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.

Integration problem! Help! (1 Viewer)

EpikHigh

Arizona Tears

Nws m8

Banned

Attachments

Trebla

Administrator

EpikHigh

Arizona Tears

Nws m8

Banned

iSplicer

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)