$I=\int_0^{2 \pi} \frac{x \cos x}{1+\cos x} d x$
On applying property $\int_0^a f(x) d x=\int_0^a f(a-x) d x$,
we get
$$
I=\int_0^{2 \pi} \frac{(2 \pi-x) \cos x}{1+\cos x} d x \ldots \text { (ii) }\{\because \cos (2 \pi-x)=\cos x\}
$$

On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 I & =2 \pi \int_0^{2 \pi} \frac{\cos x}{1+\cos x} d x \\
\Rightarrow I & =2 \pi \int_0^\pi \frac{\cos x}{1+\cos x} d x=2 \pi \int_0^\pi\left(1-\frac{1}{1+\cos x}\right) d x \\
& =2 \pi \int_0^\pi\left(1-\frac{\sec ^2 \frac{x}{2}}{2}\right) d x=2 \pi\left[x-\tan \frac{x}{2}\right]_0^\pi
\end{aligned}
$$