Integration problem! Help! (1 Viewer)

EpikHigh · Feb 27, 2013

Hey guys, can someone help me solve this question with a solution please.

Integrate (1-tanX)/(1+tanX)

Nws m8 · Feb 27, 2013

Trebla · Feb 27, 2013

EpikHigh said:
Hey guys, can someone help me solve this question with a solution please.

Integrate (1-tanX)/(1+tanX)

Alternatively, note that

$\tan\left(\dfrac{\pi}{4}-x\right) = \dfrac{\tan\frac{\pi}{4}-\tan x}{1 + \tan\frac{\pi}{4}\,\tan x} = \dfrac{1 - \tan x}{1 + \tan x}\\\\\therefore \displaystyle\int \dfrac{1 - \tan x}{1 + \tan x}\,dx = \displaystyle\int \tan\left(\dfrac{\pi}{4}-x\right)\,dx = \ln\left[\cos\left(\dfrac{\pi}{4}-x\right) \right] + c$

EpikHigh · Feb 27, 2013

Nws m8 said:

Cheers man, wow I was stupid :L Thanks again.

Nws m8 · Feb 27, 2013

EpikHigh said:
Cheers man, wow I was stupid :L Thanks again.

Nws m8

hahaha

iSplicer · Feb 28, 2013

Trebla said:
Alternatively, note that

$\tan\left(\dfrac{\pi}{4}-x\right) = \dfrac{\tan\frac{\pi}{4}-\tan x}{1 + \tan\frac{\pi}{4}\,\tan x} = \dfrac{1 - \tan x}{1 + \tan x}\\\\\therefore \displaystyle\int \dfrac{1 - \tan x}{1 + \tan x}\,dx = \displaystyle\int \tan\left(\dfrac{\pi}{4}-x\right)\,dx = \ln\left[\cos\left(\dfrac{\pi}{4}-x\right) \right] + c$

Elegant! Although I think in 3U you should expand the final line to get rid of the composite angle.

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Integration problem! Help! (1 Viewer)

EpikHigh

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