The full range is [tex]-\pi<x<\pi[/tex] (length [tex]2L=2\pi[/tex]), so the half range is [tex]L=\pi[/tex]. The half range sine series would then be given by
[tex]f(x)=\displaystyle\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L=\sum_{n\ge1}b_n\sin nx[/tex]
where
[tex]b_n=\displaystyle\frac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,\mathrm dx=\frac2\pi\int_0^\pi(\pi-x)\sin nx\,\mathrm dx[/tex]
Essentially, this is the same as finding the Fourier series for the function
[tex]\begin{cases}g(x)=\begin{cases}\pi-x&\text{for }0<x<\pi\\-\pi-x&\text{for }-\pi<x<0\end{cases}\\g(x+2\pi)=g(x)\end{cases}[/tex]
Integrating by parts yields
[tex]b_n=\dfrac2\pi\left(\dfrac\pi n-\dfrac{\sin n\pi}{n^2}\right)=\dfrac2n[/tex]
So the half range sine series for this function is simply
[tex]f(x)=\displaystyle\sum_{n\ge1}\frac{2\sin nx}n[/tex]
