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.

The indefinite integral has an elementary closed form.  Let $$f_a(\theta) = \frac{\sin^2 \theta (\cos \theta - a)}{(1 - a \cos \theta)^4},$$ and $$u = \frac{1 - a \cos \theta}{\sin \theta}, \quad du = \frac{a - \cos\theta}{\sin^2 \theta} \, d\theta,$$ and thus $$\int f_a(\theta) \, d\theta = -\int u^{-4} \, du = \frac{1}{3} u^{-3} + C = \frac{\sin^3 \theta}{3(1-a \cos\theta)^3} + C.$$  Evaluating this for $\theta = 0$ and $\theta = 2\pi$ shows that the definite integral is zero.
