Calculus of a Single Variable, Chapter 8, 8.1, Section 8.1, Problem 38

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