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I would like to compute:

$$ \int_{0}^{2\pi}\frac{\sin(nx)}{\sin(x)} \mathrm dx $$

We have:

$$ \int_{0}^{2\pi}\frac{\sin(nx)}{\sin(x)}\mathrm dx=2\int_{0}^{\pi}\frac{\sin(nx)}{\sin(x)}\mathrm dx$$

So: $ n \geq 2$ $$ \int_{0}^{\pi}\frac{\sin((n+1)x)}{\sin(x)}\mathrm dx= \ln(\sin(x))\sin(nx)\vert_0^{\pi}-n\int_{0}^{\pi} \ln(\sin(x))\cos(nx)\mathrm dx $$

$$ =-n\int_{0}^{\pi} \ln(\sin(x))\cos(nx)\mathrm dx$$

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  • 5
    Am I the only one lost as to what he did?2012-02-26
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    It's not true for even $n\ge 4$, I think.2012-02-26
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    @draks: You're right, see my answer.2012-02-26
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    @draks: it is not true of $n$ even.2012-02-26
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    @Chon : It looks to me as if you may be attempting to integrate by parts, and possibly you're under an impression that $(d/dx)\ln\sin x = 1/\sin x$, and that $(d/dx)\sin(nx) = \cos(nx)$. But it's hard to tell what you actually did, so I can only venture guesses like this. At any rate, but identities are wrong since they neglect the chain rule.2012-02-27

4 Answers 4