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?
 A: 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.
A: Your strategy is nice and correct.
If you want you can use some complex analysis.I do not know if this is simple.It depends on your knowledge.
Put $x=\frac{y}{2}$ so the endpoints will become $0,2 \pi$
Then :
Put $z=e^{it} ,dz=ie^{it}dt \Longleftrightarrow \frac{1}{zi}dz=dt$ and relate the trigonometric function with the exponentials with the appropriate identities and then  apply the residue theorem to the unit circle.
A: 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
$$\int\limits_{0}^{2\pi}\frac{dx}{(a+b \cos x)^3}=\frac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$$
