<s>[INST] Simplifying $\cos(2\arcsin(x))$ using only pythagorean trigonometric identity I know that one can simplify $\cos(2\arcsin(x))$ using $\cos(a+b)=\cos(a)\cdot\cos(b)-\sin(a)\cdot\sin(b)$:
\begin{alignat}{1}
\cos(2\arcsin(x))&=\cos^2(\arcsin(x))-\sin^2(\arcsin(x))
\\&=1-2\sin^2(\arcsin(x))
\\&=1-2x^2
\end{alignat}

I tried to make this simplification using only $\sin^2(x)+\cos^2(x)=1$:
\begin{alignat}{1}
\cos^2(2\arcsin(x))&=1-\sin^2(2\arcsin(x))
\\ \left|\cos(2\arcsin(x))\right|&=\sqrt{1-\sin^2(2\arcsin(x))}
\\ &=\sqrt{1-\left(2x\sqrt{1-x^2}\right)^2}
\\ &=\sqrt{1-4x^2 |1-x^2|}
\\ &=\sqrt{1-4x^2(1-x^2)}
\\ &=\sqrt{1-4x^2+4x^4}
\\ &=\sqrt{\left(2x^2-1\right)^2}
\\ &=|2x^2-1|
\end{alignat}
And then I could not figure out how to proceed. So, how to get rid of $|\cdot|$ in:
$$\left|\cos(2\arcsin(x))\right|=|2x^2-1|$$
and get $1-2x^2$ ?

Note: this is a part of my attempt to solve the integral $\int x^2\cdot\sqrt{1-x^2}\,\,\mathrm dx$ by trigonometric substitution.
 [/INST] $\cos(2\arcsin x)$ will be $\ge0,$ if
$-\dfrac\pi2\le2\arcsin x\le\dfrac\pi2$
$\iff -\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$
In that case $$|2x^2-1|=-(2x^2-1)$$
Check if $x>\dfrac1{\sqrt2}$ 
or $x<-\dfrac1{\sqrt2}$
</s>