$ I(r) = \int_0^{2\pi}\frac{\cos(t) - r}{1 - 2r\cos t + r^2}\,dt$ is always zero for $r\in[0,1)$. Why?
Solution 1:
An alternative proof using complex methods:
For $0< r < 1$ let $$ f_r \colon B_\frac{1}{r} (0) \to \mathbb{C} \, , \, f_r(z) = \frac{- \ln(1-rz)}{z} \, , $$ where $f_r(0) = r $ . Then $f_r$ is holomorphic, so $$ I(r) \equiv - \int \limits_0^{2\pi} \ln(1-r \mathrm{e}^{\mathrm{i}t}) \, \mathrm{d} t = - \mathrm{i} \int \limits_{S^1} f_r(z) \, \mathrm{d} z = 0 $$ holds by Cauchy's theorem. If you are not familiar with complex analysis, you can also show this using the Taylor series of the logarithm: $$ I(r) = \sum \limits_{n=1}^\infty \frac{r^n}{n} \int \limits_0^{2\pi} \mathrm{e}^{\mathrm{i} n t} \, \mathrm{d} t = 0 \, . $$ This implies \begin{align} \int \limits_0^{2\pi} \frac{\cos(t) - r}{1 - 2 r \cos(t) + r^2} \, \mathrm{d} t &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} r} \int \limits_0^{2\pi} \ln(1 - 2 r \cos(t) + r^2) \, \mathrm{d} t \\ &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} r} \int \limits_0^{2\pi} \ln[(1 -r \mathrm{e}^{\mathrm{i} t})(1 -r \mathrm{e}^{-\mathrm{i} t})] \, \mathrm{d} t \\ &= \frac{\mathrm{d}}{\mathrm{d} r} I(r) = \frac{\mathrm{d}}{\mathrm{d} r} 0 = 0 \end{align} as desired.
Solution 2:
(Since it is not mentioned yet) The function
$$
P_r(t) = \sum_{n=-\infty}^\infty r^{|n|} e^{int} =\frac{1-r^2}{1-2r\cos t +t^2},\quad0\le r<1, 0\le t\le 2\pi
$$ is called the Poisson kernel. For any continuous function $f:[0,2\pi]\to \Bbb C$ the Poisson integral
$$
u(r,\theta) = \frac{1}{2\pi i}\int_0^{2\pi} f(t)P_r(\theta-t)\mathrm{d}t\tag{*}
$$ gives the unique solution of the Dirichlet problem $\triangle u =0, \lim\limits_{r\to 1^-}u(r,\theta)=f(\theta)$.
Let $u(r,\theta)=r\cos \theta$. Since it is a real part of the analytic function $z=r e^{i\theta}$, we find that $$\triangle u=0,\ \ \ \lim\limits_{r\to 1^-}u(r,\theta)=\cos \theta.$$ By $\text{(*)}$ it follows that $$ r\cos \theta =\frac{1}{2\pi}\int_0^{2\pi} \frac{\cos t(1-r^2)}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t $$ for all $0\le r<1$ and $0\le \theta\le 2\pi$. Using the fact $$ \frac{1}{2\pi}\int_0^{2\pi} \frac{1-r^2}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t=\frac{1}{2\pi}\sum_{n=-\infty}^\infty r^{|n|} \int_0^{2\pi}e^{in(t-\theta)}\mathrm{d}t =1, $$ it follows $$ \frac{1}{2\pi}\int_0^{2\pi} \frac{\left(\cos t-r\cos \theta\right)(1-r^2)}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t =0. $$ By letting $\theta =0$, we obtain $$ \frac{1-r^2}{2\pi }\int_0^{2\pi} \frac{\cos t-r}{1-2r\cos t+r^2}\mathrm{d}t =0,$$ or equivalently $$ \int_0^{2\pi} \frac{\cos t-r}{1-2r\cos t+r^2}\mathrm{d}t =0.$$
Solution 3:
It seems to me that this result is a particular case (or rather 1D adaptation) of Newton's shell theorem. In 2D, the theorem states that no net gravitational force is perceived by objects located inside a uniform hollow sphere.
Here, consider the integral $$ \vec I(M) = \int_{X\in C}\frac{\vec{MX}}{MX^2}\mathrm ds $$ where $M$ is a point at a distance $r<1$ away from the center $O$ of the unit circle $C$, and $\mathrm ds$ is a length element of the circle. What you are asking for is a component of $\vec I$, the one directed along $\vec{OM}$. In fact, we have $\vec I=\vec 0$. There is a nice geometric proof of this in 2D due to Newton; see here. You can try to adapt it to 1D.
Solution 4:
With $\gamma$ being the counter-clockwise unit circle and $\bar\gamma$ being the clockwise unit circle,
$$
\begin{align}
\int_0^{2\pi}\frac{r-\cos(t)}{1-2r\cos(t)+r^2}\,\mathrm{d}t
&=\int_0^{2\pi}\frac12\left(\frac1{r-e^{it}}+\frac1{r-e^{-it}}\right)\mathrm{d}t\tag1\\
&=\frac1{2i}\int_\gamma\frac{\,\mathrm{d}z}{z(r-z)}-\frac1{2i}\int_{\bar\gamma}\frac{\,\mathrm{d}z}{z(r-z)}\tag2\\
&=\frac1i\int_\gamma\frac{\,\mathrm{d}z}{z(r-z)}\tag3\\
&=\frac1{ir}\int_\gamma\left(\frac1z-\frac1{z-r}\right)\mathrm{d}z\tag4\\
&=\left\{\begin{array}{}
0&\text{if }|r|\lt1\\
\frac{2\pi}r&\text{if }|r|\gt1
\end{array}\right.\tag5
\end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: $z=e^{it}$ in the left integral and $z=e^{-it}$ in the right integral
$(3)$: $\bar\gamma$ is in the opposite direction from $\gamma$
$(4)$: partial fractions
$(5)$: $2\pi i$ times the sum of the residues inside $\gamma$