\begin{align*}
\int_0^{2\pi} {\cos\theta\over 5+4\cos\theta}\ d\theta & = \int_0^{2\pi}{1\over 4}{\cos\theta\over (5/4)+\cos\theta}\ d\theta\\
& = {1\over 4}\int_0^{2\pi} 1-{5/4\over 5/4+\cos\theta}\ d\theta\\
& = {\pi\over 2} -{5\over 16}\int_0^{2\pi}{1\over 5/4+\cos\theta}\ d\theta.\\
\int_0^{2\pi}{d\theta\over 5/4+\cos\theta} & = \int_{|z| = 1}{1\over 5/4+(z+z^{-1})/2}{1\over i z}\ dz\\
& = {4\over i}\int_{|z|=1}{1\over 2z^2+5z+2}\ dz\\
& = {4\over i} 2\pi i {2\over 3}\\
& = {16\pi\over 3}.\\

Step 1: Define $F_n:= f_n - g^+$ so that $F_n+g^-\ge 0$.
Step 2: Define $E=$ the set where lininf $F_n$ is non negative.
Step 3: Apply Fatou's lemma on non negative measurable function $F_n+g^-$. LHS thus obtained is $\int \liminf (F_n+g^-)=\int (\liminf (F_n) +g^-)=\color{blue}{\int_E (\liminf (F_n) +g^-)}+\color{green}{\int_{E^c} (\liminf (F_n) +g^-)}$, where the second equality holds because of non negativity and measurability of $F_n+g^-$
Step 4: Showing that the highlighted integrals can be written as sum of integrals.