I was trying to show that for suitable t: $$ 2\pi(1+t/(\sqrt{(1-t)(3-t)})=\sum_{0}^{\infty}(t^n\int_0^{2\pi}1/(2-cos(\theta))^nd\theta $$
By uniqueness this is clearly the Taylor series about $0$ and the form of $c_n $ suggests the use of Cauchy's formula so we get $c_n=\frac{1}{2i\pi}\int_{\gamma(0,r)}\frac{2\pi(1+t/(\sqrt{(1-t)(3-t)})}{t^{n+1}}dt $ for $n\geq1$ $$ =-i\int_{\gamma(0,r)}\frac{1}{t^{n}\sqrt{(1-t)(3-t)}}dt $$ as $ \int_{\gamma(0,r)}\frac{1}{t^{n+1}}dt=0 \forall n\geq 1$
Curiously the substitution $t=2-cos(\theta)$ for $\theta\in [0,2\pi]$ gives $$ \frac{dt}{d\theta}=sin(\theta)=\sqrt{1-cos(\theta)^2}=\sqrt{-3+4t-t^2}=i\sqrt{(1-t)(3-t)} $$ $$ \therefore c_n=\int_0^{2\pi} \frac{1}{(2-cos(\theta))^n}d\theta $$ and so solves the problem, except for the small issue that it is not a valid substitution.
I was wondering if anyone has any insight into why this gives the correct answer and if there are other integrals where invalid substitutions can give the correct answer.
Here's an explanation for the invalid substitution.
First, we note that $$f(z)=\frac1{z^n\sqrt{(1-z)(3-z)}}$$ is a well-defined holomorphic function on $\mathbb C_\infty\setminus(\{0\}\cup[1,3])$ (choose a branch for $\sqrt{(1-z)(3-z)}$), where $\mathbb C_\infty$ is the Riemann sphere $\mathbb C\cup\{\infty\}$
For $0<r,\epsilon<1/4$, as you did, you can draw a circle $C_r(0)$ around $0$. Moreover, you can picture a contour $\Gamma$ around $[1,3]$ such that for each point $z\in\Gamma$, the minimal distance between $z$ and $[1,3]$ is $\epsilon$. You can easily see that $\Gamma$ is made up of two straight line segments and two half-circles.
By an appropriate choice of orientation, $\int_{C_r(0)}f(z)dz=\int_\Gamma f(z)dz$. It's because that $f$ is holomorphic. (It's important that $f$ is holomorphic at $\infty$. If you don't want to utilize this term, you can draw a large $C_R(0)$ and get $\int_{C_r(0)}-\int_\Gamma=\int_{C_R(0)}$, and let $R\to\infty$)
Now let $\epsilon\to0^+$ and $\int_\Gamma$ tends to the value you calculated by the invalid substitution.