1022. By expanding the integrand into a Maclaurin series and integrating it term by term, find the series expansions of the following integrals:
1) $\int \sin x^{2} d x$; 2) $\int \sqrt{x} e^{x} d x$; 3) $\int \sqrt{1-x^{3}} d x$.
Solution. 1) Using the Maclaurin series for $\sin x$, replacing $x$ with $x^{2}$, we have

$$
\sin x^{2}=x^{2}-\frac{x^{8}}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+\ldots+(-1)^{n-1} \frac{x^{4 n-2}}{(2 n-1)!}+\ldots
$$

Integrating term by term, we obtain the desired expansion

$$
\int \sin x^{2} d x=\frac{x^{3}}{3}-\frac{x^{7}}{3 \cdot 7}+\frac{x^{11}}{5! \cdot 11}-\frac{x^{15}}{7! \cdot 15}+\ldots+C
$$

which is valid for any value of $x$.

2) Replacing the function $e^{x}$ with its Maclaurin series, multiplying it term by term by $\sqrt{x}$, and integrating, we get

$$
\begin{gathered}
\int \sqrt{x} e^{x} d x=\int x^{\frac{1}{2}}\left(1+\frac{x}{1!}+\frac{x^{2}}{2!}+\ldots+\frac{x^{n}}{n!}+\ldots\right) d x= \\
=\int\left(x^{\frac{1}{2}}+\frac{x^{\frac{3}{2}}}{1!}+\frac{x^{\frac{5}{2}}}{2!}+\ldots+\frac{x^{\frac{2 n+1}{2}}}{n!}+\ldots\right) d x= \\
=\frac{2}{3} x^{\frac{3}{2}}+\frac{2}{1 \cdot 5} x^{\frac{5}{2}}+\frac{2}{2! \cdot 7} x^{\frac{7}{2}}+\ldots+\frac{2}{n!(2 n+3)} x^{\frac{2 n+3}{2}}+\ldots+C
\end{gathered}
$$

The obtained series converges to the desired integral for $x \geqslant 0$.

3) Expanding the integrand function into a binomial series (B)

$$
\begin{aligned}
\sqrt{1-x^{3}}=\left(1-x^{3}\right)^{\frac{1}{2}}= & 1-\frac{1}{1!2} x^{3}-\frac{1}{2!2^{2}} x^{6}-\frac{1 \cdot 3}{3!2^{3}} x^{9}-\ldots \\
& -362
\end{aligned}
$$

and integrating term by term, we obtain the desired series

$$
\int \sqrt{1-x^{3}} d x=x-\frac{x^{4}}{1!2 \cdot 4}-\frac{x^{7}}{2!2^{2} \cdot 7}-\frac{1 \cdot 3 x^{10}}{3!2^{3} \cdot 10}-\ldots+C
$$

which converges for $|x|<1$.