Evaluate , if possible in a closed form, the integral:
$$\int_0^{\pi} \left( \frac{\pi}{2} - x \right) \frac{\tan x}{x} \, {\rm d}x$$
Basically, I have not done that much. I broke the integral
\begin{align*} \int_{0}^{\pi} \left ( \frac{\pi}{2}-x \right ) \frac{\tan x}{x} \, {\rm d}x &= \int_{0}^{\pi/2} \left ( \frac{\pi}{2} - x \right ) \frac{\tan x}{x} \, {\rm d}x + \int_{\pi/2}^{\pi} \left ( \frac{\pi}{2} - x \right ) \frac{\tan x}{x} \, {\rm d}x\\ &\!\!\!\!\!\!\overset{u=\pi/2-x}{=\! =\! =\! =\! =\! =\!} \int_{0}^{\pi/2} \frac{u \cot u}{\frac{\pi}{2}-u} \, {\rm d}u + \int_{-\pi/2}^{0} \frac{u \cot u}{\frac{\pi}{2}-u} \, {\rm d}u\\ &= \int_{-\pi/2}^{\pi/2} \frac{u \cot u}{\frac{\pi}{2}-u} \, {\rm d}u\\ &\approx 2.13897 \end{align*}
I have no idea how to evaluate this. I was thinking of IBP and then some kind of Fourier , but I cannot get it to work. Any ideas?