<s>[INST] Prove that $\int_0^{2\pi}{\frac {\sin^2\theta (cos\theta - a)}{(1-a \,cos \theta)^4}}d \theta = 0$ for $0<=a<1$ I am trying to prove that for $0<=a<1$
$$\int_0^{2\pi}{\frac {\sin^2\theta (cos\theta - a)}{(1-a \,cos \theta)^4}}d \theta = 0$$
I know that 
$$\int_0^{2\pi}{\frac {\sin\theta \,cos\theta}{(1-a \,cos \theta)^4}}d \theta = 0$$
and $$\int_0^{2\pi}{\frac {3a\sin\theta}{(1-a \,cos \theta)^4}}d \theta = 
\int_0^{2\pi}{\frac {cos\theta}{(1-a \,cos \theta)^3}}d \theta$$
I have tried substituting this last equation into the first but it doesn't seem to produce any obvious simplifications.
 [/INST] You can apply a neat complex analysis trick to integrals of the form
$$ \int_0^{2\pi} F(\cos \theta, \sin \theta)\ \mathrm{d}\theta. $$
Make the substitution $z = e^{i\theta} = \cos \theta + i \sin \theta$. This transforms the contour of integration from the interval $[0,2\pi]$ into the unit circle $C_1(0)$ lying in the complex plane. We can obtain $\mathrm{d}\theta$ as
$$ \mathrm{d}z = i e^{i\theta} \mathrm{d}\theta \implies \mathrm{d}\theta = \frac{\mathrm{d}z}{iz}. $$
Furthermore, observe that $\frac{1}{z} = e^{-i\theta} = \cos \theta - i \sin \theta$, so we can obtain $\cos \theta$ and $\sin \theta$ as 
$$ \cos \theta = \frac{1}{2} \left( z + \frac{1}{z} \right) \hspace{1cm} \sin \theta = \frac{1}{2i} \left( z - \frac{1}{z} \right). $$
For your integral, this substitution produces
$$ \int_0^{2\pi} \frac{\sin^2 \theta (\cos \theta - a)}{(1 - a \cos \theta)^4} \mathrm{d}\theta = \int_{C_1(0)} \frac{2 i \left(z^2-1\right)^2 \left(z^2 - 2az +1\right)}{\left(a z^2 - 2z +a\right)^4} \mathrm{d}z.$$
Can you proceed from here?
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