How do you integrate ∫4sec4x⋅tan4x⋅sec44x using substitution?
1 Answer
Aug 16, 2016
Well, let's rewrite this, and you may see something.
Let
∫4sec4xtan4xsec4(4x)dx
=∫secutanusec4(u)du
So now... let
⇒∫(secu)4secutanudu
=∫v4dv
=v55
Back-substitute to get:
=sec5(u)5
=sec5(4x)5+C