<s>[INST] Prove that $2\sin^{-1}\sqrt x - \sin^{-1}(2x-1) = \frac{\pi}{2}$. Prove that $2\sin^{-1}\sqrt x - \sin^{-1}(2x-1) = \dfrac\pi2$.
Do you integrate or differentiate to prove this equality? If so, why?
 [/INST] I thought it might be instructive to present a way forward without the use of calculus, but rather on the use of standard trigonometric identities.  
Proceeding, we let $f(x,y) =2\arcsin(x)-\arcsin(y)$ and note that
$$\begin{align}
\sin(f(x,y))&=\sin(2\arcsin(x))\cos(\arcsin(y))-\sin(\arcsin(y))\cos(2\arcsin(x))\\\\
&=2x\sqrt{1-x^2}\sqrt{1-y^2}-y(1-2x^2) \tag 1
\end{align}$$
Substituting $x\to \sqrt{x}$ and $y\to 2x-1$ into $(1)$ reveals 
$$\begin{align}
\sin\left(2\arcsin(\sqrt{x})+\arcsin(2x+1)\right)&=2\sqrt{x}\sqrt{1-x}2\sqrt{x}\sqrt{1-x}-(2x-1)(1-2x)\\\\
&=1
\end{align}$$
Therefore, $2\arcsin(\sqrt{x})-\arcsin(2x-1)=\pi/2 +2n\pi$.  Inasmuch as the arcsine is bounded in absolute value by $\pi/2$, then we conclude immediately that $n=0$ and 
$$\bbox[5px,border:2px solid #C0A000]{2\arcsin(\sqrt{x})-\arcsin(2x-1)=\pi/2 }$$
</s>