∫xcos2xdx用分部积分怎么解

Question

如题所述

Answer 1

∫ xcos2x dx = (1/2)∫ xcos2x d(2x) = (1/2)∫ x d(sin2x)
= (x/2)sin2x - (1/2)∫ sin2x dx
= (x/2)sin2x - (1/2)(1/2)∫ sin2x d(2x)
= (x/2)sin2x + (1/4)cos2x + C本回答被提问者采纳