I reached the following expression after carrying out the integral
$$\sum_{k=0}^{\infty}\int_0^L \cos(\frac{k\pi x}{L})\sin(\frac{n\pi x}{L})\mathrm{d}x=\sum_{k=0}^{\infty}\frac{nL((-1)^{n+k}-1)}{\pi(k^2-n^2)}$$
Can anything be said about the summation on the RHS, like which terms woud vanish?