Let's say I have a periodic function
$f(x) = (1-x) , 0 \leq x \leq 1$ with a period of $p=1$. So $ f(x+1*n) = f(x) $ for any integer $n$.
To compute the Fourier Series coefficients, I use Eqn. 15 from here : Fourier Series wolfram
$$\begin{align}a_n &= \frac{1}{p/2} \int_{-p/2}^{p/2} f(x) \cos( \frac{n\pi x}{p/2})dx \\& = 2 \int_{-1/2}^{1/2} f(x) \cos( 2n\pi x)dx \end{align} $$
So far everything makes sense. However, what does not make sense to me is the next equation:
$$ a_n= 2 \int_{0}^{1} f(x) \cos( 2n\pi x)dx$$
Why is the above statement equal to the previous statement (i.e. if you change the bounds from $[-\frac{1}{2}, \frac{1}{2}] ->[0,1]$ why is it equal to the previous statement)?
I know $f(x)$ is periodic with period 1 , so I understand $2 \int_{-1/2}^{1/2} f(x)dx = 2 \int_{0}^{1} f(x)dx $. But when you throw in the $cos(2n \pi x)$ term in the integrand, how do you know if both statements are still equal? How about for the $b_n$ term where you are supposed to throw in a $sin(2n \pi x)$ multiplied by $f(x)$ ? Could you still change the integration bounds the same way? Why?
Since $f(x)$ is periodic with period $T=1$, you can change the integrand in any way as long as $a-b=T=1$ in
$$ \int^a_bf(x)\cos(2n\pi x)dx $$ which should give the same answer as long as the $f(x)$ agrees with $a$ and $b$.
In fact, you can also change the integrand of $\displaystyle{\int^a_bf(x)\sin(2n\pi x)dx}$ as long as $a-b=T=1$ and $f(x)$ is set to give the values in the interval $(b,a)$
Note that the statements above is referring to the context in obtaining the Fourier coefficients without considering any symmetry (the integrand also changes with respect to the symmetry of the signal to be decomposed)