If $g(x)=\int_{\sin x}^{\sin 2 x} \sin ^{-1}(t) d t$, then\

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Professor PC Thomas Classes · Answer

$\left({ }^{*}ight): g(x)=\int_{\sin x}^{\sin 2 x} \sin ^{-1}(t) d t$\  We have, $g^{\prime}(x)=2 \sin ^{-1}(\sin 2 x) \cos 2 x-\sin ^{-1}(\sin x) \cos x$\ $g^{\prime}(-\pi / 2)=2 \sin ^{-1}(-\sin \pi) \cos (-\pi)-\sin ^{-1}(\sin (-\pi / 2)) \cos (-\pi / 2)=0$\ $g^{\prime}\left(\frac{\pi}{2}ight)=2 \cos \pi \sin ^{-1}(\sin \pi) -\cos \frac{\pi}{2} \sin ^{-1}\left(\sin \frac{\pi}{2}ight)=0 $\ None of the choices match.