How do you integrate #intsec3x#?

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Answer 1 · 2017-02-08T21:55:44

#intsec(3x)dx=1/3lnabs(sec(3x)+tan(3x))+C#

#intsec(3x)dx#

Use the substitution #u=3x#, implying that #du=(3)dx#. Then:

#=1/3intsec(3x)(3)dx=1/3intsec(u)du#

This is a common integral: #intsec(u)du=lnabs(sec(u)+tan(u))+C#

We can derive this integral by multiplying the integrand by #(sec(u)+tan(u))/(sec(u)+tan(u))#:

#=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 #v=sec(u)+tan(u)#. This implies that #dv=(sec(u)tan(u)+sec^2(u))du#. This is the numerator:

#=1/3int(dv)/v=1/3lnabsv+C=1/3lnabs(sec(u)+tan(u))+C#

Finally:

#=1/3lnabs(sec(3x)+tan(3x))+C#