Answer
$ \dfrac{3}{2} \sin (1)$
Work Step by Step
$Jacobian, J(x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=(-1) \cdot (1) -(1)(1)=-2$
and $J(u,v)=|-\dfrac{1}{2}|=\dfrac{1}{2}$
Let us consider that $u=y-x$ and $v=x+y$
$\iint_R \cos (\dfrac{y-x}{y+x}) dA=\int_1^{2} \int_{-v}^{v} (\dfrac{1}{2} )\cos (\dfrac{u}{v}) du dv$
or, $=(\dfrac{1}{2}) \int_1^2 \sin \dfrac{u}{v}|_{-1}^1 dv$
or, $= \int_1^2 (v) (2) (\dfrac{1}{2})[ \sin (1)] dv $
Thus, we have
$\iint_R \cos (\dfrac{y-x}{y+x}) dA= (\dfrac{3}{2}) \sin (1)$
