<s>[INST] What is the expansion in power series of ${x \over \sin x}$ How can I expand in power series the following function:
$$
{x \over \sin x}
$$
? I know that:
$$
\sin x = x - {x^3 \over 3!} + {x^5 \over 5!} - \ldots,
$$
but a direct substitution does not give me a hint about how to continue.
 [/INST] You can have  the expansion to any order you please using  polynomial division along increasing powers. Finding a general formula for the coefficients is another problem.
I'll show how to obtain the development at order $6$, for instance: start from the development of $\sin x$ at order $7$ (this is because there will be a simplifacation by $x$):
$$\frac x{\sin x}=\frac x{x-\dfrac{x^3}6+\dfrac{x^5}{120}-\dfrac{x^7}{5040}+o(x^8)}=\frac 1{1-\dfrac{x^2}6+\dfrac{x^4}{120}-\dfrac{x^6}{5040}+o(x^7)}$$
Division along increasing powers up to order $6$:
$$\begin{array}{rcr@{}r@{}r@{}l@{\,}}
           & & 1 + &\!\!\!\! \dfrac{x^2}{6} +  & \hskip-12mu\dfrac{7x^4}{360}+ & \hskip-12mu\dfrac{31x^6}{15120}\\[-18mu]
                     1-\dfrac{x^2}6+\dfrac{x^4}{120}-\dfrac{x^6}{5040} & \biggl( & 1\hphantom{{}+{}}\\[-18mu]
         & & -1 + &\hskip-12mu\dfrac{x^2}6-&\hskip-12mu \dfrac{x^4}{120} + & \hskip-12mu\dfrac{x^6}{5040} \\
  & & &\hskip-12mu \dfrac{x^2}6-&\hskip-12mu \dfrac{x^4}{120} + &\hskip-12mu \dfrac{x^6}{5040} \\
 & &- &\hskip-12mu\dfrac{x^2}6 +  &\hskip-12mu \dfrac{x^4}{36} - &\hskip-12mu \dfrac{x^6}{720} \\
 & & & & \hskip-12mu\dfrac{7x^4}{360} -&\hskip-12mu \dfrac{x^6}{840} \\
 & & &- & \hskip-12mu\dfrac{7x^4}{360} + & \hskip-12mu\dfrac{7x^6}{6\cdot360} \\
 & & & & &\hskip-12mu \dfrac{31x^6}{15120}
       \end{array}$$
Thus we've found that 
$$\frac x{\sin x}=1 + \dfrac{x^2}{6} + \dfrac{7x^4}{360}+\dfrac{31x^6}{15120}+o(x^7)$$
</s>