How do you integrate #int cos^2(x) tan^3(x) dx#?
1 Answer
Jul 23, 2016
I got:
#cos^2x/2 - ln|cosx| + C#
and Wolfram Alpha agrees.
Note that
#color(blue)(int cos^2xtan^3xdx)#
#= int (sin^3x)/cosxdx#
Then, you can use
#= 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)#