Previously in this series, I showed that
So far, I have shown that
where and (and is a certain angle that is now irrelevant at this point in the calculation).
We now employ the substitution
Before going much further, let’s take a closer look at and to make sure that is positive (so that the square root is defined).
First, both and $R$ are clearly positive, and so .
Next, notice that
So as long as . Therefore, since :
So, since and , we have , and so the above substitution is well-defined.
We now employ the above substitution. The endpoints of integration remain unchanged, and so
In the above calculation, I used the fact that , which was derived above. Also, I was careful to avoid a common algebraic mistake.