∫[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
