Let $D := \{(x,y) \in \mathbb R^2∶ \lVert (x,y) \rVert \in ]1, 2[ \}$. We want to find : $$\iint_D \frac{\sin(x^2+y^2)}{2+\cos(x^2+y^2)} \ dxdy $$ I already know a first way to do it, that is using polar coordinates:
$$\iint_D \frac{\sin(x^2+y^2)}{2+\cos(x^2+y^2)} dxdy = $$$$\int_0^{2\pi}d\theta\int_1^2\frac{\sin(r^2)}{2+\cos(r^2)}r d r=$$$$-\pi\int_{\cos1}^{\cos4}\frac{dp}{2+p}$$ $$\implies \iint_D \frac{\sin(x^2+y^2)}{2+\cos(x^2+y^2)} dxdy = \pi \ln \bigg(\frac{2+\cos1}{2+\cos4}\bigg)$$
I wanted to do it by using Fubini's theorem. I tried to find the integral on the upper half of $D$ that is $D' := \{(x,y) \in \mathbb R^2∶ \lVert (x,y) \rVert \in ]1, 2[, y\geq0 \}$ ($D'$ is half a donut of internal radius $1$ and external radius $2$). What I did is the following : $$\iint_{D'} \frac{\sin(x^2+y^2)}{2+\cos(x^2+y^2)}\ dxdy = $$$$\int_{-2}^2 \bigg(\int_0^{\sqrt{4-x^2}} \frac{\sin(x^2+y^2)}{2+\cos(x^2+y^2)} dy\bigg)dx - \int_{-1}^1 \bigg(\int_0^{\sqrt{1-x^2}} \frac{\sin(x^2+y^2)}{2+\cos(x^2+y^2)} dy\bigg)dx $$ Here I compute the integral on the semi-disk of radius $2$ then I subtract the integral on the integral on the semi-disk of radius $1$ to find the integral on $D'$. I don't know if I am allowed to do this and will it work? I think not because I heard that $\int \frac{\sin(x^2+c)}{2+\cos(x^2+c)}dx$ is undefined.