How do you find the antiderivative of #1/(1-cosx)#?

Question

Scarlett Allison · Accepted Answer

#int1/(1-cos(x)) dx#

Let's start by performing some trigonometric transformations:

Remember that #=>sin^2(x) = 1/2(1-cos(2x))#

(From #sin(a)*sin(b)# formula with #a = b#)

#=>2sin^2(x) = 1-cos(2x)#

#=>2sin^2(1/2x) = 1-cos(x)#

In order for us to write:

#int1/2*1/(sin^2(1/2x))dx#

Divise numerator and denominator by #cos^2(1/2x)#

#int1/2*(1/cos^2(1/2x))/(sin^2(1/2x)/cos^2(1/2x))dx =int1/2*(1/cos^2(1/2x))/tan^2(1/2x)dx#

let's #t = tan(1/2x)#

#=> dt = 1/2*1/cos^2(1/2x)#

Just have a look we have #dt# in the integral !

It's ideal:

#int1/t^2dt = [-1/t]#

Replace with t = tan(1/2x).

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

Samantha Adkison · Answer

undefined To find the antiderivative of $ \frac{1}{1 - \cos(x)} $, you can use the following steps:

1. Start by expressing $ \frac{1}{1 - \cos(x)} $ in terms of $ \tan\left(\frac{x}{2}\right) $. You can use the half-angle identity for tangent: $ \tan\left(\frac{x}{2}\right) = \frac{1 - \cos(x)}{\sin(x)} $.

2. Rewrite $ \frac{1}{1 - \cos(x)} $ as $ \frac{\sin(x)}{1 - \cos(x)} \cdot \frac{1}{\sin(x)} $.

3. Replace $ \frac{\sin(x)}{1 - \cos(x)} $ with $ \frac{1}{\tan\left(\frac{x}{2}\right)} $.

4. Integrate $ \frac{1}{\tan\left(\frac{x}{2}\right)} $ with respect to $ x $. This can be done by substituting $ u = \tan\left(\frac{x}{2}\right) $ and using the integral of $ \sec^2(u) $.

5. After integrating, don't forget to reverse the substitution to obtain the antiderivative in terms of $ x $.