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

1 Answer
mason m
Feb 25, 2017

#1/3tan^3x+C#

Explanation:

Look for a function and its derivative inside the integrand. Here, it's very useful to know that the derivative of #tanx# is #sec^2x#.

Here, we have the #tanx# function squared times the derivative of #tanx#.

So, let #u=tanx#, implying that #du=(sec^2x)dx#. Then:

#I=intunderbrace(tan^2x)_(u^2)overbrace((sec^2x)dx)^(du)=intu^2du#

Which is a much simpler problem:

#I=1/3u^3+C=1/3tan^3x+C#

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