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

Answer 1

Christopher Agresta

Just do some basic math first:

#int(1-cos(x))/(1+cos(x))dx = (-1-cos(x)+2)/(1+cos(x))dx#

We can now factor the numerator as follows:

#=>int(-(1+cos(x))+2)/(1+cos(x))dx#

#=>int(-(1+cos(x)))/(1+cos(x))dx+2int1/(1+cos(x))dx#

#=>int -1dx+2int1/(1+cos(x))dx#

Remember that #cos^2(x) = 1/2(1+cos(2x))#

(From #cos(a)cos(b)# formula with #a = b#)

#=>2cos^2(x) = (1+cos(2x))#

#=>2cos^2(1/2x) = 1+cos(x)#

#=>int -1dx+2int1/2*1/cos^2(1/2x)dx#

#1/2*1/cos^2(1/2x)dx# is exactly the derivate of #tan(1/2x)#

#=>[2tan(1/2x) - x] + C#

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Answer 2

Samantha Adkison

To integrate ( \frac{{1 - \cos(x)}}{{1 + \cos(x)}} ), you can use a trigonometric substitution method.

First, let ( u = \tan\left(\frac{x}{2}\right) ), then ( \cos(x) = \frac{1-u^2}{1+u^2} ) and ( dx = \frac{2}{1+u^2} , du ).

Substituting these into the integral, you get:

[ \int \frac{1 - \frac{1-u^2}{1+u^2}}{1 + \frac{1-u^2}{1+u^2}} \cdot \frac{2}{1+u^2} , du ]

Simplify the expression to integrate. Then integrate to find the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.