Calculate: $\iint_{D}\cos\left(\frac{x+2y}{-x+y}\right)dx\,dy$

Question

Let D be the triangle region in $\mathbb{R}^2$ having vertices $(0,0), (-2,4), (-3,3).$ Calculate: $$\iint_{D}\cos\left(\frac{x+2y}{-x+y}\right)dx\,dy$$

I use change variable $x+2y=u$, $-x+y=v$. Finally, it leads to calculate: $\int\limits_0^6\int\limits_{\frac1{3}u}^{\frac2{3}u}\frac13\cos(\dfrac{u}{v})dudv$. I cannot finish with this final step. Could you please help me?

Answer 1

Let $x=(u+v)/2$ and $y=(u-v)/2$. This takes the region to the right triangle (that's the real point of this substitution) with vertices $(0,0)$, $(0,-6)$ and $(2,-6)$ in the $uv$-plane. The Jacobian is $\frac{1}{2}$. This is just the typical substitution taking an ugly triangle to something more tractable (namely a right triangle).

Choosing the order in which we integrate correctly (we can do that here): $$\iint_{D}\cos\left(\frac{x+2y}{-x+y}\right)\,dx\,dy=\frac{1}{2}\int_{-6}^{0}\int_{0}^{-v/3}\cos\left(\frac{v-3 u}{2 v}\right)\,du \,dv\ldotp$$

This is $$\frac{1}{2}\int_{-6}^{0}\frac{2}{3} v \left(\sin \left(\frac{1}{2}\right)-\sin (1)\right)\,dv=6 \left(\sin (1)-\sin \left(\frac{1}{2}\right)\right)$$