How do you integrate #int sec^6(3x)#?

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Answer 1 · 2017-01-10T22:11:58

#tan(3x)+2/3tan^3(3x)+1/5tan^5(3x)+C#

First let #u=3x# so that #du=3dx#. Then we have:

#I=intsec^6(3x)dx=1/3intsec^6(3x)3dx=1/3intsec^6(u)du#

We will now make use of the identity #tan^2(theta)+1=sec^2(theta)#.

#I=1/3intsec^4(u)sec^2(u)du=1/3int(sec^2(u))^2sec^2(u)du#

#I=1/3int(1+tan^2(u))^2sec^2(u)du#

Now we should perform the substitution #v=tan(u)#. Note that #dv=sec^2(u)du#, which is why we left a #sec^2(u)# floating around in the integrand.

#I=int(1+v^2)^2dv#

Expand #(1+v^2)^2=(1+v^2)(1+v^2)# and then integrate term by term:

#I=int(1+2v^2+v^4)dv=v+2/3v^3+1/5v^5#

From #v=tan(u)#:

#I=tan(u)+2/3tan^3(u)+1/5tan^5(u)#

From #u=3x#:

#I=tan(3x)+2/3tan^3(3x)+1/5tan^5(3x)+C#