Fourier Sine Series Exapnsion

The question ask to expand the following function of period $\displaystyle 2L$ on the interval $\displaystyle (0,L)$ as a Fourier sine series.

$\displaystyle \[f(x)=sin(\frac{\pi}{2L}x)\]

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This function is periodic on a period of $\displaystyle 4L$ and thus the coefficients are given by

$\displaystyle \[b_{n}=\frac{1}{2L}\int_{0}^{2L}sin(\frac{\pi x}{2L})sin(\frac{n\pi x}{2L})dx\]

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Upon evaluating the integral, I split the sine terms into two seperate cosine terms using a trig identity. After integration I am unable to get a single cosine as given in the correct answer

$\displaystyle \[f(x)=\frac{8}{\pi}\sum_{n=1}^{\infty}\frac{ncos(n\ pi)}{1-4n^{2}}sin(\frac{n\pi}{L}x)\]

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Have I made an error or am I on the correct track?