I am trying to find the Fourier expansion of $f$, only knowing that $f$ has period $2$ and $f(x) = |x|$ if $-1 < x < 1$.
Since $f$ is even, there is no $b_n$ term, which is given by $$b_n = \frac{2}{T} \int _0 ^T f(x) \sin (2\pi n \frac{x}{T} ) \, dx .$$ However, I am having trouble setting up the $a_n$ term $$a_n = \frac{2}{T} \int _0 ^T f(x) \cos (2\pi n \frac{x}{T} ) \, dx .$$ I do not think it is correct to immediately substitute $f(x) = |x|$ in the integral because if we let $T = 2$ in our case, we are integrating on $[0,2]$, and $|x|$ would only be on $[0,1]$. For $[1,2]$ it looks like the function $f(x) = -x$, but I am not sure how to write this out properly as a sum of two integrals. For example, if we did go with this, we would get $$\frac{2}{2} \left[ \int _0 ^1 x \cos (2\pi n \frac{x}{T} ) \, dx + \int _1 ^2 -x \cos (2\pi n \frac{x}{T} ) \, dx \right]. $$ Any suggestions?
If $g$ has period $p$ then $\int_a^{a+p} g(x) dx$ does not depend on $a$. In our case, taking $T=2$ the function $g(x)=f(x) \cos (\frac {2 \pi n x} T)$ has period $2$ so its interval from $0$ to $T=2$ is same as its integral from $-1$ to $+1$. So take the limits as $-1$ to $1$ and put $f(x)=|x|$.