Taylor expansion of $\arccos(1-x)$ around $x=0$ to two terms

Let $y=f(x)=\arccos(1-x)$. Then $$1-x=\cos y=1-\frac{y^2}2+\frac{y^4}{24}+O(y^6),$$ so $$x=\frac{y^2}2-\frac{y^4}{24}+O(y^6).$$ Now clearly there is no actual Taylor series for $y$ about $x=0$ because $f'(0)$ does not exist. However, a generalized power series solution can be written down, known variously as the Frobenius method or the asymptotic expansion of $y=f(x)$ near $x=0$. Solving this equation formally:

$$2x=y^2\left(1-\frac{y^2}{12}+O(y^4)\right)\Rightarrow y=\sqrt{2x}\left(1-\frac{y^2}{12}+O(y^4)\right)^{-1/2}$$

Since $0<y\ll1$, $\frac{y^2}{12}\ll1$ so that we can use the binomial theorem $(1+x)^p=1+px+\cdots$ to get the next-leading order term:

$$y=\sqrt{2x}+\sqrt{2x}\frac{y^2}{24}+O(y^4)=\sqrt{2x}+\frac{\sqrt{2x}}{24}\left(\sqrt{2x}+\frac{\sqrt{2x}}{24}y^2+O(y^4)\right)^2+O(y^4)$$ $$=\sqrt{2x}+\frac{(2x)^{3/2}}{24}\left(1+\frac{1}{12}y^2+O(y^4)\right)+O(y^4)=\sqrt{2x}+\frac{(2x)^{3/2}}{24}+\frac{(2x)^{3/2}}{24\cdot 12}y^2+O(x^{3/2}y^4)+O(y^4)$$

Now, since $y=O(\sqrt x)$ (which follows from the leading order term), we can simplify all that to get $$y=\sqrt{2x}+\frac{(2x)^{3/2}}{24}+O(x^2).$$

Start from the derivative of the function

$$\frac{1}{\sqrt{x}\sqrt{2-x}}= \frac{1}{\sqrt{2x}}\frac{1}{\sqrt{1-x/2}} = \frac{1}{\sqrt{2x}}(1+\frac{1}{4}x+\frac{3}{32}x^2+\dots )=\dots\,. $$

Now, integrate the above series to get your Taylor series.