How do you integrate #int cos^2(x) tan^3(x) dx#?

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Answer 1 · 2016-07-23T07:08:15

I got:

#cos^2x/2 - ln|cosx| + C#

Note that #tan^3x = sin^3x/cos^3x#. Therefore:

#color(blue)(int cos^2xtan^3xdx)#

#= int (sin^3x)/cosxdx#

Then, you can use #u#-substitution. When #u = cosx#, #du = -sinxdx#. To do that, you need to turn #sin^2x# into #1-cos^2x#.

#= int ((1-cos^2x)sinx)/(cosx)dx#

Hence:

#= -int ((1-cos^2x)(-sinx))/(cosx)dx#

#= -int (1-u^2)/(u)du#

#= int (u^2 - 1)/(u)du#

#= int u - 1/udu#

#= u^2/2 - ln|u|#

Re-substitute to get:

#=> color(blue)(cos^2x/2 - ln|cosx| + C)#