I spent 3 hours today on the following integral, (I am trying to verify why my variable substitution fails):
$$\int\int\frac{x^2\sin(xy)}{y}dxdy$$ over $$\Omega=\{x^2<y<2x^2, y^2<x<2y^2\}$$ which is (everything is positive): $$\Omega=\{1<y/x^2<2, 1<x/y^2<2\}$$
So, I deduced that for a proper transform $T$ we will get that: $$T\Omega=\{(u,v)=T(x,y):1<u<2, 1<v<2\}$$ So we want the opposite of $T(x,y)=(y/x^2,x/y^2)$.
I will skip some of the calculations, but I got that: $$|J_T|=3x^-2y^-2\Longrightarrow |J_{T^-1}|=3^{-1}x^2y^2$$ and that $$T^{-1}(x,y)=(y^{-\frac{1}{3}}x^{-\frac{2}{3}},y^{-\frac{2}{3}}x^{-\frac{1}{3}})$$
and that
$$\int_\Omega f=\int_{T\Omega}f\circ T^{-1}|J_{T^{-1}}|$$
which is after putting all of the information:
$$\frac{1}{3}\int_1^2\int_1^2xy^2\sin(y^{-1}x^{-1})$$
And this is not leading to anywhere, and I also compared to a full solution that does somewhat similar variable substitution, and after that, he got similar, but solveable integral: $$\frac{1}{3}\int_{\frac{1}{2}}^1\int_{\frac{1}{2}}^1u\sin(uv)$$
I will appreciate any kind of help, I spent too much time on this
$$u=\frac x{y^2},v=\frac y{x^2} \implies x=\frac1{u^{\frac23}v^{\frac13}},y=\frac1{u^{\frac13}v^{\frac23}}$$
Hence
$$J=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}=\begin{bmatrix}-\frac2{3u^{\frac53}v^{\frac13}}&-\frac1{3u^{\frac23}v^{\frac43}}\\-\frac1{3u^{\frac43}v^{\frac23}}&-\frac2{3u^{\frac13}v^{\frac53}}\end{bmatrix} \implies |\det J| = \frac1{3u^2v^2}$$
Then the integral becomes
$$\iint_\Omega\frac{x^2\sin(xy)}y\,\mathrm dx\,\mathrm dy = \frac13\int_1^2\int_1^2 \frac1{u^2v^3} \sin\left(\frac1{uv}\right)\,\mathrm du\,\mathrm dv$$