How do you integrate #intsec3x#?
1 Answer
Feb 8, 2017
Explanation:
#intsec(3x)dx#
Use the substitution
#=1/3intsec(3x)(3)dx=1/3intsec(u)du#
This is a common integral:
We can derive this integral by multiplying the integrand by
#=1/3intsec(u)(sec(u)+tan(u))/(sec(u)+tan(u))du=1/3int(sec^2(u)+sec(u)tan(u))/(sec(u)+tan(u))du#
Now let
#=1/3int(dv)/v=1/3lnabsv+C=1/3lnabs(sec(u)+tan(u))+C#
Finally:
#=1/3lnabs(sec(3x)+tan(3x))+C#