So I'm working on a orbital mechanics project and need to find the area of a conic section with an arbitrary eccentricity greater than 1. I have a general polar formula, $p=\frac{l}{wcos\theta+1}$, where $p$ is Semi Latus Rectum, $w$ is eccentricity, and $\theta$ is angle of around the origin (true anomaly).
To get area, I'm using the formula $A=\frac{1}{2}\int_{\theta_1}^{\theta_2}r^2d\theta$
plugging in my radius formula and using an online integration calculator (cause I am not talented enough to do it myself), I got this: $$A=\frac{1}{2}\int_{\theta_1}^{\theta_2}(\frac{l}{wcos\theta+1})^2d\theta$$ $$A=\frac{l^2}{2}[\dfrac{w\sin\left({\theta}\right)}{\left(w^2-1\right)\left(w\cos\left({\theta}\right)+1\right)}+\dfrac{2\sqrt{1-w}\arctan\left(\frac{\sqrt{1-w}}{\sqrt{1+w}}\tan\left(\frac{{\theta}}{2}\right)\right)}{\left(w-1\right)^2\left(w+1\right)^\frac{3}{2}}]_{\theta_1}^{\theta_2}$$ When I go ahead and plug this into desmos, choosing an arbitrary number for $\theta$ (normally $\frac{\pi}{2}$), I get a graph that suddenly stops at 1. I recognize it's because there's a lot of $\sqrt{1-w^2}$ in there, but I was wondering if there's any way to fix that because clearly for conic sections with a eccentricity greater than one, there's a sector you can get an area from, but I'm just stumped.
The result of your integral can also be expressed as follows, avoiding complex numbers: $$ A={l^2\over2}\left[ \frac{w \sin\theta}{\left(w^2-1\right) (w \cos\theta+1)} -\frac{\ln \left(\frac{\sqrt{w^2-1} \sin\theta+\cos\theta+w}{w \cos\theta+1}\right)}{\left(w^2-1\right)^{3/2}} \right]_{\theta_1}^{\theta_2} $$