Question

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

Answer 1

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

First, let's do some trigonometric transformation :

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)`

So we can 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 perfect :

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

Substitute back for t = tan(1/2x)

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