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?
 A: The function $f$ given by
$$f(x)=2\arcsin \sqrt x-\arcsin(2x-1)$$ is constant. Try to prove this by differentiating the function $f$. To find the constant, calculate $f(x)$ for some nice $x$. 
A: 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 
$$\bbox[5px,border:2px solid #C0A000]{2\arcsin(\sqrt{x})-\arcsin(2x-1)=\pi/2 }$$
A: alternative to differentiating, let $$\phi=2\sin^{-1}\sqrt{x}$$
$$\implies x=\sin^2(\phi/2)=\frac 12(1-\cos \phi)$$
$$\implies \cos\phi=1-2x$$
$$\implies\phi=\cos^{-1}(1-2x)=\frac{\pi}{2}-\sin^{-1}(1-2x)=\frac{\pi}{2}+\sin^{-1}(2x-1)$$
A: Let $\arcsin y=u\implies y=\sin u$
Using the definition of principal values $-\dfrac\pi2\le u\le\dfrac\pi2$
Now $\arcsin(2y^2-1)=\arcsin(-\cos2u)=-\arcsin(\cos2u)$ as $\arcsin(-v)=-\arcsin v$
and we know $\arcsin(\cos2u)=\dfrac\pi2-\arccos(\cos2u)$
finally $\arccos(\cos2u)=\begin{cases} 2u &\mbox{if } 0\le2u\le\pi\iff0\le u\le\dfrac\pi2 \\-2u & \mbox{if } -\pi\le2u<0 \end{cases}$
Here $y=\sqrt x\ge0\implies0\le u\le\dfrac\pi2$
Can you take it from here?
