Is the solution of $ u_{tt}=c^2u_{xx}+xt $ correct? Consider the following $ u_{tt}=c^2u_{xx}+xt,\\ u(x,0)=0,\\ u_t(x,0)=\sin (x)$ 
and find the solution.
Solution.
We have that $u(x,t)$ is given by $$u(x,t)=\frac{1}{2}(g(x+ct)+g(x-ct))+\frac{1}{2c}\int_{x-ct}^{x+ct}h(u)du+\frac{1}{2c}\iint_\Delta f(y,s)dA,\cdots\cdots (1)$$ where $\Delta $ is a triangle with vertex $(x-ct,0),(x+ct,0),(x,t)$.
Thus in our case, $u(x,t)=\frac{1}{2c}\int_{x-ct}^{x+ct}\sin(u)du+\frac{1}{2c}\iint_\Delta xt \ dA.$
Using Green Theorem we have that $\frac{1}{2c}\iint_\Delta xt \ dA=\frac{1}{2}[\cos(x_0+ct)-\cos(x_0-ct)-2cu(x_0,t_0)],$ where I took a point $(x_0,t_0)$ to be able to integrate  using the theorem.
and 
$\frac{1}{2c}\int_{x-ct}^{x+ct}\sin(u)du=\frac{1}{2}[-\cos(x_0+ct_0)+\cos(x_0-ct_0)]$
Hence the final solution is $u(x,t)=\frac{1}{2}[\cos(x_0+ct_0)-\cos(x_0-ct_0)]+\frac{1}{2}[\cos(x_0+ct)-\cos(x_0-ct)-2cu(x_0,t_0)]$
Is it correct?
This is the first time I use formula (1), so if something can be improved don't hesitate to suggest. 
Edit 
Apparently the solution is wrong because I had to calculate the double integral over the area of the triangle and not the contour integral, that's what my professor said, but I don't know why, I didn't understand why. Could someone explain why is that? 
I still don't see why application of Green theorem it's not correct in the double integral.
 A: There's a lot of steps missing in your work, so I can't reliably tell you exactly what you did wrong. But I can tell you that most of it is wrong. 
Here's the "usual" way to evaluate the area integral
$$ \frac{1}{2c}\iint_{D} ys \ dyds $$
The sides of the triangle $D(y,s)$ are given by
\begin{align} 
&S_1\{(x-ct,0)\to(x+ct,0)\} &&: s = 0 \\
&S_2\{(x+ct,0)\to (x,t)\} &&: y = x + c(t-s) \\
&S_3\{(x,t)\to(x-ct,0)\} &&: y = x - c(t-s) \end{align}
Then 
\begin{align} 
\frac{1}{2c}\iint_{D} ys \ dyds &= \frac{1}{2c}\int_{0}^{t} \int_{x-c(t-s)}^{x+c(t-s)} ys\ dyds \\
&= \frac{1}{2c}\int_0^t \frac{\big[(x+c(t-s)\big]^2 - \big[x-c(t-s)\big]^2}{2}s\ ds \\
&= \frac{1}{2c}\int_0^t \big[{2c}x(t-s)\big]s\  ds \\
&= \int_0^t x(ts-s^2)\ ds \\
&= x\left(\frac{t^3}{2} - \frac{t^3}{3}\right) \\
&= \frac{xt^3}{6}
\end{align}

Green's theorem method: You need to find two functions $M$ and $N$ such that
$$ \frac{\partial M}{\partial x} - \frac{\partial N}{\partial t} = xt $$
These functions need not be unique, so we can arbitrarily choose
$$ M(x,t) = 0, \quad N(x,t) = -\frac{xt^2}{2} $$
Then
$$ \iint_D \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial s} \right) dyds = \int_{\partial D} Ndy + Mds = \int_{\partial D} -\frac{ys^2}{2}dy $$
Following the boundary of the triangle in the counter clockwise direction, we have to split the integral into
\begin{align} 
\int_{\partial D} -\frac{ys^2}{2}dy &= \int_{S_1} -\frac{ys^2}{2}dy + \int_{S_2}-\frac{ys^2}{2}dy + \int_{S_3}-\frac{ys^2}{2}dy \\
&= 0 + \int_{x+ct}^{x} -\frac{\big[x+c(t-s)\big]s^2}{2}(-c\ ds) + \int_{x}^{x-ct} -\frac{\big[x-c(t-s)\big]s^2}{2}(c\ ds) \\
&= \frac{c}{2}\int_{x+ct}^{x} \big[x+c(t-s)\big]s^2\ ds - \frac{c}{2}\int_{x}^{x-ct} \big[x-c(t-s)\big]s^2\ ds
\end{align}
You can finish the integral if you want to, but this shows why using Green's theorem here is not a good idea. You get a bigger mess than the one you started with

