Question

How do you find the integral of `int 1/(1 + sec(x))`?

Answer 1

`int dx/(1+secx ) = x -tan(x/2)+C`

Explanation:

Note that:

`1/(1+secx ) = 1/(1+1/cosx)`

use now the parametric formula:

`cosx = (1-tan^2(x/2))/(1+tan^2(x/2))`

`1/(1+secx ) = 1/(1+(1+tan^2(x/2))/(1-tan^2(x/2)))`

`1/(1+secx ) = (1-tan^2(x/2))/((1-tan^2(x/2))+(1+tan^2(x/2)))`

`1/(1+secx ) = (1-tan^2(x/2))/2`

`1/(1+secx ) = 1/2 - 1/2(sec^2(x/2) -1)`

`1/(1+secx ) = 1 - 1/2sec^2(x/2)`

Then:

`int dx/(1+secx ) = int (1 - 1/2sec^2(x/2))dx`

`int dx/(1+secx ) = int dx - int sec^2(x/2)d(x/2)`

`int dx/(1+secx ) = x -tan(x/2)+C`