How do you integrate #int4sec4xtan4xsec^4 4x# using substitution?

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Answer 1 · 2016-08-16T23:16:08

Well, let's rewrite this, and you may see something.

Let #u = 4x#, and you have #du = 4dx#.

#color(blue)(int 4sec4xtan4xsec^4 (4x)dx)#

#= int secutanusec^4(u)du#

So now... let #v = secu#, and you get #dv = secutanudu#:

#=> int (secu)^4secutanudu#

#= int v^4dv#

#= v^5/5#

Back-substitute to get:

#= sec^5(u)/5#

#= color(blue)(sec^5(4x)/5 + C)#

How do you integrate #int4sec4x*tan4x*sec^4 4x# using substitution?