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