Trigonometrical integral of $1/(b+\cos\theta)$ Show that: if $b>1$ $$2\int_0^\pi \frac{1}{b+\cos\theta} \, d\theta= \int_0^{2\pi}\frac{1}{b+\cos\theta} \, d\theta$$ Thanks for your help!!

Well,

${\cos\theta}={\cos(-\theta)}={\cos(2\pi-\theta)}$, where ${\theta}\in[0,2\pi]$

which means ${\cos\theta}$ is symmetry besides ${\theta}=\pi$

therefore , ${b+cosθ}={b+cos(2\pi-\theta)}$

The requirement b>1 exists in order to avoid poles for $\frac{1}{b+cosθ}$, since the min of $cos{\theta}$ is -1

$$\int_\pi^{2\pi} \frac{1}{b+cosθ} d\theta=\int_\pi^{2\pi}\frac{1}{b+cos(2\pi-\theta)} d\theta=\int_\pi^0\frac{1}{b+cos\alpha} {-d\alpha}=\int_0^\pi \frac{1}{b+cosθ} d\theta$$

$$\int_0^{2\pi}\frac{1}{b+cosθ} d\theta=\int_0^\pi \frac{1}{b+cosθ} d\theta+\int_\pi^{2\pi} \frac{1}{b+cosθ} d\theta=2\int_0^\pi \frac{1}{b+cosθ} d\theta$$