I'm working through Griffith's Electrodynamics book during the winter break, and I'm having trouble on evaluating this integral from problem 2.7 of Introduction to Electrodynamics 4th edition.
I have my electric field right?
$$ \frac{1}{4\pi\epsilon_{0}} \sigma R^{2} 2\pi \int_{0}^{\pi} \frac{z - R\cos\theta}{(R^2+z^2 -2Rz\cos\theta)^{3/2}} \sin\theta \ d\theta $$
The integral i want to evaluate is (obviously) $$ \int_{0}^{\pi} \frac{z - R\cos\theta}{(R^2+z^2 -2Rz\cos\theta)^{3/2}} \sin\theta \ d\theta $$
The solution manual says to use partial fractions, but I feel like that would take quite a bit of working out to do. I want to avoid using a lot of algebra as possible (to reduce possible hiccups). Also, looking around, a lot solution of the internet just go the route of plugging it in mathematica or maple. I'm trying to avoid using those tools.
Is there an integral table that that has something of this form or is there a much clever way of evaluating this?
Thank you.
Edit: If it helps, this is the integral evaluated from $0$ to $\pi$ (based off of the solution manual).
$$ \frac{1}{4\pi\epsilon_{0}} \frac{\sigma R^{2} 2\pi}{z^2} \Bigg\{ \frac{(z-R)}{|z-R|} - \frac{(-z-R)}{|z+R|} \Bigg\} $$
Their will conditions of course, such as, when $z>R$ and $z<R$, but they can be applied after evaluating the integral.
Putting $$R^2+z^2-2Rz\cos\theta=t^2$$ one gets $$\int_0^\pi \frac {(z-R\cos\theta)}{(R^2+z^2-2Rz\cos\theta)^{3/2}}\,\sin\theta \,d\theta=$$$$\frac 1{2Rz^2} \int _{|R-z|}^{R+z}\left [1-\frac {R^2-z^2}{t^2} \right]\, dt=$$$$\frac 1{2Rz^2} \left(2R-\frac {R^2-z^2}{|R-z|}-|R-z| \right)=\begin {cases} 0\qquad \text{if}\quad z<R\\ \\\dfrac 2{z^2}\quad\; \text{if}\quad z>R \end {cases}$$