How do you integrate #int tan^3xsec^2x# using substitution?

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Answer 1 · 2018-03-14T07:45:23

Use the substitution #u=tanx#.

Let

#I=inttan^3xsec^2dx#

Apply the substitution #u=tanx#:

#I=intu^3du#

Integrate directly:

#I=1/4u^4+C#

Reverse the substitution:

#I=1/4tan^4x+C#