int 1/(cos(theta) -1) d theta Find the indefinite integral

Explanation for cos(x) = (1-u^2)/(1+u^2)
before that , we know
cos(2x)= cos^2(x) -sin^2(x)
as cos^2(x) can be written as 1/(sec^2(x))
and we can show sin^2(x) = ((sin^2(x))/(cos^2(x) ))/(1/(cos^2(x)))
 = tan^2(x)/sec^2x
so now ,
cos(2x)= cos^2(x) -sin^2(x)
= (1/sec^2(x)) - (tan^2(x)/sec^2(x))
=(1-tan^2(x))/(sec^2(x))
but sec^2(x) = 1+tan^2(x) ,as its an identity
so,
=(1-tan^2(x))/(sec^2(x))
=(1-tan^2(x))/(1+(tan^2(x)))
 
so ,
cos(2x) = (1-tan^2(x))/(1+(tan^2(x)))
so,
then
cos(x) = (1-tan^2(x/2))/(1+(tan^2(x/2)))
as before we told to assume that u= tan(x/2),
so,
cos(x) = (1-u^2)/(1+u^2)
Hope this helps to understand better

Given to solve
int 1/(cos(theta) - 1) d theta
For convenience, let  theta = x
=>
int 1/(cosx - 1) dx
let u = tan(x/2) => dx = (2/(1+u^2)) du
so ,
cos(x) = (1-u^2)/(1+u^2) (See my reply below for an explanation)
so,
int 1/(cos(x) - 1) dx
= int 1/((1-u^2)/(1+u^2) - 1) (2/(1+u^2)) du
= int 1/(((1-u^2)-(1+u^2))/(1+u^2) ) (2/(1+u^2)) du
=int (1+u^2)/(((1-u^2)-(1+u^2)) ) (2/(1+u^2)) du
=int (2)/(((1-u^2)-(1+u^2)) ) du
=int (2)/(((1-u^2)-1-u^2)) ) du
= int (2)/(-2u^2) du
= -int(1/u^2) du
= -[u^(-2+1)/(-2+1)]
= u^-1
= 1/u
= 1/tan(x/2)
= cot(x/2)+c
But x= theta
so,
int 1/(cos(theta) - 1) d theta = cot(theta/2)+c