Question

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

• A. $1$
• B. $0$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

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

let $I={\int }_{-\pi }^{\pi }\frac{co{s}^{2}x}{1+a^x}dx$

$I={\int }_{-\pi }^{\pi }\frac{co{s}^2(-x)}{1+a^{-x}}dx$

adding both

$2I={\int }_{-\pi }^{\pi }co{s}^{2}x\frac{1+a^{-x}+1+a^x}{(1+a^{-x})(1+a^x)}dx$

$={\int }_{-\pi }^{\pi }co{s}^{2}xdx$ ( integrating after breaking $co{s}^{2}x$ to $cos2x$)

$=\pi$

$I=\frac{\pi }{2}$