f (x) = ∫[a sin(ln x) + b cos(ln x)]dx

1个回答

  • 设lnx=y则x=e^y

    asin(lnx)dx积分=asinyd(e^y)=asiny*e^y-ae^yd(siny)=asiny*e^y-acosyd(e^y)

    asiny*e^y-acosy*e^y+ae^yd(cosy)=a*e^y(siny-cosy)-asiny*e^ydy

    2asiny*d(e^y)=a*e^y(siny-cosy)

    asiny=a*e^y(siny-cosy)/2

    asin(lnx)dx积分=a*x{sin(lnx)-cos(lnx)}/2

    同理bcos(lnx)dx积分=b*x{sin(lnx)+cos(lnx)}/2

    f(x)=[ax{sin(lnx)-cos(lnx)}+bx{sin(lnx)+cos(lnx)}]/2