Evaluate $\int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2 t} dt$ \begin{align}
  \int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2t} dt
\end{align}
I am trying to evaluate above integral. The results is $2\pi$ according to Mathematica. I want to obtain this result by integrating properly 
Can this integral be evaluated using simple trigometric identities? 
Do I have to use complex analysis i.e., $\cos(\theta) = \frac{z+\frac{1}{z}}{2}$ and do residue calculus? 

Rewrite,
\begin{align}
  I=\int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2t} dt =
2\int_0^{\pi} \frac{2-\cos t}{3\cos^2 t -4\cos t +2} dt 
\end{align}
and then substitute $\cos t = \frac{1-x^2}{1+x^2}$, along with $dt = \frac{2dx}{1+x^2}$,
$$I= 4\int_0^{\infty} \frac{1+3x^2}{9x^4 -2x^2 +1} dx $$
$$= 4\int_0^{\infty} \frac{\frac{1}{x^2}+3}{9x^2 +\frac{1}{x^2}-2} dx $$
$$=4\int_0^{\infty} \frac{d\left(3x-\frac{1}{x}\right)} {\left(3x-\frac{1}{x}\right)^2+4}
$$
$$=2\tan^{-1}\left[\frac12\left(3x-\frac{1}{x}\right)\right]_0^{\infty}=2\pi$$
