$$u(x,t) = \frac{1}{2}U(x+ct)+\frac{1}{2}U(x-ct)+\frac{1}{2}W(x+ct)-\frac{1}{2}W(x-ct) $$\par$$ =\frac{1}{2}U(x+ct)+\frac{1}{2}U(x-ct)+\frac{1}{2c}\int_{x-ct}^{x+ct}V(z)dz$$
I'm trying to move from the first expression to the second. I've managed to arrive at the first expression from the linear second-order wave equation. Here, I can see that the second expression involves the fundamental theorem of calculus. What stumps me is the constant 'c' in the denominator, otherwise, my attempt would have correct.
It's been a long day, could someone help me?
I don't know what you are doing exactly, but let me guess. If I'm wrong, please clarify what your notations are, and what you are doing.
I guess you are solving the wave equation $$ u_tt=c^2u_xx,\quad u(x,0)=U(x),\quad u_t(x,0)=V(x). $$ Maybe you have solved the differential equation and found out that $$ u(x,t)=f(x-ct)+g(x+ct) $$ for some functions $f$ and $g$?
Then, matching with the initial conditions, $$ U(x)=u(x,0)=f(x)+g(x),\quad V(x)=u_t(x,0)=-cf'(x)+cg'(x) $$ Integrating the second condition (divided by $c$) from some constant $a$ (which will play no role in the end) to $x$, $$ \frac{1}{c}\int_a^x V(z)\,dz = -f(x)+g(x). $$ Now, we have a linear system to solve for $f$ and $g$. Subtracting the second equation from the first, we find that $$ 2f(x)=U(x)-\frac{1}{c}\int_a^x V(z)\,dz. $$ Adding the equations we find that $$ 2g(x)=U(x)+\frac{1}{c}\int_a^x V(z)\,dz. $$ Thus, dividing by $2$ and summing, $$ \begin{aligned} u(x,t)&=f(x-ct)+g(x+ct)\\ &=\frac{1}{2}\bigl(U(x-ct)+U(x+ct)\bigr)+\frac{1}{2c}\int_{x-ct}^{x+ct}V(z)\,dz. \end{aligned} $$ In the last step, we used that $\int_a^{x+ct}V(z)\,dz-\int_a^{x-ct}V(z)\,dz=\int_{x-ct}^{x+ct}V(z)\,dz$.