If I have even piecewise periodic function ($T=6$) $$x(t)=\begin{cases} 0 &-3\leq t \leq-2 \\ 2+t &-2\leq t \leq-1 \\ 1 &-1\leq t \leq 1 \\ -t+2 &1\leq t \leq 2 \\ 0 &2 \leq t \leq 3\end{cases}$$
Find cosine fourier series for this function. So I need to calculate the fourier coefficient $a_0$ and $a_n$ because $b_n=0$ for even function.
$a_0$ is easy $$ a_0=\frac 1T \int_T x(t)dt = \frac 12$$
I have seen in many books/sources that $a_n=\frac 2 T\int_T x(t)\cos(n2\pi t/T) dt$. BUT is it times $2$ because for even and periodic function you can do $\int_{T}f(x)dx=2\int_0^{T/2} f(x)dx$ so it the correct formula $a_n=\frac 2 T\int_0^{T/2} x(t)\cos(n2\pi t/T) dt$ or Is this for special case where the period of $x(t)$ is $2\pi$?
Well I calculated $a_0$ like this for my function.
\begin{align} a_n&=\frac 2 T\int_T x(t)\cos(n2\pi t/T) dt \\ &=2 \frac 2T \int_0^T x(t)\cos(n2\pi t/T) dt\\ &=\frac{2 \sin\left(\frac{n\pi}{3}\right)}{n\pi}-\frac{2\sin\left(\frac{n\pi}{3}\right)}{n\pi}+\frac{6\cos\left(\frac{n \pi}{3}\right)-6\cos\left(\frac{2n\pi}{3}\right)}{n^2 \pi^2} \\ &=\frac{6\cos\left(\frac{n \pi}{3}\right)-6\cos\left(\frac{2n\pi}{3}\right)}{n^2 \pi^2} \end{align}
Correct? So my series is $x(t)\sim a_0+ \sum_{n=1}^{\infty} a_n \cos \left(\frac{n2\pi t}{T}\right)$
$$x(t) \sim \frac 12 + \sum_{n=1}^{\infty} \left( \frac{6\cos\left(\frac{n \pi}{3}\right)-6\cos\left(\frac{2n\pi}{3}\right)}{n^2 \pi^2}\right) \cos \left(\frac{n\pi t}{3}\right)$$
Did I do my calculation correctly?