Briefly, in class we proved Kepler's $2$nd law like this:
We've some random trajectory and two position vectors, $\mathbf r$ and $\mathbf{r}+d\mathbf{r}$. Supposing $dr$ is small, we can approximate it to be the arc length, $ds$. Thus, we've got that the differential area swept would be a triangle with height $r$ and base $dr$: $$dA=\dfrac{1}{2}rdr\approx\dfrac{1}{2}rds=\dfrac{r^2}{2}d\theta.$$
We then "divide" by $dt$ and make use of conservation of angular momentum $\left(\dot \theta=\frac{L}{mr^2}\right)$ to finally prove it: $$\dfrac{dA}{dt}=\dfrac{r^2}{2}\dot \theta=\dfrac{L}{2m}=\text{ct.}$$
However, I'm not satisfied with this proof since there are some unrigorous steps and I was wondering whether maybe it could be done using double integrals and differentiation wrt time under the integral sign. If not, what other ways to prove it would be correct and rigorous too?
Conservation of areal velocity (Kepler's 2nd Law) is seen immediately by writing Hamilton's equations for a central force potential in polar coordinates. Our Hamiltonian (the total energy) in this case is:
$$H =\frac{1}{2}(\dot r^2 + r^2\dot\theta^2) + V(r) $$
So Hamilton's equations for the coordinate $\theta$ are: $$\frac{\partial H}{\partial p_\theta} = \dot\theta \quad \frac{\partial H}{\partial \theta} = -\dot p_{\theta}$$
From the first equation, we see that $p_{\theta} = \frac{1}{2}r^2\dot\theta$. This is the momentum $conjugate$ to the $\theta$ coordinate and also happens to be the areal velocity. From the second equation, we immediately get that $\dot p_{\theta} = 0$, i.e. that areal velocity is a constant of motion. Also, it's worth mentioning that areal velocity is a constant of motion no matter what the potential energy $V(r)$ is, as long as the potential energy is not a function of $\theta$.
It sounds like you may not already be with the Hamiltonian formalism. If this is so, you can get $F = ma$ from Hamilton's equations without too much difficulty to convince yourself of their equivalence.