In a example problem I am trying to solve, it says the right side of the equals sign of the next equation is a Fourier series: $$ -{1\over EA}\sum_{n = 1}^{n_1}{\cos(n\pi)\over n_1}{\Biggl(c^2{n\pi\over L}P(t) +{L\over n\pi}{d^2P(t)\over dt^2}\Biggl)\sin\biggl({n\pi x\over L} \biggl)}=\sum_{m=1}^{\infty}\Big(c^2w_m^2f_m(t)+{d^2f_m(t)\over dt^2}\Big)\sin(w_mx) $$
where ${w_m}={{2m-1}\over2L}\pi $.
It also says that the coefficient of that Fourier series can be calculated as:
$$ c^2w_m^2f_m(t) + {d^2f_m(t)\over dt^2} =\\ {2\over L}{\sum_{n=1}^{n_1}\int_{0}^L} \Biggl(\biggl( {\cos(n\pi)\over n_1} c^2{n\pi\over L}P(t)+{L\over n\pi}{d^2P(t)\over dt^2}\biggl) \sin\biggl({n\pi x\over L}\biggl)\sin(w_mx)\Biggl)dx $$
I do not understand how is the equation above obtained. I know the general sinus form of a Fourier series is:
$$ g(x)=a_0+\sum_{k=1}^\infty b_k\sin\biggl({k\pi x\over L} \biggl) $$
Here is what I have tried: I assumed $g(x)$ to be:
$$ g(x)=-{1\over EA}\sum_{n = 1}^{n_1}{\cos(n\pi)\over n_1}{\Biggl(c^2{n\pi\over L}P(t) +{L\over n\pi}{d^2P(t)\over dt^2}\Biggl)\sin\biggl({n\pi x\over L} \biggl)} $$
and I also assumed that $a_0=0$ and assumed that:
$$ sin\biggl({k\pi x\over L}\biggl)=sin\biggl(w_mx\biggl) $$
I also assumed that:
$$ b_k= c^2w_m^2f_m(t) + {d^2f_m(t)\over dt^2} $$
I tried calculating $b_k$ as:
$$ b_k={2\over L}\int_0^L g(x)sin\biggl({k\pi x\over L}\biggl) $$
and got:
$$ b_k=\\ -{1\over EA}{2\over L}\sum_{n = 1}^{n_1}\int_0^L{ {\Biggl({\cos(n\pi)\over n_1}}{\biggl(c^2{n\pi\over L}P(t) +{L\over n\pi}{d^2P(t)\over dt^2}\biggl)\sin\biggl({n\pi x\over L} \biggl)}sin(w_mx)\Biggl)}dx $$
which means:
$$ c^2w_m^2f_m(t) + {d^2f_m(t)\over dt^2} =\\ -{1\over EA}{2\over L}\sum_{n = 1}^{n_1}\int_0^L{ {\Biggl({\cos(n\pi)\over n_1}}{\biggl(c^2{n\pi\over L}P(t) +{L\over n\pi}{d^2P(t)\over dt^2}\biggl)\sin\biggl({n\pi x\over L} \biggl)}sin(w_mx)\Biggl)}dx $$
As you can see, the equation I obtained is not the same as the second equations I have shown in this post. I am trying to understand where did I make a mistake. Here is the source where I found this equation:
I kindly request your help. Thank you in advance.
I wish you a good day.