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How do you integrate #∫sec^2 (x) tan^3 (x)dx#?

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Mar 6, 2015

Derivative of #tanx# is #sec^2x# , so use substitution.

#intsec^2xtan^3xdx=int(tanx)^3sec^2xdx#.

You can probably figure it out from there, but just in case:

Let #u=tanx#, then #du=sec^2xdx#, so the integral becomes #intu^3du#, which, of course, evaluates to #1/4tan^4x+C#.

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