Question

Evaluate : ${\int }_0^{\pi }\log (1+\cos x)dx$

Answer 1

$I={\int }_0^{\pi }\log (1+\cot x)dx$
$I={\int }_0^{\pi }\log (x+2{\cos }^2\frac{x}{2}-x)dx$
$I={\int }_0^{\pi }\left[\log \right(2)+\log {\left(\cos \frac{x}{2}\right)}^2]dx$
$I={\int }_0^{\pi }{\log }_{2}xdx+{\int }_0^{\pi }2\log \left(\cos \frac{x}{2}\right)dx$
$I={\log }_2(x{)}_0^{\pi }+2{\int }_0^{\pi }\log \left(\cos \frac{x}{2}\right)dx$
$I_1={\int }_0^{\pi }\log \left(\cos \frac{x}{2}\right)dx$
$I_1={\int }_0^{\pi }\log \left(\cos \frac{(\pi -x)}{2}\right)dx$
$I_1={\int }_0^{\pi }\log \left(\sin \frac{x}{2}\right)dx$
So, $I_1={\int }_0^{\pi }\log \left(\sin \frac{x}{2}\right)dx$
let $\frac{x}{2}\to t$
$I_1={\int }_0^{\pi }\log \left(\sin t\right)dt$
$=\frac{-\pi }{2}\log 2$
So, $I=\log 2(\pi -0)+2\left[\frac{-\pi }{2}\log 2\right]$
$=\pi \log 2-\pi \log 2$
$=0$