Could someone explain how to the integral is obtained from the following equation.
$$\sum_{n=1}^{\infty}B_n \sin\left(\frac{n \pi v}{l}\right)= x(x-l)$$ where $ 0 \le x \le l$.
Integrate the above with limits $0$ to $l$
The next line of working after the one above was, $$B_n = \frac{2}{l}\int_{0}^{l} v(v-l)\sin\left(\frac{n \pi v}{l}\right) \operatorname d v$$
I dont understand how they get that line, it looks like they did a subsitution of $x = v$ ? But what about the summation and why is the sin not dividing the $v(v-l)$.
Could someone explain the steps between starting from that equation to getting that integral please.
Thank you
The key point is that (Prove it by yourself!)
$$\int_{-l}^{l}{\sin\frac{m\pi x}{l}\sin\frac{n\pi x}{l}dx}=\begin{cases} 0, & m\neq n; \\ l, & m=n.\end{cases}$$
Now you fix a positive integer, say $m$. Then in the integral
$$\int_{-l}^{l}{\sin\frac{m\pi x}{l}\left(\sum_{n=1}^{\infty}{B_n\sin\frac{n\pi x}{l}}\right)dx}=\sum_{n=1}^{\infty}{\int_{-l}^{l}{B_n\sin\frac{m\pi x}{l}\sin\frac{n\pi x}{l}dx}}$$
all the terms with $n\neq m$ are equal to $0$! That is, it collapses to
$$\sum_{n=1}^{\infty}{\int_{-l}^{l}{B_n\sin\frac{m\pi x}{l}\sin\frac{n\pi x}{l}dx}}=\int_{-l}^{l}{B_m\sin^2\frac{m\pi x}{l}dx}=B_m\int_{-l}^{l}{\sin^2\frac{m\pi x}{l}dx}=B_m\cdot l.$$
Note that
$$\sum_{n=1}^{\infty}{B_n\sin\frac{n\pi x}{l}}=x(l-x).$$
It follows that
$$\int_{-l}^{l}{x(l-x)\sin\frac{m\pi x}{l}dx}=\int_{-l}^{l}{\sin\frac{m\pi x}{l}\left(\sum_{n=1}^{\infty}{B_n\sin\frac{n\pi x}{l}}\right)dx}=B_m\cdot l.$$
Therefore,
$$B_m=\frac{1}{l}\int_{-l}^{l}{x(l-x)\sin\frac{m\pi x}{l}dx}=\frac{2}{l}\int_{0}^{l}{x(l-x)\sin\frac{m\pi x}{l}dx},$$
as $x(l-x)\sin(m\pi x/l)$ is odd.