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

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Answer 1 · 2017-01-11T00:26:09

#int tan^3x dx= 1/(2cos^2x) +ln abs cos x + C#

You can write #tan^3x# as:

#tan^3x = (sin^3x)/(cos^3x) = sinx (1-cos^2x)/(cos^3x)#

So that:

#int tan^3x dx= int sinx (1-cos^2x)/(cos^3x)dx#

Now substitute:

#t=cosx#
#dt = -sint dt#

and you have:

#int tan^3x dx= -int (1-t^2)/(t^3)dt#

Separating the sum:

#int tan^3x dx= -int 1/(t^3)dt + int (dt)/t = 1/(2t^2)+ln abs(t)#

and substituting back #t=cosx#

#int tan^3x dx= 1/(2cos^2x) +ln abs cos x + C#