Confirm definite integral equals zero $\frac{\sin(x)}{(1-a\cos(x))^{2}}$ Is this statement about the definite integral of a particular function $F$ true?
$$\int_0^{2\pi}F(x)\, \mathrm{d}x =
\int_0^{2\pi}\frac{\sin(x)}{(1-a\cos(x))^2}\, \mathrm{d}x = 0 \ \text{    for }\ 0<a<1$$
I have evaluated this expression (in WolframAlpha) for various values of a and they all give the value zero.  I have read that integrals of the form
$$\int G(\cos(x))\sin(x)\, \mathrm{d}x$$  where $G$ is some continuously integrable function are zero over the range $-\pi/2$ to $\pi/2$.
(Edited after comment from Andrey)
It seems possible to proceed from here to confirm the postulated statement by symmetry.
The function $F$ to be integrated is cyclic with period 2$\pi$ such that $F(x-2\pi) = F(x) =F(x+2\pi)$.  Then we just need to prove that the two integrals:- (1) between $-\pi$ and $-\pi/2$, and (2) between $\pi/2$ and $\pi$ are equal in magnitude and opposite in sign.
This would be the case if $F(x) = -F(-x)$.  Such is actually the case because the denominator in F() expands to $(1-2a\cos(x)+a^2\cos^2x)$ and has the same values for $(+x)$ and $(-x)$.  Whereas the numerator $\sin(x)$ is such that $\sin(x) = -\sin(-x)$.
However I would still like to find an analytical solution.
 A: Using the general rule (link)
$$ \int_a^bF(x)dx = \int_a^bF(a+b-x)dx,$$
we have in this case
$$ \int_0^{2\pi}F(x)dx = \int_0^{2\pi}F(2\pi-x)dx,$$
and, from knowledge of the symmetries of the sin and cos functions, we know in this case that
$$F(x) = - F(2\pi-x),$$
so, with $I =\int_0^{2\pi}F(x)dx$, we have 
$$I=-I,$$
which can only be true if $$I = 0$$
A: Let $u=1-a\cos x$, $du=a\sin x dx$ to get $\displaystyle\frac{1}{a}\int_{b}^{b}\frac{1}{u^2}du=0$ $\;\;\;$ (where $b=1-a$).
A: This is a special case of a general fact about $u$-substitution. If $G(x)$ is integrable on the interval $[a,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then $$\int_a^b G(u(x))\,u'(x)\, dx = g(u(b))-g(u(a)).$$
If $u(a)=u(b)$, the integral is zero.
The integrand in your example has this form, where $u(x)=\cos(x)$ and $G(u)=\frac{-1}{(1-a\cos(x))^2}$, and $\cos(x)$ has the same value at both limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\cos x$ to show it.)
You can write down lots of messy-looking integrals that turn out to be zero because they have this form for some $u(x)$ and $G(x)$.
$$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$
$$\int_0^{2\pi} (\pi-x)\log(2+\sin^2(x-\pi)^2)\, dx$$
$$\int_{\pi/2}^{3\pi/2} (\cos^2 x)^{\sin^2 x}\sin2x\,dx$$
