求定积分∫[-11]x2(sin2x+x2)dx,迷惑中,

2个回答

  • ∫x^2(six2x+x^2)dx=[∫x^2sin2xdx+x^4dx]|[-1,1].

    =[(1/3)∫sin2xd(x^3)+∫x^4dx]|[-1,1]

    ={[(1/3)sin2x*x^3|[-1,1]-∫x^3(1/2)d(-cos2x)+(1/5)∫d(x^5)}|[-1,1].

    ={[(1/3)sin2x*x^3|[-1,1]+(1/2)cos2x*x^3|[-1,1]+(1/5)x^5|[-1,1].

    =(1/3)sin2*1*(1)^3-(1/3)sin2*(-1)*(-1)^3+(1/2)cos2*1*(1^3)-(1/2)cos2*(-1)*(-1^3)+(1/5)[1^5-(-1^5)]

    =(1/3)sin2-1/3sin2+(1/2)cos2+(1/2)cos2+2/5.

    =2/5+cos2.