Question

Find:${\int }_{-\pi }^{\pi }\frac{{\cos }^{2}xdx}{1+a^x}$ where $a>0$

Answer 1

${\int }_a^{b}f\left(x\right)dx={\int }_0^b[f\left(x\right)+f(-x)]dx$

${\int }_{-\pi }^{\pi }\frac{{\cos }^{2}xdx}{1+a^x}$

$={\int }_0^{\pi }(\frac{{\cos }^{2}x}{1+a^x}+\frac{{\cos }^{2}x}{1+a^{-x}})dx$

$={\int }_0^{\pi }(\frac{{\cos }^{2}x}{1+a^x}+\frac{{\cos }^{2}x}{1+\frac{1}{a^x}})dx$

$={\int }_0^{\pi }\frac{{\cos }^{2}x+a^x{\cos }^{2}x}{1+a^x}dx$

$={\int }_0^{\pi }\frac{{\cos }^{2}x(1+a^x)}{1+a^x}dx$

$={\int }_0^{\pi }{\cos }^{2}xdx$

$=\frac{1}{2}{\int }_0^{\pi }2{\cos }^{2}xdx$

$=\frac{1}{2}{\int }_0^{\pi }(\cos 2x+1)dx$ using $2{\cos }^{2}x=\cos 2x+1$

$=\frac{1}{2}{[\frac{\sin 2x}{2}+x]}_0^{\pi }$

$=\frac{1}{2}[(\frac{\sin 2\pi }{2}+\pi )-(\frac{\sin 0}{2}+0)]$

$=\frac{1}{2}[0+\pi ]$

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