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?

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?
