∫(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.
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