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I'm trying to evaluate the following:

$$\frac{\sin x}{1-\cos\beta\cos x} - 2\cot\beta \arctan\left(\sin \left(\frac{\beta-x}{2}\right) \csc \left(\frac{\beta+x}{2}\right)\right)$$

This is the result of an integral and I need to evaluate at $x = \pi$ and $x=0$.

The first part of the expression evaluates to zero for both $x = \pi$ (upper limit) and $x=0$ (lower limit). For $x=0$, I think the expression yields $0.5\pi \cot\beta$. I'm getting stuck when $x=\pi$ although I suspect that this also evaluates to $0.5\pi \cot\beta$

rdemo
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  • If you're looking for a simplified answer instead of solving a complicated integral, just use WolframAlpha, Mathematica, Maple, etc. – DecarbonatedOdes Feb 18 '24 at 23:26
  • Yes, i did use WolframAlpha to get the indefinite integral, but it would not provide me with a solution for the definite integral and I have the Pro version. – rdemo Feb 18 '24 at 23:29

1 Answers1

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As a hint:$$\sin (a+b)=\sin a \cos b + \cos a \sin b $$so when you put $x=\pi$ you will have \begin{align*}\sin (\frac{\beta-x}{2}) \csc (\frac{\beta+x}{2})& = \frac{\sin (\frac{\beta-x}{2})}{\sin (\frac{\beta+x}{2})}= \frac{\sin (\frac{\beta-\pi}{2})}{\sin (\frac{\beta+\pi}{2})}\\& =\frac{\sin \frac{\beta}{2}\cos \frac{\pi}{2}-\cos \frac{\beta}{2}\sin \frac{\pi}{2} }{\sin \frac{\beta}{2}\cos \frac{\pi}{2}+\cos \frac{\beta}{2}\sin \frac{\pi}{2}} \\& =\frac{0-\cos \frac{\beta}{2}\sin \frac{\pi}{2}}{0+\cos \frac{\beta}{2}\sin \frac{\pi}{2}}= -1\end{align*} now you deal with $\arctan(-1)$

Gary
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Khosrotash
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    And the arctan of - 1 is $\frac{-\pi}{4}$ which then results in $0.5\cot\beta$. So the answer then is $\pi\cot\beta$. Thanks. P.S. I wonder why Wolfram Alpha couldn't evaluate this definite integral. It seems straightforward now that you showed me the piece of the puzzle I was missing. – rdemo Feb 18 '24 at 23:34