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 transformation. I'll give you a hint as what the integral should look like when you're done $$\frac12\int_2^4 dv \int_0^v du \sin(\frac uv)$$ from here it's elementary calculus.

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!