Integral in spherical coordinates, $\Omega$ is the unit sphere, of $\iiint_\Omega 1/(2+z)^2dx\ dy\ dz$ $$\iiint_\Omega \frac{1}{(2+z)^2}dx\ dy\ dz$$
There is a VERY similar question How to integrate $\iiint\limits_\Omega \frac{1}{(1+z)^2} \, dx \, dy \, dz$ here
But this is different.
I like my spherical coordinates to have the angle in the x/z plane taken from "3 oclock" as normal, rather than from 12. So anyway, I got this:
$$\iiint_\Omega \frac{1}{(2+p\sin(\theta))^2}p^2 \cos(\theta)dp\ d\theta\ d\psi$$
Over
$$\Omega = \lbrace(p,\theta,\psi)|\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],\psi\in\left[0,2\pi\right],p\in[0,1]\rbrace$$
I'm not sure what to do, I want to do a substitution but I've been explicitly told to use spherical coords, and it'd be a multivariate substitution (involving p and $\psi$)
Thanks.
Addendum:
I'm hoping you guys will make me feel silly and it'll be obvious, but I've been looking at it for a while, I can't see partial fractions helping,I can't get the p out of the denominator, I'm stumped, sadly.
Thoughts:
It actually seems quite easy in Cartesian. Also why can't I substitute, I'm going to have a go, treating it first as integral involving 2+something, then deal with the sin.

Doing this in Cartesian would be hellish; the limits would be a nightmare.
It may be easier to write the integral using $u = \sin \theta$ instead.
The resulting integral is
$$2\pi \int_{-1}^1 \int_0^1 \frac{p^2}{(2+pu)^2} \, dp \, du$$
A table of integrals helps here.  Consulting wikipedia tells us that
$$\int \frac{x^2}{(ax+b)^2} \, dx = \frac{1}{a^3} \left( ax -2b \ln (ax + b) - \frac{b^2}{ax+b} \right) + C$$
That helps us take care of the radial integral:
$$\int_0^1 \frac{p^2}{(2+pu)^2} \, dp = \frac{1}{u^3} \left(u - 0 - 4 \ln (u+2) + 4 \ln 2 - \frac{4}{u+2} + \frac{4}{2} \right)$$
You can probably finish the problem from here.
