If $a^2>b^2$ prove that $\int\limits_0^{\pi} \frac{dx}{(a+b\cos x)^3}=\frac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$. Problem: If $a^2>b^2$ prove that $\int\limits_0^\pi \dfrac{dx}{(a+b\cos x)^3} = \dfrac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$.
My effort:
If we choose $$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+1} \, dx\;\;,\;\;\cos\theta=\frac{1-x^2}{1+x^2}$$ then the integral becomes critical. What is the simplest way to solve?

As question is closed let me bring short answer here:
Having $I_n=\int\frac{dx}{(a+b\cos x)^n}$ partial integration gives
$$I_2(a^2-b^2) = aI_1-\frac{b\sin x}{a+b\cos x}$$
$$I_3 2(a^2-b^2) = 3aI_2 - I_1 -\frac{b\sin x}{a+b\cos x}$$
So $I_3$ can be represented as sum of constant member with $I_1$ plus function members.
Knowing, for $0\leqslant \varepsilon < 1$
$$\int\limits_{0}^{2\pi}\frac{dx}{1+\varepsilon \cos x} = \frac{2\pi}{\sqrt{1-\varepsilon^2}}$$
we can obtain
$$\int\limits_{0}^{2\pi}\frac{dx}{(a+b \cos x)^3}=\frac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$$
