Calculus.6

Changing variables to $u = x-y$ and $v = x+y$ , we have the Jacobian $\frac{\partial(u,v)}{\partial(x,y)} \; =\; \left| \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right| \; = \; 2$ and we note that the region $\mathfrak{R}$ is defined by the inequalities $0 \le v \le 1$ , $-v \le u \le v$ . Thus $\begin{array}{rcl} \displaystyle \iint_{\mathfrak{R}} \cos\left(\frac{x-y}{x+y}\right)\,dx\,dy & = & \displaystyle \tfrac12\int_0^1\,dv \int_{-v}^v \cos\left(\frac{u}{v}\right)\,du \; = \; \tfrac12 \int_0^1 \,dv\, \left[ v \sin \left(\frac{u}{v}\right)\right]_{u=-v}^v \\ & = & \displaystyle \tfrac12 \int_0^1 2v\sin 1\,dv \; = \; \tfrac12\sin 1 \; = \; \boxed{0.4207354924} \end{array}$