Definite Integral: Proving $\int_{0}^{2\pi}\frac{x \cos(x)}{1+\cos(x)}dx = 2\pi^{2}$

Question

There's a question, to prove that: $$\int_{0}^{2\pi}\frac{x \cos(x)}{1+\cos(x)}dx = 2\pi^{2}$$ I want to know where am I going wrong? Here's my approach:
Taking Left Hand Side $$\int_0^{2\pi}\frac{x \cos(x)}{1+\cos(x)}dx = \int_{0}^{\pi}\frac{x\cos(x)}{1+\cos(x)}dx+\int_{0}^{\pi}\frac{(2\pi-x)\cos(2\pi-x)}{1+\cos(2\pi-x)}dx$$ $$=\int_0^{\pi}\frac{x\cos(x)}{1+\cos(x)}dx+\int_{0}^{\pi}\frac{(2\pi-x)\cos(x)}{1+\cos(x)}dx$$ $$=\int_0^\pi\frac{x\cos(x)+(2\pi-x)\cos(x)}{1+\cos(x)}dx$$ $$=\int_0^\pi\frac{\cos(x)(x+2\pi-x)}{1+\cos(x)}dx$$ $$=2\pi\int_0^\pi\frac{\cos x}{1+\cos x}dx$$ $$=2\pi\int_0^\pi\frac{\cos x+1-1}{1+cos x} \, dx = 2\pi\int_0^\pi \bigg(1-\frac{1}{1+\cos x}\bigg)dx$$ $$=2\pi\bigg[x-\tan\frac{x}{2}\bigg]_0^\pi$$ Now if we evaluate this we'll get an undefined result, which is definitely $\neq 2\pi^2$. So, exactly where am I going wrong?
Please pardon me if I'm making the silliest mistake of all time, but please notify where am I wrong.

Answer 1

This integral diverges. $1+\cos x = 1 - \cos(\pi -x) $, hence around $x=\pi$ the denominator behaves like $\frac 12(\pi-x)^2$, which gives us a non-integrable singularity.