If g(x)=∫sinxsin(2x)sin-1(t)dt, then

g'x=sin−1sin2x· 2cos2x−sin−1sinx·cosx   by Newton's Leibnitz rule=4xcos2x−xcosx⇒g'π2=2π.−1−π2.0=−2πand g'−π2=−2π−1=2π