$$\int \sum_{n=0}^\infty \frac{(-t^2)^n}{n!} dt$$
$$= \sum_{n=0}^\infty \int \frac{(-t^2)^n}{n!} dt$$
$$= \sum_{n=0}^\infty \int \frac{(-1)^n}{n!} t^{2n}dt$$
$$= \sum_{n=0}^\infty \left(\frac{(-1)^n}{n!} \int t^{2n}dt\right)$$
$$= \sum_{n=0}^\infty \left(\frac{(-1)^n}{n!} \frac{t^{2n+1}}{2n+1}\right)$$
Let $f(x)=\arcsin(1-x)$ for $x\in[0,2]$.
Since the derivative of $f(x)=O\left( x^{-1/2}\right)$ for $x\sim 0$, we let $t=x^{1/2}$ and $g(t)=\arcsin(1-t^2)$.
We will now develop the first few terms of the Taylor series for $g(t)$ around $t=0$.
We have for the first derivative $g^{(1)}(t)$
$$\begin{align}
g^{(1)}(t)&=-\frac{2t}{\sqrt{1-(1-t^2)^2}}\\\\
&=-\frac{2}{\sqrt{2-t^2}}\tag 1
\end{align}$$
Differentiating the right-hand side of $(1)$, we obtain the second derivative, $g^{(2)}(t)$
$$\begin{align}
g^{(2)}(t)&=-\frac{2t}{(2-t^2)^{3/2}}\tag 2
\end{align}$$
Continuing, we have for $g^{(3)}(t)$
$$\begin{align}
g^{(3)}(t)&=-\frac{4(t^2+1)}{(2-t^2)^{5/2}}\tag 3
\end{align}$$
And finally, we have for $g^{(4)}(t)$
$$\begin{align}
g^{(4)}(t)&=-\frac{12t(t^2+3)}{(2-t^2)^{7/2}}\tag 4
\end{align}$$
We evaluate $(1)-(4)$ at $t=0$ and form the expansion
$$\bbox[5px,border:2px solid #C0A000]{\arcsin(1-x)=\frac{\pi}{2}-\sqrt{2}x^{1/2}-\frac{\sqrt{2}}{12}x^{3/2}+O\left(x^{5/2}\right)}$$
Best Answer
Hint:
Set $t=1+u\enspace(u\to 0)\,$ first. The integrand becomes $$\frac{\mathrm e^t-\mathrm e}{t-1}=\mathrm e\,\frac{\mathrm e^u-1}u.$$ Can you take it from here?