Ways to prove the area of an ellipse formula One can prove the ellipse area formula $A=\pi a b$ ($a$, $b$ the major and minor semi-axis) either by integration or by the stretched-circle argument. See for instance here: https://proofwiki.org/wiki/Area_of_Ellipse
Are there any other proofs of this formula?

Let's get creative. For any $R>1$ we have
$$ f_R(\theta)=\frac{1}{R+e^{i\theta}} = \sum_{n\geq 0}(-1)^n \frac{e^{ni\theta}}{R^{n+1}} $$
hence by Parseval's identity
$$\int_{0}^{2\pi}\frac{d\theta}{(R^2+1)+2R\cos\theta}= \int_{0}^{2\pi}f_R(\theta)\overline{f_R}(\theta)\,d\theta = 2\pi\sum_{n\geq 0}\frac{1}{R^{2n+2}}=\frac{2\pi}{R^2-1}$$
and by multiplying both sides by $(R^2+1)$ we get
$$ \int_{0}^{2\pi}\frac{d\theta}{1+\frac{2R}{R^2+1}\cos\theta}=2\pi\frac{R^2+1}{R^2-1}. $$
Through a substitution we get that for any $B>1$ the identity
$$ \int_{0}^{2\pi}\frac{d\theta}{B+\cos\theta} = \frac{2\pi}{\sqrt{B^2-1}} $$
holds. By differentiating both sides with respect to $B$ we get
$$\forall B>1,\quad \int_{0}^{2\pi}\frac{d\theta}{(B+\cos\theta)^2}=\frac{2\pi B}{(B^2-1)^{3/2}}.\tag{1} $$
Let us assume that an ellipse has semiaxis $a>b$. How can we write its polar equation with respect to a focus?
By imposing that the property $PF_1+PF_2=2a$ holds, i.e. by imposing that
$$ \rho(\theta) + \sqrt{\rho^2(\theta)\sin^2(\theta)+(\rho(\theta)\cos(\theta)-2c)^2} = 2a, $$
$$ \rho^2(\theta)-4c\rho(\theta)\cos(\theta)+4c^2 = 4a^2-4a\rho(\theta)+\rho^2(\theta), $$
$$ -c\rho(\theta)\cos(\theta) = b^2-a\rho(\theta), $$
$$ \rho(\theta) = \frac{b^2/a}{1-e\cos\theta} \tag{2}$$
where $e=\frac{c}{a}=\sqrt{1-\frac{b^2}{a^2}}$. The area enclosed by the ellipse is 
$$ A(a,b)=\frac{1}{2}\int_{0}^{2\pi}\rho^2(\theta)\,d\theta = \frac{b^4}{2a^2}\int_{0}^{2\pi}\frac{d\theta}{(1-e\cos\theta)^2}\stackrel{\theta\mapsto\pi+\theta}{=}\frac{b^4}{2a^2 e^2}\int_{0}^{2\pi}\frac{d\theta}{\left(\frac{1}{e}+\cos\theta\right)^2}\tag{3}$$
and by invoking $(1)$ we have
$$ A(a,b) = \frac{2\pi b^4}{2a^2(1-e^2)^{3/2}}=\large{\color{red}{\pi a b}}. $$
