Example 2. Expand the function $f(x)=\arcsin x$ into a Maclaurin series.
Solution. Substitution in (4) $\alpha=-\frac{1}{2}$ and replacement of $x$ with $-x^{2}$ leads to the expansion

$$
\begin{aligned}
& \frac{1}{\sqrt{1-x^{2}}}=1+\frac{1}{2} x^{2}+\frac{1 \cdot 3}{2^{2} \cdot 2!} x^{4}+\frac{1 \cdot 3 \cdot 5}{2^{3} \cdot 3!} x^{6}+\ldots+ \\
& \quad+\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1)}{2^{n} \cdot n!} x^{2 n}+\ldots, \quad x \in(-1 ; 1)
\end{aligned}
$$

Since $\quad(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}}, \quad$ taking into account Theorem 17, we can integrate the obtained series: $\arcsin x=\int_{0}^{x} \frac{d t}{\sqrt{1-t^{2}}}, x \in(-1 ; 1)$. We obtain: $\quad \arcsin x=x+\frac{x^{3}}{2 \cdot 3}+\frac{1 \cdot 3}{2^{2} \cdot 2!} \cdot \frac{x^{5}}{5}+\frac{1 \cdot 3 \cdot 5}{2^{3} \cdot 3!} \cdot \frac{x^{7}}{7}+\ldots+$ $+\frac{1 \cdot 3 \cdot 5 \cdot \ldots \cdot(2 n-1)}{2^{n} \cdot n!} \cdot \frac{x^{2 n+1}}{2 n+1}+\ldots, x \in(-1 ; 1)$.