Solution
= Solution
The function is an <even function>, so every sine <Fourier coefficient> vanishes. Direct integration gives
$$
a_0=\frac2\pi\int_0^\pi x\,dx=\pi,
\qquad
a_n=\frac2\pi\int_0^\pi x\cos(nx)\,dx
=\frac{2}{\pi n^2}((-1)^n-1).
$$
Thus
$$
\boxed{|x|=\frac\pi2-\frac4\pi
\sum_{k=0}^{\infty}
\frac{\cos((2k+1)x)}{(2k+1)^2}}
$$
for the $2\pi$-periodic extension.