An integral with $\cos x$

Evaluate $$\int_{-\pi}^\pi \frac{3\; dx}{5-4\cos x}.$$

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Method I: Real analysis
We use the half angle substitution $u=\tan \dfrac{x}{2}$. Then $du=\dfrac{1}{2}\sec^2 \dfrac{x}{2}\;dx\Rightarrow du=\dfrac{1}{2}(1+u^2)\;dx$. We evaluate the indefinite integral for illustration: $$\begin{align}\int \dfrac{3 \dfrac{2}{1+u^2}}{5-4\dfrac{1-u^2}{1+u^2}}\;du&=\int \dfrac{6\; du}{5(1+u^2)-4(1-u^2)}\\&=6\int \dfrac{du}{1+9u^2}\\&=2\int \;dt\\&=2\tan^{-1}(3u)+C\\&=2\tan^{-1}(3\tan \dfrac{x}{2})+C\end{align}$$

Let $F(x)=2\tan^{-1}(3\tan \dfrac{x}{2})$. This function is undefined for all real $x=(2k+1)\pi, k\in \Bbb Z$ because $\tan \dfrac{x}{2}$ is undefined at those points. We know the graph of $F(x)$ looks like that of a tangent function except that $F(x)$ is bounded. Since the range of $\tan^{-1} x$ is $(\dfrac{-\pi}{2},\dfrac{\pi}{2})$, that of $F(x)$ is $(-\pi,\pi)$. One cycle of $F(x)$ occurs between $(-\pi,\pi)$ just like the case in $\tan \dfrac{x}{2}$. Note also that

$$\lim_{x\to (2k+1)\pi^-} F(x)=2\cdot \dfrac{\pi}{2}=\pi\\ \lim_{x\to (2k+1)\pi^+} F(x)=2\cdot \dfrac{-\pi}{2}=-\pi.$$ Finally, to evaluate the given definite integral, we can use the continuous antiderivative $$\begin{align}\bar{F}(x)=\begin{cases}-\pi,&\quad x=-\pi^+\\2\tan^{-1}(3\tan \dfrac{x}{2}),&\quad -\pi<x<\pi\\ \pi,&\quad x=\pi^-\end{cases}\end{align}.$$ As a result, $$\int_{-\pi}^\pi \frac{3\; dx}{5-4\cos x}=\bar{F}(\pi^-)-\bar{F}(-\pi^+)=\pi-(-\pi)=2\pi.$$
Method II: Residue Theorem
Since the integrand is symmetric in $[0,2\pi]$ about $\theta=\pi$, the integral is the same as $$\int_0^{2\pi} \frac{3\; dx}{5-4\cos x}.$$Transforming this integral into a contour integral, we have $$\int_C \frac{3}{5-4\frac{1}{2}(z+z^{-1})}\dfrac{dz}{iz},$$ where $C$ is the unit circle. After some simplifications, we have $$\dfrac{-3}{i}\int_C \dfrac{dz}{(2z-1)(z-2)}=\dfrac{-3}{i} 2\pi i\bigg(\dfrac{1}{4z-5}\bigg|_{z=\frac{1}{2}}\bigg)=2\pi.$$ The singularity at $z=2$ is ignored because it is outside the unit circle.

Reference:
Improper Riemann Integrals Book by Ioannis Markos Roussos