I would like to solve the following equation for $b$. As it is now, Wolfram Alpha is not giving an answer.
$$\frac{\pi\cot B}{A}=\int^{\pi}_0\left(\frac{\frac{b\cos x\sin^2B}{R}+\sqrt{-\frac{b^2}{R^2}\sin^2B\sin^2x+sin^2x+cos^2x\sin^2B}}{sin^2x+cos^2xsin^2B}\right)^2\cos xdx$$
As you can probably guess, most of the integrand came from the quadratic formula and I've simplified from there. $A, R$, and $B$ are constants. $b$ is also constant but is the variable I am trying to determine.
ADDITIONAL INFORMATION: From a technical paper, if $A = 1$, then
$$\frac{b}{R} = cotB$$
I don't know how that was determined, but it is correct. I have verified that the two sides of the integral are the same if when $A=1$, $\frac{b}{R} = cotB$. I want to point out that the authors of that paper never included the variable $A$ and therefore by default it was equal to 1.
I thought maybe the answer to my problem might be as simple as:
$$\frac{b}{R} = \frac{cotB}{A}$$
but I can't get that to work if I substitute it into the integral, plus numerical integration indicates that while close, it is not the correct solution. Any suggestions are appreciated.
