Hello !
- The Maclaurin serie for \text{arctan}(t) is
\text{arctan}(t) = \sum_{n=0}^{\infty} (-1)^n \frac{t^{2n+1}}{2n+1}.
Proof. \text{arctan}'(t) = \frac{1}{1+t^2} = (1 - (-t^2))^{-1} and we know that (1-X)^{-1} = \sum_{n=0}^\infty X^n (for |X|<1). Actually,
\text{arctan}'(t) = \sum_{n=0}^\infty (-t^2)^n = \sum_{n=0}^\infty (-1)^n t^{2n}
After integration, you get
\text{arctan}(t)-\text{arctan}(0) = \sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{2n+1}
The proof is ended because \text{arctan}(0)=0.
- Because x^3\to 0 when x\to 0, you can compose and write
\text{arctan}(x^3) = \sum_{n=0}^{\infty} (-1)^n \frac{(x^3)^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+3}}{2n+1}
If you don't like \sum symbol,
\text{arctan}(x^3) = x^3 - \frac{x^9}{3} + \frac{x^15}{5} - \frac{x^{21}}{7} + \ldots
