I was asked to show the equality of these integrals $$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)dydx =\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)dxdy\tag{1}$$ Which can be answered by using Fubini's theorem but in order to use Fubini here is what I did
Let $f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)$. Now
\begin{align*} |f(x,y)|&= \biggl|\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\log(4+\sin x)\biggl|\\ &\leq \log5\biggl|\frac{x^2-y^2}{(x^2+y^2)^{3/2}}\biggl|\\ &\leq\log 5\biggl(\frac{x^2+y^2}{(x^2+y^2)^{3/2}}\biggl)=\frac{\log5}{(x^2+y^2)^{1/2}}:=g(x,y) \end{align*} To apply Tonelli Theorem on a non negative function $$\int_0^1\int_0^1\frac{\log5}{(x^2+y^2)^{1/2}}dxdy\tag{2}$$ This is the part I get stuck. If the above integral in $(2)$ is finite, then we can apply Fubini and $(1)$ follows immediately. Is my choice of $g(x,y)$ fine? If so, how can I evaluate $(2)$. But if not, what is the best choice to bound $|f(x,y)|$?
Hint: change to polar coordinates.