∫(cosx)^2/(cosx-sinx)dx

1个回答

  • 2(cosx)^2-1=cos(2x)=(cosx)^2-(sinx)^2

    cos(x)^2=[cos(2x)+1]/2

    ∫(cosx)^2/(cosx-sinx)dx

    =∫[cos(2x)+1]/[2(cosx-sinx)]dx

    =∫cos(2x)/[2(cosx-sinx)]dx+∫1/[2(cosx-sinx)]dx

    =∫[(cosx)^2-(sinx)^2]/[2(cosx-sinx)]dx+∫1/[2(cosx-sinx)]dx

    =0.5∫(cosx+sinx)dx-sqr(0.5)∫1/sin(x-pi/4)dx

    (下面用万能公式sinA=(2tg(A/2))/(1+(tgA/2)^2)

    =0.5sinx-0.5cosx-sqr(1/2)∫(1/2tg((x-pi/4)/2))(1+tg(x/2-pi/8)^2)dx*2

    =0.5sinx-0.5cosx-sqr(1/2)∫1/{tg[(x-pi/4)/2]}d(tg(x-pi/4)/2))

    =0.5sinx-0.5cosx-sqr(1/2)ln(tg(x/2-pi/8))+C

    sqr(A)是根号A.