We are looking for solution to the Neumann problem, so by D'A formula
$$
u(x,t)=\frac{1}{2}\bigl( \Phi(x+ct) +\Phi (x-ct)\bigr) +\frac{1}{2c}\int_{x-ct}^{x+ct} \Psi (y)\,dy
$$
where right-hand expression was $0$ and $\Phi$, $\Psi$ are even continuations of $\phi$, $\psi$.
(a) As $x>ct$ ($t>0)$ we just plug $\phi,\psi$ instead of them, getting the same solution as of Cauchy problem.
(b) As $0<x<ct$ we get $x+ct<0$ (so $\Phi(x+ct)= \phi(x+ct)$) and $x-ct<0$ so $\Phi (x-ct)=\phi(-x+ct)$; we get the first term as below. WQith $\Psi$ it is a bit more complicated: we integrate $\Psi(y)$ from $x-ct$ to $0$ and from $0$ to $x+ct$ and the latter is
$\int_0^{x+ct} \psi(y)\,dy$ while the former is $\int_0^{-x+ct} \psi(y)\,dy$ and we get the second and the third terms below. This is a correct formula generally
$$
u(x,t)=\frac{1}{2}\bigl( \Phi(x+ct) +\Phi (-x+ct)\bigr) +\frac{1}{2c}\int_0^{x+ct} \psi (y)\,dy +\frac{1}{2c} \int_0^{x-ct}\psi(y)\dy.
$$
Case $t<0$, $x+ct >0$ is like (a) and $t<0$, $x+ct >0$ is like (b).
