Yeah if you put $t = \tan\frac{x}{2}$, then you have $dt = \frac{1}{2}\cdot \sec^{2}\frac{x}{2} \ dx$. And from here note that $dx = \frac{2 \: dt}{1+t^{2}}$. And when $x = 0$ you have $t = \tan(0)=0$. And when $x=2\pi$ you have $t = \tan(\pi) = 0$.
I think the value of your integral will be $0$. But in case you want to evaluate something like \begin{align*} \int \frac{1}{2-\cos{x}} \ dx &= \int\frac{1}{2 - \frac{1-\tan^{2}x/2}{1+\tan^{2}x/2}} \ dx \\ &= \int \frac{1}{2 - \frac{1-t^{2}}{1+t^{2}}} \cdot \frac{2}{1+t^{2}} \ dt \\ &=\int\frac{2}{2+2t^{2}-1+t^{2}} \ dt \\ &= \int\frac{2}{3t^{2} +1} dt = \frac{2}{3} \int \frac{1}{t^{2}+\frac{1}{3}} \ dt \end{align*}
I guess you can evaluate the integral now by putting $t =\frac{1}{\sqrt{3}}\:\tan{v}$. This is the way how one generally evaluates integrals of the form $\int \frac{dx}{a+b\cos{x}} \qquad \text{and} \qquad \int \frac{dx}{a+b\sin{x}}$ While evaluating integrals of the form $a+b\sin{x}$ one uses the formula $\sin{2x} = \frac{2\tan{t}}{1+\tan^{2}{t}}$