Transform the problem into the first quadrant of the characteristic coordinates $(\xi,\eta)$.

\begin{align}

-4c^2\tilde{u}_{\xi \eta} &= \tilde{f}(\xi, \eta) && \xi > 0, \eta > 0 \label{a} \\

\tilde{u}|_{\xi=0} &= \tilde{g} \left( -\frac{\eta}{2c} \right) && \eta > 0 \\

\tilde{u}|_{\eta=0} &= \tilde{h} \left( \frac{\xi}{2c} \right) && \xi > 0

\end{align}

The solution to $(\ref{a})$ is as follows.

\begin{align}

\tilde{u}(\xi,\eta)= -\frac{1}{4c^2} \int_0 ^\xi

\int_0 ^\eta \tilde{f}(\xi',\eta' )\,d\eta' d\xi' + \psi(\eta) + \phi(\xi)

\end{align}

The domain of dependence is a rectangle defined as $\tilde{R}(\xi,\eta) = \{ (\xi',\eta') \vert 0< \xi' < \xi,\, 0< \eta' < \eta\}$.

Assume $\psi(0) = \phi(0) = \frac{1}{2}g(0) = \frac{1}{2}h(0)$.

\begin{align}

\phi(\xi) &= h \left( \frac{\xi}{2c} \right) - \frac{1}{2}h(0)\\

\psi(\eta) &= g \left( -\frac{\eta}{2c} \right) - \frac{1}{2}g(0)

\end{align}

Translate back to the $(x,t)$ coordinates. Use the hint provided and the fact that the Jacobian is equal to $2c$.

\begin{align}

u(x,t) = -\frac{1}{2c}\iint _{R(x,t)} f(x',t')\,dx'dt' + h \left( \frac{x+ct}{2c} \right) + g \left( -\frac{x-ct}{2c} \right) - h(0)

\end{align}

where $R(x,t)=\{ (x',t'):\, 0< x'-ct' < x-ct,\, 0< x'+ct' < x+ct\}$.