I've been calculating the Fourier series of $f: [-\pi, \pi]$, $$ f(x) = |\cos^3(x)| - \frac{4}{3\pi} $$
But I've got strange result,
$f$ is even so $b_n = 0$
First, let's compute $a_0$,
\begin{align*} a_0 &= \frac{2}{\pi} \int_0^{\pi} \cos^3(x) - \frac{4}{3\pi} dx \\ &= \frac{2}{\pi} \left( \int_0^{\pi} cos(x)(1-\sin^2(x)) dx - \int_0^{\pi} \frac{4}{3\pi} dx \right) \\ &= \frac{2}{\pi} \left( \int_0^{0} (1-u^2) du - \left[ \frac{4x}{3\pi} \right]_0^{\pi} \right) \\ &= - \frac{8}{3\pi} \end{align*}
Next, we can compute $a_n$,
\begin{align*} a_n &= \frac{2}{\pi} \int_0^{\pi} (\cos^3(x) - \frac{4}{3\pi}) cos(nx) dx \\ &= \frac{2}{\pi} \int_0^{\pi} \cos^3(x) cos(nx) - \frac{4}{3\pi} cos(nx) dx \\ &= \frac{2}{\pi} \int_0^{\pi}\cos(x)\cos(nx)(\frac{1 + cos(2x)}{2})^2 dx - \left[ \frac{4}{3\pi} \frac{sin(nx)}{n} \right]_0^{\pi} \end{align*}
If we take a look at $\int_0^{\pi}\cos(x)\cos(nx)(\frac{1 + cos(2x)}{2})^2 dx$ (maybe I'm wrong here ?),
\begin{align*} \int_0^{\pi}&\cos(x)\cos(nx)(\frac{1 + cos(2x)}{2})^2 dx \\ &= \frac{1}{2} \int_0^{\pi} \cos(x)\cos(nx) dx + \frac{1}{2} \int_0^{\pi} \cos(x)\cos(nx)\cos(2x) dx \\ &= \frac{1}{4} \int_0^{\pi} \cos(x + nx)\cos(x - nx) dx + \frac{1}{4} \int_0^{\pi} [\cos(x+nx) + cos(x-nx)] \cos(2x) dx \\ &= \frac{1}{4} \left[ \frac{sin(x+nx)}{n+1} + \frac{\sin(x-nx)}{n-1} \right]_0^{\pi} + \frac{1}{8} \int_0^{\pi} cos(3x+nx) + cos(nx-x) dx \\ & + \frac{1}{8} \int_0^{\pi} cos(3x-nx) + cos(-x-nx) dx \\ &= \frac{1}{4} \left[ \frac{\sin(x+nx)}{n+1} + \frac{\sin(x-nx)}{n-1} \right]_0^{\pi} + \left[ \frac{1}{8} \ \frac{sin(3x+nx)}{3+n} + \frac{\sin(nx-x)}{n-1} \right]_0^{\pi} \\ & + \frac{1}{8} \left[ \frac{\sin(3x-nx)}{3-n} + \frac{\sin(x+nx)}{n+1} \right]_0^{\pi} \\ &= 0 \quad \textit{because $\sin(n\pi) = 0$} \end{align*}
In consequence, $$ a_n = 0 $$
So the Fourier series would be
$$ S(f(x)) = \frac{a_0}{2} = - \frac{8}{6\pi} $$
I'm pretty sure I'm wrong.