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I'm trying to integrate $\int^{\theta_{2}}_{\theta_{1}}\frac{\sin m \theta}{2\sin(\frac{\theta}{2})} -\frac{1}{2}d\theta$ where m is a constant.

The answer should be $$\frac{\theta_{1} -\theta_{2}}{2} + \frac{\cos m\theta_{1}}{2m\sin(\frac{\theta_{1}}{2})} - \frac{\cos m\theta_{2}}{2m\sin(\frac{\theta_{2}}{2})} - \frac{1}{4m} \int^{\theta_{2}}_{\theta_{1}}\frac{\cos m\theta \cos(\frac{\theta}{2})}{\sin^{2}(\frac{\theta}{2})} d\theta$$

I can get this answer but I end up with $-\frac{1}{2m}$ outside the integral.

Let $u= 2\csc(\frac{\theta}{2})$ $du = \frac{-\cos(\frac{\theta}{2})}{2\sin^{2}(\frac{\theta}{2})}$ $dv=\sin m\theta$ $v=-\frac{1}{m}\cos m\theta$

Integrating by parts I get as written above but with the $1/2m$ instead of $1/4m.$

Can anyone help, thanks.

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There's already $1/2$ in front of the integral. But you pick up another $1/2$ with the chain rule when you differentiate $\left(\sin(\frac{\theta}2)\right)^{-1}$. You seem to have gotten in trouble when you let $u=2\text{ cosec}(\theta/2)$ rather than $u=\frac12\text{cosec}(\theta/2)$. Word of general advice: Factor out numerical constants at the beginning of the computation, rather than trying to keep track of them through all the steps.

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  • $\begingroup$ Ahh, it's been a long day. Thanks. $\endgroup$ – Numbertheorylearner Jan 30 '17 at 22:32
  • $\begingroup$ You're welcome. Please accept the answer when you're satisfied so that this question will go off the "need to be answered" list :) $\endgroup$ – Ted Shifrin Jan 30 '17 at 22:36

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