integrate 1+tanx /1-tanx
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both answers are correct
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(1 + tanx) / (1 - tanx) = tan(x + π/4)
Int [(1 + tanx) / ( 1 + tanx) dx
= Int tan(x + π/4) dx
= log sec(x + π/4)
Int [(1 + tanx) / ( 1 + tanx) dx
= Int tan(x + π/4) dx
= log sec(x + π/4)
Answered by
14
see the attachment for the solution...
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