Let $D\subset\mathbb R^2$ be given by $$ y=0,y=x,x+2y=8.\,\,(1) $$ Let $$ u=x+2y,v=x-y. $$
We can find a diffeomorphism $(u,v)\mapsto T(u,v)=(x,y)$. The diffeomorphism is given by $$ T(u,v)=\begin{pmatrix}1/3u+2/3v\\1/3u-1/3v\end{pmatrix}. $$ I want to find $D^*$ in the $u,v$-plane such that $T(D^*)=D$. I have drawn the triangle $D$ in the $x,y$-plane, but I’m not sure how to find $D^*$. I tried writing (1) in terms of $u$ and $v$: $$ \begin{align} y=0&\implies v=u\\ y=x&\implies v=0\\ x+2y=8&\implies u=8 \end{align} $$ The point of this transformation was to compute the integral $$ \int\int_D\cos(x+2y)\sin(x-y)=\int\int_{D*}3\cos(u)\sin(v), $$ so I was expecting to find a rectangular $D^*$ in order to make the integral easier with by a change of coordinates. Is my $D^*$ correct, and if so, how should I proceed computing the integral? Could I make use of a symmetric argument, and integrate over $(0,8)^2$, and then divide the integral by 2?