答：原积分=1/2∫x^2sin2x dx=-1/4∫x^2 d(cos2x)=-1/4x^2cos2x+1/4∫cos2x d(x^2)=-1/4x^2cos2x+1/2∫xcos2x dx=-1/4x^2cos2x+1/4∫x d(sin2x)=-1/4x^2cos2x+1/4xsin2x-1/4∫sin2x dx=-1/4x^2cos2x+1/4xsin2x+1/8co...