Well, I want to solve:
$$\int_0^x\frac{1}{\left(1+\epsilon\cos\left(t\right)\right)^2}\space\text{d}t\tag1$$
Using a Taylor series.
I know that, the taylor series looks like (series expansion at $t=0$):
$$\frac{1}{\left(1+\epsilon\cos\left(t\right)\right)^2}=\frac{1}{\left(1+\epsilon\right)^2}+\frac{\epsilon t^2}{\left(1+\epsilon\right)^3}+\dots\tag2$$
But that series has to be convergent for a value that is at least equal to $x$, to be able to solve the integral using that series.
But how can find if the series is convergent for a value that is al least equal to $x$, or should I pick another value to find the expansion at (now I've $t=0$ but what is the consequence, for the convergence, when I choose $t=1$ or $t=\frac{\pi}{2}$)? For $\epsilon\approx0.0167086$
We know that
$$\frac1{(1-r)^2}=1+2r+3r^2+4r^3+\dots=\sum_{k=0}^\infty(k+1)r^k$$
converges for $|r|<1$, and since $|\epsilon\cos(t)|<1$ for $|\epsilon|<1$, we have
$$\frac1{(1+\epsilon\cos(t))^2}=1-2\epsilon\cos(t)+3\epsilon^2\cos^2(t)-\dots=\sum_{k=0}^\infty(k+1)\epsilon^k\cos^k(t)$$
Thus,
$$\int_0^x\frac1{(1+\epsilon\cos(t))^2}\ dt=\sum_{k=0}^\infty(k+1)\epsilon^k\int_0^x\cos^k(t)\ dt$$
In terms of the hypergeometric function,
$$\int_0^x\cos^k(t)\ dt=\frac{-x|\sin(x)|\cos^{k+1}(x)\csc(x)_2F_1\left(\frac12,\frac{1+k}2;\frac{3+k}2;\cos^2(x)\right)}{1+k}$$
At least according to WolframAlpha.