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).
$\int_0^{2\pi}\frac {dx}{2+\cos x}=\int_0^{\pi}\frac {dx}{2+\cos x}+\int_{\pi}^{2\pi}\frac {dx}{2+\cos x}$
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