Guidance or advice with $I=\int_0^{2\pi}\frac{1}{4+\cos t}dt$ Let
$$
\begin{align}
I=\int_0^{2\pi}\frac{1}{4+\cos t}dt
\end{align}
$$
I would like to evaluate this integral using cauchhy's Integral formula, I understand that I have to convert this into a form like $\int_{\gamma}\frac{f(z)}{z-z_0}dz$ .
I tried using $\cos t=\frac{e^{it}+e^{-it}}{2}$, but I didn't get anywhere. I haven't seen an example like this.

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$\ds{I\equiv\int_{0}^{2\pi}{1 \over 4 + \cos\pars{t}}\,\dd t
     :\ {\large ?}}$.

\begin{align}
I&=\int_{-\pi}^{\pi}{1 \over 4 - \cos\pars{t}}\,\dd t
=2\
\overbrace{\int_{0}^{\pi}{1 \over 4 - \cos\pars{t}}\,\dd t}
^{\ds{\mbox{Set}\ x \equiv \tan\pars{t \over 2}}}\ =\
\\[3mm]&=2\int_{0}^{\infty}{1 \over 4 - \pars{1 - x^{2}}/\pars{1 + x^{2}}}\,
{2\,\dd x \over 1 + x^{2}}
=4\int_{0}^{\infty}{1 \over 5x^{2} + 3}\,\dd x
\\[5mm]&={4 \over 3}\,\root{3 \over 5}
\int_{0}^{\infty}{1 \over \pars{\root{5/3}x}^{2} + 1}\,\root{5 \over 3}\dd x
={4 \over 3}\,{\root{15} \over 5}\
\overbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}^{\ds{=\ {\pi \over 2}}}
=\color{#66f}{\large{2\root{15} \over 15}\,\pi}
\\[3mm]&\approx {\tt 1.6223}
\end{align}

