How do you integrate #arcsec(x)#?

1 Answer
Mar 9, 2015

Method: To integrate #arc sec (x)#, substitution, then integrate by parts.

You'll also need #int secu du#, which can be done by substitution and partial fractions.
Here's a nice explanation: http://socratic.org/questions/what-is-the-integral-of-sec-x .

Details:#int arcsec(x) dx#

Let #y=arc sec(x)#, so #x=secy# and #dx = secy tany dy#.

With this substitution, the integral becomes:

#inty secy tany dy#.

Integrate this by parts:
Let #u=y# and #dv=secytany dy#.
Then #du=dy# and #v=secy#.

#u v - int v du=ysecy-intsecy dy#
#=ysecy-ln abs(secy+tany)+C#.

With #y=arc sec(x)# we get #x=secy#, and #tany=sqrt (x^2-1)#.

The integral becomes:

#int arcsec x dx = (arc sec(x))x-ln abs(x+sqrt(x^2-1))+C#.

This is more easily read is we write it as:

#x (arc sec(x)) - ln abs(x+sqrt(x^2-1))+C#.