∫cosx cos2xcos3xdx

Question

如题所述

Answer 1

【1】积化和差：2cosxcos3x=cos4x+cos2x4cosxcos2xcos3x=2cos2x[cos4x+cos2x]=cos6x+cos2x+cos4x+1【2】可设原式=y4y=∫[cos6x+cos4x+cos2x+1]dx=(1/6)∫cos6xd(6x)+(1/4)∫cos4xd(4x)+(1/2)∫cos2xd(2x)+∫dx=[(sin...