Let $f(x)=\sin(\pi x)$ and let $0<x<1$
Find the Fourier series of the above function.
Now at first we should rescale $x$ to a symmetric interval : Let $s=x- 1/2 $
Thus $-1/2<s<1/2$
Now the task is to find the Fourier series of $f(s)=\sin(\pi (s+1/2))$
Which is very long to do but at the end I hopefully got the right answer (using the rescaled inner product to find the coefficients)
$FS(\sin(\pi x))= 2/\pi + \sum_{k=1}^\infty \frac{8}{\pi(1-4k^2)} \cos2k\pi x$
Now my first question is , is this right, and for the second, was there another way to compute its Fourier series because this method took 2 pages.
It is not necessary to rescale the whole setup. Extending your $f$ to a periodic function of period $1$ gives $$f(t)=\bigl|\sin(\pi t)\bigr|\qquad(-\infty<t<\infty)\ ,$$ which is an even function. Its Fourier series is therefore of the form $$f(t)\rightsquigarrow{a_0\over2}+\sum_{k=1}^\infty a_k\,\cos(2k\pi t)\ ,$$ whereby the $a_k$ are given by $$a_k=2\int_0^1 f(t)\,\cos(2k\pi t)\>dt=2\int_0^1 \sin(\pi t)\,\cos(2k\pi t)\>dt\qquad(k\geq0)\ .$$ Now use the formula $$2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$$ to "linearize" the integrand on the RHS.