作业帮请问1/（1+tanx）的不定积分怎么求?
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请问1/（1+tanx）的不定积分怎么求?
请问1/（1+tanx）的不定积分怎么求?
请问1/（1+tanx）的不定积分怎么求?
∫1/(1+tanx)dx
=∫1/(1+sinx/cosx)dx
=∫cosx/(cosx+sinx)dx
=∫cosx(cosx-sinx)/(cosx+sinx)(cosx-sinx)dx
=∫(cos²x-sinxcosx)/(cos²x-sin²x)dx
=[∫(1+cos2x-sin2x)/cos2xdx]/2
=[∫(1+cos2x-sin2x)/cos2xd2x]/4
=(∫sec2xd2x+∫d2x+∫tan2xd2x)/4
=ln|sec2x+tan2x|/4+x/2+ln|cos2x|/4+C
=x/2+ln|cos2x(sec2x+tan2x)|/4+C
=x/2+ln(1+sin2x)/4+C
令1+tanx=u x=arctan(u-1) dx=du/(1+(u-1)^2)
原式=∫du/u(u^2-2u+2)
=1/2*∫1/u-(u-2)/(u^2-2u+2)du
=1/2*ln|u|-1/2*∫(u-2)du/[(u-2)^2+2(u-2)+2]
令u-2=t
=1/2*ln|u|-1/2*∫tdt/(t^2+2t+2)
=...
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令1+tanx=u x=arctan(u-1) dx=du/(1+(u-1)^2)
原式=∫du/u(u^2-2u+2)
=1/2*∫1/u-(u-2)/(u^2-2u+2)du
=1/2*ln|u|-1/2*∫(u-2)du/[(u-2)^2+2(u-2)+2]
令u-2=t
=1/2*ln|u|-1/2*∫tdt/(t^2+2t+2)
=1/2*ln|u|-1/2*[∫(t+1)dt/(t^2+2t+2)-∫dt/(t^2+2t+2)]
=1/2*ln|u|-1/4*∫d(t^2+2t+2)/(t^2+2t+2)+1/2*∫d(t+1)/[(t+1)^2+1]
=1/2*ln|u|-1/4*ln|(t^2+2t+2)|+1/2*arctan(t+1)+C
=1/2*ln|u|-1/4*ln|u^2-2u+2|+1/2*arctan(u-1)+C
=1/2*ln|1+tanx|-1/4*ln[1+(tanx)^2]+x/2+C
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