Expand in Taylor series $\frac{1}{1-\sin{x}}$ Expand in Taylor series $\frac{1}{1-\sin{x}}$
I have an idea that $\frac{1}{1-\sin{x}} = 1 + \sin {x} + \sin^2 {x} + \sin^3 {x} + \dots$
But I don't know what to do next. Every sine expands in infinity series...
Can anybody help?

If you're looking for a closed form, see OEIS sequence A099612.
The coefficient of $x^n$ is 
$$ \dfrac{(\cos(\pi n/2) - \sin(\pi n/2))\; (4^{n+2}-2^{n+2}) }{n!}
\left( \dfrac{\zeta(-n-1,3/4) - \zeta(-n-1,1/4)}{2^{-n-1}-2} - \zeta(-n-1)\right)$$
where the two-argument $\zeta$ is the Hurwitz zeta function.
But if this is a homework exercise, it's likely just asking you to compute the first few terms.
