I have two questions: one related to evaluating the integral and the other related to how area in general is evaluated in polar coordinates.
(1) The area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ written in Cartesian Coordinates is 4 times the integral $\int_0^a \frac{b}{a}\sqrt{a^2-x^2}dx$ and is equal to $\pi a b$.
The standard equation of an ellipse in polar coordinates is $r=\frac{a(1-e^2)}{1+ e\cos \theta}$, ($e$ is the eccentricity). The area (which is given by $\int_0^{2\pi}\frac{1}{2}r^2 d\theta$) would then be: $$\int_0^{2\pi}\frac{1}{2}\frac{a^2(1-e^2)^2}{(1+e \cos \theta)^2} d\theta$$ This might seem to diverge but (fortunately) it does not as $0<e<1$ for an ellipse. But I don't seem to be able to evaluate this integral and I'm interested in knowing how.
(2) I put it in integral-calculator and it says "integral is divergent" because I don't know how to put the condition $0<e<1$. I tried replacing $e$ with $\sqrt{1-a^2}$ to satisfy that condition (as we're doing it in real domain) and integrated $\dfrac{1}{\left(1+ \sqrt{1-a^2}\cos\left(\theta\right)\right)^2}$ w.r.t $\theta$ and found the antiderivative to be: $$\dfrac{2\sqrt{1-\sqrt{1-a^2}}\sqrt{\sqrt{1-a^2}+1}\ \arctan\left(\frac{\sqrt{1-\sqrt{1-a^2}}\tan\left(\frac{\theta}{2}\right)}{\sqrt{\sqrt{1-a^2}+1}}\right)}{a^4}-\dfrac{\sqrt{1-a^2}\sin\left(\theta\right)}{a^2\cdot\left(\sqrt{1-a^2}\cos\left(\theta\right)+1\right)}\tag1$$
Now, from my understanding, in polar coordinates we don't need to split the area into different quadrants like we do in Cartesian (to avoid the negative areas canceling positive ones) due to the way the area integral is derived in polar coordinates. But, when we put in the bounds $0$ and $2\pi$ in expression (1), we get zero.
Please explain.