What is the integral of #sec(x)#?

1 Answer
Apr 18, 2018

Answer:

#intsecxdx=ln|secx+tanx|+C#

Explanation:

Integrating the secant requires a bit of manipulation.

Multiply #secx# by #(secx+tanx)/(secx+tanx)#, which is really the same as multiplying by #1.# Thus, we have

#int((secx(secx+tanx))/(secx+tanx))dx#

#int(sec^2x+secxtanx)/(secx+tanx)dx#

Now, make the following substitution:

#u=secx+tanx#

#du=(secxtanx+sec^2x)dx=(sec^2x+secxtanx)dx#

We see that #du# appears in the numerator of the integral, so we may apply the substitution:

#int(du)/u=ln|u|+C#

Rewrite in terms of #x# to get

#intsecxdx=ln|secx+tanx|+C#

This is an integral worth memorizing.