What is the integral of #int tan^3(x) dx#?

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Answer 1 · 2017-01-14T00:49:11

#tan^2(x)/2+ln(abscos(x))+C#

Split up #tan^3(x)# into #tan^2(x)tan(x)# then rewrite #tan^2(x)# using the identity #tan^2(theta)+1=sec^2(theta)=>tan^2(theta)=sec^2(theta)-1#.

#inttan^3(x)dx=inttan^2(x)tan(x)dx=int(sec^2(x)-1)tan(x)dx#

Distribute:

#=intsec^2(x)tan(x)dx-inttan(x)dx#

For the first integral, apply the substitution #u=tan(x)=>du=sec^2(x)dx#, both of which are already in the integral.

#=intucolor(white).du-inttan(x)dx#

#=u^2/2-inttan(x)dx#

#=tan^2(x)/2-inttan(x)dx#

Now rewrite #tan(x)# as #sin(x)/cos(x)# and apply the substitution #v=cos(x)=>dv=-sin(x)dx#.

#=tan^2(x)/2-intsin(x)/cos(x)dx#

#=tan^2(x)/2+int(-sin(x))/cos(x)dx#

#=tan^2(x)/2+int(dv)/v#

This is a common integral:

#=tan^2(x)/2+ln(absv)+C#

#=tan^2(x)/2+ln(abscos(x))+C#