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

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`

Comments

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