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#.
