0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.