Consider the following boundary value problem: \begin{equation} \begin{cases} u_{tt}-u_{xx}=x, &0\leq x \leq \pi,t\geq0 \\ u(0,t)=u_{x}(\pi,t)=0 \\ u(x,0)=\sin(\frac{3}{2}x), & u_{t}(x,0)=0 \end{cases} \end{equation}
Normally, I would have look for substitution of $u(x,t)$ such that the PDE would be equal to zero, so that I could separate the variables. But, my instructor choose instead to proceed as if the variables could be separated anyway. So, assuming the solution is on the form $u(x,t)=X(x)T(t)$, we find that the eigenfunctions $X(x)$ of the problem satisfy:
\begin{equation} \frac{X''(x)}{X(x)}=-\lambda, \quad \quad X(0)=X'(\pi)=0 \tag{*} \end{equation} Solving $(*)$ we obtain $X_{n}(x)=\sin(\frac{2n+1}{2}x)$ for $n=0,1,2,\dots$. So far I follow his reasoning; however, his next move is where my trouble begins. He claims we should expand the function (right hand side of the PDE) with respect to the eigenfunctions, yielding (according to him):
\begin{equation} x \sim \sum_{n=0}^{\infty}c_{n}\sin(\frac{2n+1}{2}x), \quad \text{where} \quad c_{n}=\frac{\int_{0}^{\pi}x\sin(\frac{2n+1}{2}x)dx}{\int_{0}^{\pi}\sin^{2}(\frac{2n+1}{2}x)dx}=\frac{8(-1)^{n}}{\pi(2n+1)^{2}} \end{equation}
What exactly are we doing when we expand the RHS of the PDE with respect to the eigenfunctions, and why is $c_{n}=\frac{\int_{0}^{\pi}x\sin(\frac{2n+1}{2}x)dx}{\int_{0}^{\pi}\sin^{2}(\frac{2n+1}{2}x)dx}?$
The system $\{\sin((2\,n+1)x)/2$ is orthogonal on $[0,\pi]$: $$ \int_0^\pi\sin\Bigl(\frac{2\,m+1}{2}x\Bigr)\sin\Bigl(\frac{2\,n+1}{2}x\Bigr)\,dx=0\quad\text{if}\quad m\ne n. $$ Assume $$ x=\sum_{m=0}^{\infty}c_{n}\sin\Bigl(\frac{2\,m+1}{2}x\Bigr). $$ Multiply by $\{\sin((2\,n+1)x)/2$ and integrate between $0$ and $\pi$. Assuming that the integral of the sum is the sum of the integrals, and using the orthogonality, you get $$ \int_0^\pi x\sin\Bigl(\frac{2\,n+1}{2}x\Bigr)\,dx=c_n\int_0^\pi\sin^2\Bigl(\frac{2\,n+1}{2}x\Bigr)\,dx. $$