∫[0,2π][x(cosx)^2]dx

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∫[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