$\int \left(\cos x\right)\mathrm{logcot}\left(\frac{x}{2}\right)dx=$

Applying integration by parts

$\int \cos x·\log \cot \left(\frac{x}{2}\right)dx=\log \cot \left(\frac{x}{2}\right)\int \cos xdx-\int \left(\frac{d\log \cot \left(\frac{x}{2}\right)}{dx}\times \int \cos x dx\right)dx$

$=\log \cot \left(\frac{x}{2}\right)\int \cos xdx-\int \left[\frac{1}{cot{\displaystyle \frac{x}{2}}}\times \frac{d}{dx}\left(cot\frac{x}{2}\right)\times \int \cos x dx\right]dx$

$=\sin x·\log \cot \left(\frac{x}{2}\right)-\int \left[\frac{1}{cot{\displaystyle \frac{x}{2}}}\times \left(-\cos e{c}^2\frac{x}{2}\right)\times \frac{1}{2}\times \int \cos x dx\right]dx$

$=\sin x·\log \cot \left(\frac{x}{2}\right)+\int \left(\frac{1}{2·\sin {\displaystyle \frac{x}{2}\cos \frac{x}{2}}}\times \int \cos x dx\right)dx$

$=\sin x·\log \cot \left(\frac{x}{2}\right)+\int dx$

$=\sin x·\log \cot \left(\frac{x}{2}\right)+x+c$