If g(x) = ∫ _ x ^ 2x ^ - 1 tdt, then

Given,$g\left(x\right) = {∫}_{\sin x}^{\sin 2x}\sin^{- 1}\left(t\right)dt . . . . . . . . ( i )$

D.w.r.to x both side $g\left(x\right) = \mathop{∫}_{sin x}^{\sin 2x}\sin^{- 1} tdt ⇒ {g}^{'}\left(x\right) = 2\sin^{- 1} \left(\sin 2x\right) \times \cos 2x - \sin^{- 1} \left(sin x\right)cos x$

$⇒ {g}^{'}\left(\dfrac{\pi}{2}\right) = 0and{g}^{'}\left(- \dfrac{\pi}{2}\right) = 0$