Note:

For a piecewise continuous function of period $2\pi$,

$${\int_{-\pi}^{\pi} f(x) \ dx \ = \ \int_{a}^{a + 2\pi} f(x) \ dx \ \ \ {\mbox for \ all \ } a \in {\mathbb{R}}. }$$

(For the sake of completeness, we include a proof here. By the Fundamental Theorem of Calculus, the function $h(a) := \int_{a}^{a + 2\pi} f(x) \ dx$ has the derivative $h'(a) = f(a+2\pi)-f(a)$. Since $f$ is $2\pi$-periodic, we see that $h'(a)=0$ for all $a \in {\mathbb{R}}$, whence there is a constant $c \in {\mathbb{R}}$ such that $h(a)=c$ for all $a \in {\mathbb{R}}$. In particular, it follows that $c=h(-\pi)=\int_{-\pi}^{\pi} f(x) \ dx$. Thus we obtain $\int_{a}^{a + 2\pi} f(x) \ dx=h(a)=c=\int_{-\pi}^{\pi} f(x) \ dx$ for all $a \in {\mathbb{R}}$, which finishes the proof.)

Thus, the integral over some interval of length $2\pi$ is equal to the integral over any such interval. In case $f$ is given on the interval $[0, 2\pi]$ rather than $[-\pi, \pi]$, we can calculate the Fourier coefficients as follows:

$a_n = \displaystyle{\frac{1}{\pi} \int_{0}^{2\pi}} f(x)\cos nx dx, b_n = \frac{1}{\pi} \displaystyle{\int_{0}^{2\pi}} f(x)\sin nx dx$.