what is the solid angle subtended by a sweeping/precessing cone? I am able to calculate the solid angle subtended by a cone with apex opening angle $\theta$ via the following:
$$d \Omega = \mathrm{sin} (\theta) d \theta d \phi$$
$$\Omega = \int_{0}^{2\pi} \int_{0}^{\theta/2} sin(\theta) d \theta d \phi = 2 \pi \left[ 1-cos(\theta/2) \right]$$
dividing $\Omega$ by the solid angle of a sphere ($4\pi$) gives us the percentage of the sphere that is subtended by the cone.
I wish to now know what the solid angle would be if the cone is precessing about some angle $\Delta i$ over one whole cycle.
Here's an awful diagram explaining showing the area I want to calculate
 A: Let $$\Omega(\theta) = 2\pi \Biggr( 1 - \cos\left(\frac{\theta}{2}\right) \Biggr)$$ be the solid angle subtended by a cone with aperture $\theta$.  If you have a cone precessing at angle $\phi \gt \theta/2$ (with respect to the axis), then the solid angle is $$\Omega_p = \Omega\left(2 \phi + \theta\right) - \Omega\left(2 \phi - \theta\right)$$
where the first term is the cone corresponding to the outer edge of the solid angle covered by the precessing cone, and the second term is the cone corresponding to the inner edge of the solid angle covered by the precessing cone. ("Inner" and "Outer" considered with respect to axis of precession.)
This simplifies to $$\Omega_p = 2 \pi \left( \cos\left(\phi - \frac{\theta}{2}\right) - \cos\left(\phi + \frac{\theta}{2}\right) \right)$$
and using $\cos(x-y) - \cos(x+y) = 2 \sin(x) \sin(y)$ to
$$\Omega_p = 2 \pi \sin\left(\phi\right) \sin\left(\frac{\theta}{2}\right)$$
If $\phi \le \theta/2$, the precessing cone always includes the axis, and there is no "inner uncovered cap"; then, $$\Omega_p = 2\pi \Biggr( 1 - \cos\left(\phi + \frac{\theta}{2}\right) \Biggr)$$
Therefore, we can combine the two for any $0 \le \phi \le 90°$, $0 \le \theta \le 180°$ as
$$\Omega_p = \begin{cases}
2 \pi \sin\left(\phi\right) \sin\left(\displaystyle \frac{\theta}{2}\right), & 0 \le 2 \phi \le \theta \le 180° \\
2\pi \Biggr( 1 - \cos\left(\displaystyle \phi + \frac{\theta}{2}\right) \Biggr), & 0 \le \theta \lt 2 \phi \le 180° \\
\end{cases}$$
