Is there any closed expression of this definite integral? \begin{equation}
\int_0^\theta \frac{1}{(1+e\cos x)^2} dx = ?
\end{equation}
in which $\theta(\le\pi/2)$ and $e$ are constants.
 A: I guess you are dealing with Kepler's second law, so I am going to assume that the eccentricity $e$ belongs to the interval $(-1,1)$. As suggested in the comments, by setting $x=2\arctan\frac{t}{2}$ the integral boils down to an elementary integral, but it is probably easier to apply a geometric approach: your integral is related with the area of an elliptic sector, that can be decomposed as a triangle and a sector centered at the centre of the ellipse.  Through an affine map that becomes a circle sector, whose area is simple to compute. By computing the determinant of the involved affine map you also get the area of the original elliptic sector, namely
$$ \frac{2}{(1-e^2)^{3/2}}\,\arctan\left(\sqrt{\frac{1-e}{1+e}}\tan\frac{\theta}{2}\right)-\frac{e\sin(\theta)}{(1-e^2)(1+e\cos\theta)} $$
not that nice, indeed. This formula summarizes Kepler's relations about the mean, eccentric and true anomalies:
$$ M = E-e\sin E,\qquad (1-e)\tan^2\frac{\theta}{2}=(1+e)\tan^2\frac{E}{2}.$$
In the original problem we are interested in $M$, given $e$ and $\theta$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\int_{0}^{\theta}{\dd x \over \bracks{1 + \expo{}\cos\pars{x}}^{2}} & =
\int_{0}^{\theta}
{\dd x \over \braces{1 + \expo{}\bracks{2\cos^{2}\pars{x/2} - 1}}^{2}} =
2\int_{0}^{\theta/2}
{\dd x \over \bracks{1 - \expo{} + 2\expo{}\cos^{2}\pars{x}}^{2}}
\\[5mm] & =
{1 \over 2\expo{}^{2}}\int_{0}^{\theta/2}
{\dd x \over \bracks{\pars{1 - \expo{}}/\pars{2\expo{}} + \cos^{2}\pars{x}}^{2}} \\[5mm] & =
\left.-\,{1 \over 2\expo{}^{2}}\partiald{}{\mu}\int_{0}^{\theta/2}
{\dd x \over \mu + \cos^{2}\pars{x}}\right\vert_{\ \mu\ = \pars{1 - \expo{}}/\pars{2\expo{}}}
\\[5mm] &=
\left.-\,{1 \over 2\expo{}^{2}}\partiald{}{\mu}\int_{0}^{\theta/2}
{\sec^{2}\pars{x}\,\dd x \over \mu\sec^{2}\pars{x}  + 1}\right\vert_{\ \mu\ = \pars{1 - \expo{}}/\pars{2\expo{}}}
\\[5mm] &=
\left.-\,{1 \over 2\expo{}^{2}}\partiald{}{\mu}\int_{0}^{\theta/2}
{\sec^{2}\pars{x}\,\dd x \over \mu\tan^{2}\pars{x}  + \mu + 1}\right\vert_{\ \mu\ = \pars{1 - \expo{}}/\pars{2\expo{}}}
\end{align}

The remaining steps are a straightforward ones.

