In my previous question @JackD'Aurizio helped me to solve the problem I had. Now I have the following identity:
$$\frac{A}{\text{n}}\int_{0}^{2\pi}\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta = \int_{0}^{\theta_A}\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta$$
QUESTION: And I need to solve that equation for $\theta_\text{A}$.
What I did, is try to evalute the integral:
$$\int\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta=\frac{p^2}{2}\int\frac{1}{\left(1+\varepsilon\cos\left(\theta\right)\right)^2}\,d\theta=$$ $$\frac{p^2}{2}\int\frac{1}{1+2\epsilon\cos\left(\theta\right)+\epsilon^2\cos^2\left(\theta\right)}\,d\theta$$
No, you don't, for two reasons. One is that it is incorrect. The other is that even if it was correct, you would not need to do so in the context of your other question. The formula
$$\text{G}_\text{sc}\left(\text{n}\right)=1367\cdot\left(1+0.03\cos\left(\frac{360}{365}\cdot\text{n}\right)\right)\space\space\space\space\space\space\space\space\space\left[\text{W}/\text{m}^2\right]\tag1$$
is an approximation. It assumes
Given the somewhat rough nature of this approximation, assuming that mean motion and true motion are the same is quite fine. This, coupled with a year of 365 days and perihelion on December 31 leads to $\theta(n) = 360^\circ \frac n {365}$.
Regarding my egregious charge that $\frac{A}{\text{n}}\int_{0}^{2\pi}\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta = \int_{0}^{\theta_A}\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta$ is incorrect: The two integrals have solutions in the elementary functions. Kepler's equation is transcendental, meaning it does not have a solution in the elementary functions. Try as hard as you want, but you will not be able to integrate the correct expression.
What you can do is come up with a somewhat ugly relationship between eccentric anomaly and true anomaly and a seemingly simple relationship between eccentric anomaly and mean anomaly, the last of which marches uniformly with time. The somewhat ugly relationship between eccentric anomaly and true anomaly, $\sqrt{1-e}\tan\frac\theta2=\sqrt{1+e}\tan\frac E2$, is invertible. The seemingly simple relationship between eccentric anomaly and mean anomaly, $M = E - e\sin E$, is not invertible (in the elementary functions).