xsinx/(1 + e^x)在[-π/2,π/2]上的定积分

1个回答

  • ∫[xsinx/(1+e^x)]dx=∫[xsinx/(1+e^x)]dx+∫[xsinx/(1+e^x)]dx (分成两个积分)

    =-∫[xsinx/(1+1/e^x)]dx+∫[xsinx/(1+e^x)]dx (第一个积分用-x代换x)

    =∫[xsinx/(1+1/e^x)]dx+∫[xsinx/(1+e^x)]dx

    =∫[1/(1+1/e^x)+1/(1+e^x)]xsinxdx

    =∫[e^x/(e^x+1)+1/(1+e^x)]xsinxdx

    =∫[(e^x+1)/(1+e^x)]xsinxdx

    =∫xsinxdx

    =∫xd(-cosx)

    =(-xcosx)│+∫cosxdx (应用分部积分法)

    =-(π/2)cos(π/2)+0*cos0+(sinx)│

    =sin(π/2)-sin0

    =1.