<s>[INST] 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?
 [/INST] It is well-known that
$$\int_0^\pi\frac{dx}{t+\cos x}=\frac{\pi}{\sqrt{t^2-1}}$$
for $t>1$. Differentiating gives
$$-\int_0^\pi\frac{dx}{(t+\cos x)^2}=-\frac{\pi t}{(t^2-1)^{3/2}}.$$
Differentiating again gives
$$2\int_0^\pi\frac{dx}{(t+\cos x)^3}
%=-\frac{\pi(t^2-1)}{(t^2-1)^{5/2}}+\frac{3\pi t^2}{(t^2-1)^{5/2}}
=\frac{\pi(2t^2+1)}{(t^2-1)^{5/2}}.$$
Homogenising this gives your formula.
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