Part I
Let us say I need to determine Fourier coefficient at the end of the solution to a PDE. On applying the non-homogeneous BC, I arrive at the following
$$\sum_{n=1}^{\infty} A_{n}\sinh(n\pi c)\sin\bigg(\frac{n\pi x}{a}\bigg)=f(x)$$
We have two cases that needs to be evaluated now:
- $f(x) = 2e^{-\frac{x}{3}}$
- $f(x) = \frac{4}{3} (x-x^2)$
Part II
For $\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0$ defined on $x \in [0,a]$ and $y \in [0,b]$
with the BC as
$T(0,y) = T(a,y) = T_a$
$\frac{\partial T(x,0)}{\partial y} = k_1\bigg[e^{-s_1y}\bigg(T_1 + s_1\int_0^y e^t T(x,t)\mathrm{d}t\bigg) - T(x,0)\bigg]$
$\frac{\partial T(x,b)}{\partial x} = 0$
For Part I After applying @Mattos suggestions
For $f(x) = 2e^{-\frac{x}{3}}$
$$\sum_{n=1}^{\infty} A_{n}\sinh(n\pi c)\int_0^a\sin\bigg(\frac{n\pi x}{a}\bigg)\sin\bigg(\frac{k\pi x}{a}\bigg)\mathrm{d}x=2\int_0^ae^{-\frac{x}{3}}\sin\bigg(\frac{k\pi x}{a}\bigg)\mathrm{d}x$$
The integrand on the LHS has values only for $n=k$ (From orthogonality)
$$A_k\sinh(n\pi c)\bigg(\frac{a}{2}\bigg)=2\int_0^a e^{\frac{-x}{3}}\sin\bigg(\frac{k\pi x}{a}\bigg)\mathrm{d}x$$
Since, $k$ is a dummy integer it can be replaced by arbitrary integer $n$
$$A_n = \frac{\frac{4}{a}\bigg[\int_0^ae^{\frac{-x}{3}}\sin\bigg(\frac{n\pi x}{a}\bigg)\mathrm{d}x\bigg]}{\sinh(n\pi c)}$$
Similarly, for $f(x) = \frac{4}{3} (x-x^2)$
$$A_n = \frac{\frac{8}{3a}\bigg[\int_0^a x\sin\bigg(\frac{n\pi x}{a}\bigg)\mathrm{d}x - \int_0^a x^2\sin\bigg(\frac{n\pi x}{a}\bigg)\mathrm{d}x\bigg]}{\sinh(n\pi c)}$$