0

I have the following homework problem:

(a) Find $\int_{-\pi}^{\pi} \frac{dx}{2+ \cos x}$.
(b) Using (a), compute $\int_{0}^{2\pi} \frac{dx}{2+ \cos x}$.

I did (a) and obtained $\frac{2\pi}{\sqrt3}$. Of course, similarly I can also compute (b). However, I don't know how to do (b) using (a). If I make the substitution $y=x-\pi$, then I get

$$\int_{0}^{2\pi} \frac{dx}{2+ \cos x}=\int_{-\pi}^{\pi} \frac{dy}{2+ \cos (y+\pi)}=\int_{-\pi}^{\pi} \frac{dy}{2- \cos y}$$ and I can't compute the latter integral using (a).

xpaul
  • 39,891
  • 3
  • 59
  • 82
Guest
  • 1,515
  • 3
  • 11
  • 3
    The integrands have discontinuities, and they are divergent improper integrals. – Sungjin Kim May 25 '23 at 19:13
  • @TravisWillse Yes, apologies. – Guest May 25 '23 at 19:21
  • 3
    The integrals are equal due to periodicity. – Ryszard Szwarc May 25 '23 at 19:21
  • See [this](https://math.stackexchange.com/questions/98409/integral-of-periodic-function-over-the-length-of-the-period-is-the-same-everywhe) – Ryszard Szwarc May 25 '23 at 19:36
  • 4
    If you set $y=x-\pi$ you'll be forced to use the parity of $\cos$. It's better to work over the period like this. $\int_0^{2\pi}\frac {dx}{2+\cos x}=\int_0^{\pi}\frac {dx}{2+\cos x}+\int_{\pi}^{2\pi}\frac {dx}{2+\cos x}$ Now apply $y=x-2\pi$ to the second integral, which result in interval $[-\pi, 0]$ and conclude by stitching the two parts.. – zwim May 25 '23 at 19:52

0 Answers0