Combined with the homogeneous piece (which is trivial to evaluate), the correct solution is
$$ y(x,t) = \frac{xt^3}{6} + \frac{1}{2c}\big[\cos(x-ct) - \cos(x+ct) \big] $$
A: 2 method


*

*Let $u_1$ is solution of the problem


$$u_{tt}=c^2u_{xx}+xt,\quad u(x,0)=0,\quad u_t(x,0)=0.\qquad (1)$$
$u_2$ is solution of the problem
$$u_{tt}=c^2u_{xx},\quad u(x,0)=0,\quad u_t(x,0)=\sin(x).\qquad (2)$$
Then
$$u=u_1+u_2.$$


*

*The solution of  (1) is sought in the form $$u=v(t)\,x$$ Substitution this in $(1)$ yields
$$v''(t)=t,\;v(0)=0,\;v'(0)=0,$$
$$v(t)=\frac{t^3}{6},$$
$$u_1=\frac{t^3x}{6}$$


*

*The solution of  (2) is sought in the form $$u=v(t)\sin(x)$$ Substitution this in $(2)$ yields


$$v''(t)+c^2v(t)=0,\;v(0)=0,\;v'(0)=1,$$
$$v(t)=\frac{\sin(ct)}{c},$$
$$u_2=\frac{\sin(ct)}{c}\sin(x)$$


*

*$$u=u_1+u_2=\frac{t^3x}{6}+\frac{\sin(ct)}{c}\sin(x)$$

A: Similar to IVP Wave Equation $u_{tt} = 4u_{xx} + \sin(ct)\cos(x)$ (PDE):
Let $\begin{cases}p=x+ct\\q=x-ct\end{cases}$ ,
Then $u_x=u_pp_x+u_qq_x=u_p+u_q$
$u_{xx}=(u_p+u_q)_x=(u_p+u_q)_pp_x+(u_p+u_q)_qq_x=u_{pp}+u_{pq}+u_{pq}+u_{qq}=u_{pp}+2u_{pq}+u_{qq}$
$u_t=u_pp_t+u_qq_t=cu_p-cu_q$
$u_{tt}=(cu_p-cu_q)_t=(cu_p-cu_q)_pp_t+(cu_p-cu_q)_qq_t=c^2u_{pp}-c^2u_{pq}-c^2u_{pq}+c^2u_{qq}=c^2u_{pp}-2c^2u_{pq}+c^2u_{qq}$
$\therefore c^2u_{pp}-2c^2u_{pq}+c^2u_{qq}=c^2u_{pp}+2c^2u_{pq}+c^2u_{qq}+\dfrac{(p+q)(p-q)}{4c}$
$-4c^2u_{pq}=\dfrac{p^2-q^2}{4c}$
$u_{pq}=\dfrac{q^2-p^2}{16c^3}$
$u(p,q)=f(p)+g(q)+\dfrac{pq^3-p^3q}{48c^3}$
$u(p,q)=f(p)+g(q)+\dfrac{pq(q^2-p^2)}{48c^3}$
$u(x,t)=f(x+ct)+g(x-ct)+\dfrac{c^2xt^3-x^3t}{12c^2}$
A: see:
Dean G. Duffy, Transform methods for solving partial differential equations, 2nd edition, Example 2.3.1, page 141
Ravi P. Agarwal, Donal O’Regan, Ordinary and Partial Differential Equations, Springer, 2009, Lecture 48
Example. We solve wave equation
$u_{tt}=c^2u_{xx}+xt,\quad -\infty<x<\infty,\quad t>0$,
subject to initial conditions
$u(x,0)=0,\; u_t(x,0)=\sin (x),\quad -\infty<x<\infty$.
Taking the Laplace transform and substituting the initial conditions, we obtain
ordinary differential equation
$${{c}^{2}}\, \left( \frac{{{d}^{2}}}{d {{x}^{2}}} \operatorname{U}\left( x,s\right) \right) -{{s}^{2}} \operatorname{U}\left( x,s\right) =-\sin{(x)}-\frac{x}{{{s}^{2}}}$$
with boundary conditions
$$\lim_{x\to -\infty}|U(x,s)|<\infty,\;\lim_{x\to \infty}|U(x,s)|<\infty.\qquad(1)$$
Solving this ordinary differential equation,
$$U(x,s)=\frac{\sin{(x)}}{{{s}^{2}}+{{c}^{2}}}+\frac{x}{{{s}^{4}}}+C_1\, {{e}^{\frac{s x}{c}}}+C_2\, {{e}^{-\frac{s x}{c}}}$$
From boundary conditions $(1)$ we get $C_1=C_2=0$.
Then
$$U(x,s)=\frac{\sin{(x)}}{{{s}^{2}}+{{c}^{2}}}+\frac{x}{{{s}^{4}}}$$
Final step is the inversion of the Laplace transform $U(x,s)$. We get answer
$$u=\frac{\sin{\left( c t\right) } \sin{(x)}}{c}+\frac{{{t}^{3}} x}{6}$$
