What is the integral of #int tan^3x secxdx#?

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Answer 1 · 2016-01-18T14:37:30

#int tan^3x secx dx = 1/3 sec^3x-secx +C#. If you really want the integral of that integral ask and we'll see what we can do.

Useful facts:

#d/dx(secx) = secxtanx#

#tan^2x = sec^2x-1#

#int tan^3x secx dx = int(sec^2x-1)secxtanx dx#

Let #u = secx#, so that #du = secxtanxdx#.

The integral then is #int(u^2-1) du#.

The student should be able to finish from here.