How do you integrate $\frac{1 - tan x}{1 + tan x} d x$ $(1-tanx)/(1+tanx) dx$ ?

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How do you integrate $\frac{1 - tan x}{1 + tan x} d x$ $(1-tanx)/(1+tanx) dx$ ?

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Answer 1 · 2016-12-05T07:08:46

Rewrite as sines and cosines.

Explanation:

$\frac{1 - tan x}{1 + tan x} = \frac{1 - \frac{sin x}{cos x}}{1 + \frac{sin x}{cos x}} = \frac{cos x - sin x}{cos x + sin x} = \frac{cos x cos (\frac{π}{4}) - sin x sin (\frac{π}{4})}{cos x cos (\frac{π}{4}) + sin x sin (\frac{π}{4})} = cot (x + \frac{π}{4})$ $(1-tan x)/(1+tan x)=(1-(sin x)/(cos x ))/(1+(sin x)/(cos x)) = (cos x - sin x )/(cos x + sin x ) = (cos x cos (pi/4)-sin x sin (pi/4))/(cos x cos (pi/4) + sin x sin (pi/4) )=cot(x+pi/4)$
Note that $sin (\frac{π}{4}) = cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $sin(pi/4)=cos(pi/4)=1/sqrt 2$

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Explanation:

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