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

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Answer 1 · 2018-05-08T08:53:38

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

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 #