The three segments $AB,BC,CA$ can each be parameterized on the interval $[0,1]$ as follows:
\begin{eqnarray}
(1-t)A+tB=(1-t)\langle-2,-2\rangle+t\langle2,-2\rangle&=&\langle-2+4t,-2\rangle\\
(1-t)B+tC=(1-t)\langle2,-2\rangle+t\langle0,1\rangle&=&\langle2-2t,-2+3t\rangle\\
(1-t)C+tA=(1-t)\langle0,1\rangle+t\langle-2,-2\rangle&=&\langle-2t,1-3t\rangle
\end{eqnarray}
The differential vectors $\langle dx,dy\rangle$ for the three respective segments are $\langle4,0\rangle dt,\langle-2,3\rangle dt,\langle-2,-3\rangle dt$
Thus
\begin{equation}
\int_{AB}-y\,dx+x\,dy=\int_0^1(2)(4)+(2-4t)(0)dt=\int_0^18\,dt=8
\end{equation}
\begin{equation}
\int_{BC}-y\,dx+x\,dy=\int_0^1(2-3t)(-2)+(2-2t)(3)dt=\int_0^12\,dt=2
\end{equation}
\begin{equation}
\int_{CA}-y\,dx+x\,dy=\int_0^1(-1+3t)(-2)+(-2t)(-3)dt=\int_0^12\,dt=2
\end{equation}
Therefore,
\begin{equation}
\int_{ABC}-y\,dx+x\,dy=8+2+2=12
\end{equation}
Gimusi's answer gets you the right result, however you could use the change of variables $u = x+y , v = x-y$. Pay attention to what the domain $R$ looks like under this (linear!) transformation. I'll give you a hint as what the integral should look like when you're done
$$\frac12\int_2^4 dv \int_0^v du \sin(\frac uv)$$
where $\frac12$ is the determinant resulting from the change of variables.
From here it's elementary calculus.
Also, in your computation you can see at a glance that there's something wrong: you have the variable $y$ appearing in the upper extreme of the second integral, but you're integrating with respect to that same variable!
Best Answer
So you've got $0\leq x\leq 4$ in one variable, and $0\leq y \leq 8-2x$ on the other. You have to make sure to integrate with respect to $y$ first in this case, since $y$ depends on $x$.