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

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Answer 1 · 2017-02-25T15:10:26

$1/3tan^3x+C$

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:

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

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