[SOLVED] Integration term by term and orthogonality

$\displaystyle\int_0^L\varphi_m(x)\varphi_n(x) \ dx=0 \ \ \ \text{if} \ m\neq n$

$\displaystyle\int_0^L\varphi^2_n(x) \ dx>0$

$\displaystyle\int_0^L f(x)\sin\left(\frac{m\pi x}{L}\right) \ dx=\sum_{n=1}^{\infty}b_n\int_0^L\sin\left(\frac{n \pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right) \ dx$

Every term is 0 except for when m = n.

$\displaystyle\int_0^L f(x)\sin\left(\frac{m\pi x}{L}\right) \ dx=b_m\int_0^L\left[\sin\left(\frac{m\pi x}{L}\right)\right]^2 \ dx$

$\displaystyle f(x)=\sin\left(\frac{3\pi x}{L}\right)$

$\displaystyle\int_0^L \sin\left(\frac{3\pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right) \ dx=b_m\int_0^L\left[\sin\left(\frac{m\pi x}{L}\right)\right]^2 \ dx$

Looking at the LHS:

$\displaystyle b_m\int_0^L\left[\sin\left(\frac{m\pi x}{L}\right)\right]^2 \ dx=b_m\frac{1}{2}\int_0^L \left[1-\cos\left(\frac{2m\pi x}{L}\right)\right] \ dx$

$\displaystyle b_m\frac{1}{2}\left[x-\frac{L\sin\left(\frac{2m\pi x}{L}\right)}{2m\pi x}\right]_0^L\Rightarrow\frac{L}{2} b_m$

$\displaystyle b_n=\frac{2}{L}\int_0^L\sin\left(\frac{3\pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right) \ dx$

What do I do now? My solution I am obtain is way wrong.