1021. Find the Maclaurin series for the function $\arcsin x$, starting from its representation as an integral: $\arcsin x=\int_{0}^{x} \frac{d t}{\sqrt{1-t^{2}}}$.
Solution. As in the solution of the previous problem, we transform the integrand function $\frac{1}{\sqrt{1-t^{2}}}=\left(1-t^{2}\right)^{-\frac{1}{2}}$, expand it into a binomial series (B), setting $x=-t^{2}, m=-\frac{1}{2}$ :

$$
\begin{aligned}
\left(1-t^{2}\right)^{-\frac{1}{2}} & =1+\frac{1}{2} t^{2}+\frac{1 \cdot 3}{2 \cdot 4} t^{4}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} t^{6}+\ldots+ \\
& +\frac{1 \cdot 3 \ldots(2 n-1)}{2 \cdot 4 \ldots 2 n} t^{2 n}+\ldots
\end{aligned}
$$

ı integrating from 0 to $x$, we obtain the desired series

$$
\begin{aligned}
\arcsin x & =x+\frac{1}{2} \frac{x^{3}}{3}+\frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{x^{7}}{7}+\ldots+ \\
& +\frac{1 \cdot 3 \ldots(2 n-1)}{2 \cdot 4 \ldots 2 n} \frac{x^{2 n+1}}{2 n+1}+\ldots
\end{aligned}
$$

which converges to $\arcsin x$ for $|x| \leq 1$.