Solution of 1d wave equation by Laplace transform

Question

This is a homework problem that I can almost finish. I just can't invert the Laplace transform at the end. $$u_{xx}=u_{tt}, u(t=0)=u_t(t=0)=0, u(x=0)=\sin\omega t, u(x=2)=0.$$ Taking the Laplace transform with respect to $t$ and using the zero initial conditions, I obtain $U'' = s^2U$, where $s$ is the transform variable and $U$ is the transform of $u$. Solving this trivial ODE, I obtain $$U(x,s)=C_1(s)\cosh sx+C_2(s)\sinh sx.$$ Using the condition at $x=0$ and the transform of $\sin\omega t$, I get $$C_1(s) = \frac{\omega}{s^2+\omega^2}.$$ Using the condition at $x=2$, I get $$C_1(s)\cosh 2s+C_2(s)\sinh 2s = 0. $$ Solving for $C_2(s)$ and putting $C_1$ and $C_2$ back into the expression for $U(x,s)$, I get $$U(x,s)=\frac{\omega\cosh(sx)}{s^2+\omega^2}- \frac{\omega\coth(2s)\sinh(sx)}{s^2+\omega^2} .$$ Now I am stuck.