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?

alternative to differentiating, let $$\phi=2\sin^{-1}\sqrt{x}$$
$$\implies x=\sin^2(\phi/2)=\frac 12(1-\cos \phi)$$
$$\implies \cos\phi=1-2x$$
$$\implies\phi=\cos^{-1}(1-2x)=\frac{\pi}{2}-\sin^{-1}(1-2x)=\frac{\pi}{2}+\sin^{-1}(2x-1)$$
