$$ \begin{cases} u_{tt}-u_{xx}=sin(m\pi x)sin(\omega t),\quad0<x<1 ,t>0\\ u(x,0)=sin(x),\quad 0\leq x\leq 1\\ u_t(x,0)=1+x, \quad 0\leq x\leq 1\\ u(0,t)=u(1,0)=0,\quad t\geq0 \end{cases} $$
Where $m\in \mathbb{N}, \omega \in \mathbb{R}, \omega\neq m \pi$
For the homognous part I got:
$$u^h(x,t)=\sum_{n=1}^{\infty}sin(n\pi x)[A_ncos(n\pi t)+B_nsin(n\pi t)]$$
Using the first initial condition we get
$$sin x=\sum_{n=1}^{\infty}A_n sin(n\pi x)$$
$$sin x=A_1 sin(\pi x)$$
So it is $A_1=1$ for $x\notin\mathbb{N}$?
I tried using Fourier coefficents and got the following integral
$$A_n=2\int_{0}^{1}sin(x)sin(n\pi x)dx$$
How can it be integrated?
I would do this:
$$ A_n=2\int_{0}^{1}sin(x)sin(n\pi x)dx = \int_{0}^{1} \left(cos((1 - n\pi)x) - cos((1 + n\pi)x)\right) dx, $$ and take it from there.