How do you integrate #(1-tanx)/(1+tanx) dx#?

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How do you integrate #(1-tanx)/(1+tanx) dx#?

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Answer 1 · 2016-12-05T07:08:46

Rewrite as sines and cosines.

Explanation:

#(1-tan x)/(1+tan x)=(1-(sin x)/(cos x ))/(1+(sin x)/(cos x)) = (cos x - sin x )/(cos x + sin x ) = (cos x cos (pi/4)-sin x sin (pi/4))/(cos x cos (pi/4) + sin x sin (pi/4) )=cot(x+pi/4)#
Note that #sin(pi/4)=cos(pi/4)=1/sqrt 2#

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How do you integrate #(1-tanx)/(1+tanx) dx#?

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