If g x = ∫ sin x sin 2 x sin - 1 t d t , then

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

D.w.r.to x both side $g\left(x\right)= \underset{\mathrm{sin }x}{\overset{\mathrm{sin }2x}{\int }}{\sin }^{-1}t dt \Rightarrow g^{\text{'}}\left(x\right)=2{\sin }^{-1}\left(\mathrm{sin }2x\right)\times \mathrm{cos }2x-{\sin }^{-1}\left(\mathrm{sin }x\right)\mathrm{ cos }x$

$\Rightarrow g^{\text{'}}\left(\frac{\pi }{2}\right)=0 \mathrm{a}\mathrm{n}\mathrm{d} g^{\text{'}}\left(-\frac{\pi }{2}\right)=0$