∫[0,2π][x(cosx)^2]dx

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/18 19:42:23
∫[0,2π][x(cosx)^2]dx∫[0,2π][x(cosx)^2]dx∫[0,2π][x(cosx)^2]dx∫[x(cosx)^2]dx=(1/2)∫xcos2xdx+(1/2)∫xdx=

∫[0,2π][x(cosx)^2]dx
∫[0,2π][x(cosx)^2]dx

∫[0,2π][x(cosx)^2]dx
∫[x(cosx)^2]dx
=(1/2)∫xcos2xdx+(1/2)∫xdx
=x^2/4+(1/4)∫xdsin2x
=x^2/4+(xsin2x)/4-(1/4)∫sin2xdx
=x^2/4+(xsin2x)/4+cos2x/8+c
定积分=π^2