Confused about Fourier Series

Question

I've been told to construct a Fourier Series for the odd function that has period $2\pi$ and is equal to $\cos(x)$ for $x \in (0,\pi]$. For $f$ that is $2\pi$ period I have a formula
$$b_n=\int_{-\pi}^\pi f(x)\sin(nx) \, dx.$$
I don't know what this question is asking, I thought it wanted $f(x)=\cos(x)$ but when substituting in the equation, that would give an odd integrand so it would just be $0$.

Answer 1

Since $f(x)$ is odd and equals $\cos x$ on $(0,\pi]$, it equals $-\cos (-x)=-\cos x$ on $[-\pi,0)$. So, $$ b_n:=\int_{-\pi}^\pi f(x) \sin nx \, dx $$ $$ = \int_{-\pi}^0 (-\cos x) \sin nx \, dx + \int_{0}^\pi (\cos x) \sin nx \, dx. $$