option is C $\frac{\pi }{2}$

Let $I={\int }_{-\pi }^{\pi }\frac{{cos}^{2}x}{1+a^x}dx$ ...(1)
Using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{-\pi }^{\pi }\frac{{cos}^2(-\pi +\pi -x)}{1+a^{(-\pi +\pi -x)}}dx={\int }_{-\pi }^{\pi }\frac{a^x{cos}^{2}x}{1+a^x}dx$ ...(2)
Adding (1) and (2), we get
$2I={\int }_{-\pi }^{\pi }{cos}^{2}xdx={\left[\frac{1}{2}(x+sinxcosx)\right]}_{-\pi }^{\pi }=\frac{1}{2}(\pi +0+\pi -0)=\pi \Rightarrow I=\frac{\pi }{2}$