I've been asked to show $$ \displaystyle \int_{0}^{\pi} \dfrac{2(1+\cos x) - \cos((n-1)x) - \cos((n+1)x) - 2\cos nx}{1-\cos 2x} \ dx = n\pi $$

The integrand simplifies nicely to $$\frac{\cos nx - 1}{\cos x - 1}$$ but I'm not sure how to proceed from here, I've tried using a $x \mapsto \pi - x$ sub, but that doesn't work either. Real methods, please.

  • $\begingroup$ Off the top of my head, I would look for some sort of symmetry. That, or use contour integration over $\Bbb C$. $\endgroup$ – Omnomnomnom Feb 1 '16 at 21:23
  • $\begingroup$ @Omnomnomnom I'm still in high school, the intended approach has nothing to do with contour integration. I'll edit that in, thanks. I was hoping it would have symmetry in $x=\pi$, but apparently not. $\endgroup$ – Zain Patel Feb 1 '16 at 21:24
  • $\begingroup$ Perhaps there's symmetry about the line $x = \pi/2$? Or perhaps that's what you meant. $\endgroup$ – Omnomnomnom Feb 1 '16 at 21:25
  • $\begingroup$ Would symmetry in $x=\pi/2$ help given the limits? $\endgroup$ – Zain Patel Feb 1 '16 at 21:28
  • $\begingroup$ For example: if shifting the graph $\pi/2$ to the left makes the function odd, then that would help. $\endgroup$ – Omnomnomnom Feb 1 '16 at 21:30

HINT: Observe that $$ \int_0^{\pi}\frac{\cos nx−1}{\cos x−1}\mathrm{d}x=\int_0^{\pi}\frac{\sin^2(nx/2)}{\sin^2(x/2)}\mathrm{d}x $$ and then you can follow here.

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    $\begingroup$ I'm guessing that in a High School calc class, the student is not looking for Fourier analysis and Perseval's identity any more than he is looking for contour integration. $\endgroup$ – Mark Fischler Feb 1 '16 at 22:02
  • $\begingroup$ @MarkFischler perhaps not, but the recursion-based method looks very much like what was intended, especially given how the problem was presented. $\endgroup$ – Omnomnomnom Feb 1 '16 at 22:04
  • $\begingroup$ There isn't Fourier analysis...just trigonometry and if you follow the link just a recursion formula... $\endgroup$ – alexjo Feb 1 '16 at 22:06

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