Question with Solution

$$I={\int }_0^{2\pi }(2\pi -x)\ln \left(\frac{3+\cos x}{3-\cos x}dx\right)\Rightarrow 2I=2\pi {\int }_0^{2\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx$$

$$\Rightarrow I=2\pi {\int }_0^{\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx=2\pi {\int }_0^{\pi }\ln \left(\frac{3+\cos x}{3-\cos x}\right)dx=-I\Rightarrow I=0$$