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