$$u_{tt} = 4u_{xx},\ u(x,\ 0) = \sin(\pi x),\ u_t (x,\ 0) = 0.$$
I factored the differential operators and used undetermined coefficients (can go into more detail on this part but I think it matters not) to find the general solution $$u = h(2t + x) + g(2t - x)$$ for arbitrary functions $h$ and $g$. Applying the initial conditions yields $$\begin{cases}\sin(\pi x) = h(x) + g(-x) \\ 0 = \frac{\partial}{\partial t}[h(2t + x) + g(2t - x)]_{t = 0}\end{cases}$$ which will require the second equation to be antidifferentiated with respect to $t$ in order for the system to be solved simultaneously. Thus, I end up with $$\begin{cases}\sin(\pi x) = h(x) + g(-x) \\ f(x) = \int \frac{\partial}{\partial t}[h(2t + x) + g(2t - x)]_{t = 0}\,\mathrm dt\end{cases}$$ for an arbitrary function $f$. I'm not really sure how to finish solving this.
I could make the substitutions $y = 2t + x$ and $z = 2t - x$, leading to $$f(x) = 2 \int \left[\frac{\partial h(y)}{\partial y} + \frac{\partial g(z)}{\partial z}\right]_{t = 0}\,\mathrm dt,$$ but other than getting the $2$ from the chain rule derivatives pulled out, I'm not sure how to continue. I don't think I can just cancel the partial differentiation with the antidifferentiation, because of the $t = 0$ evaluation operation. How can I finish solving these equations?
$\def\g{\widetilde{g}}$One of the boundary condition, i.e.$$ \left. \frac{\partial}{\partial t}(h(x + 2t) + g(-x + 2t)) \right|_{t = 0} = 0, $$ can be simplified as $h'(x) + g'(-x) \equiv 0$ by the chain rule. Thus$$ \begin{cases} h(x) + g(-x) = \sin(πx)\\ h'(x) + g'(-x) = 0 \end{cases}. $$ Defining $\g(x) = g(-x)$, the above equations become$$ \begin{cases} h(x) + \g(x) = \sin(πx)\\ h'(x) - \g'(x) = 0 \end{cases}, $$ where the second equation implies that $\g(x) = h(x) + c$ for some constant $c$, and plugging into the first equation shows that$$ h(x) = \frac{1}{2} \sin(πx) + C_0,\quad \g(x) = \frac{1}{2} \sin(πx) - C_0, $$ where $C_0 = -\dfrac{c}{2}$ is a constant. Thus$$ h(x) = \frac{1}{2} \sin(πx) + C_0,\quad g(x) = -\frac{1}{2} \sin(πx) - C_0. $$