Domain of $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$ How do you find the domain of the function $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$
I know that the domain of $\arcsin$ function is $[-1,1]$
So, $-1\le{2x\sqrt{1-x^{2}}}\le1$ probably?
or maybe $0\le{2x\sqrt{1-x^{2}}}\le1$ , since $\sqrt{1-x^{2}}\ge0$  ? 
EDIT: So many people have answered that the domain would be $[-1,1]$ but my book says that its $[\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}].$ 
Can anyone explain how are those restrictions made in the given formulas? 

A proof without calculus or trigonometry.
$$\begin{aligned}
(1-2x^2)^2&\geq0
\\
(1-4x^2+4x^4)&\geq0
\\
x^2-x^4&\leq\frac{1}{4}
\\
0\leq \sqrt{x^2({1-x^2})}&\leq\frac{1}{2}
\\
0\leq |x\sqrt{1-x^2}|&\leq\frac{1}{2}
\\
-1\leq 2x\sqrt{1-x^2}&\leq1
\end{aligned}$$
Then, since $\arcsin:\,[-1,1]\mapsto\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$, the domain is $[-1,1]$.
