How do you integrate # sec^3x (tanx) dx#?

1 Answer
mason m
Oct 4, 2016

#sec^3x/3+C#

Explanation:

When working with integrals of secant and tangent, it's important to remember the following:

  • #d/dxtanx=sec^2x#
  • #d/dxsecx=secxtanx#

Here, we see that we can write #sec^3x(tanx)# as #sec^2x(secxtanx)#, which is perfect, since it composed of #sec^2x# and the derivative of secant, #secxtanx#. This indicates to us that we want to use a substitution of #u=secx#.

#intsec^3x(tanx)dx=intsec^2x(secxtanx)dx#

With #u=secx# and #du=(secxtanx)dx#:

#=intu^2du=u^3/3+C=sec^3x/3+C#

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