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

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Answer 1 · 2017-01-07T08:10:22

The answer is #=1/3ln(∣cos(3x)∣)+1/6sec^2(3x)+C#

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

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

#tanx=sinx/cosx#

#secx=1/cosx#

Let #u=3x#, #=>#, #du=3xdx#

#inttan^3(3x)dx=1/3inttan^3udu#

#=1/3int(sec^2u-1)tanudu#

#=1/3int(1/cos^2u-1)*sinu/cosudu#

#=1/3int(1-cos^2u)/(cos^3u)sinudu#

Let #v=cosu#, #=>#, #dv=-sinudu#

Therefore,

#=1/3int(1-cos^2u)/(cos^3u)sinudu=1/3int-(1-v^2)/v^3dv#

#=1/3int(v^2-1)/v^3dv#

#=1/3int(1/v-1/v^3)dv#

#=1/3(lnv+1/(2v^2))#

#=1/3(ln(cosu)+1/(2cos^2u))#

#=1/3(ln(∣cos(3x)∣)+1/(2cos^2(3x)))+C#

#=1/3ln(∣cos(3x)∣)+1/6sec^2(3x)+C#