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

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Answer 1 · 2016-08-08T15:46:13

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

Recall that the Pythagorean identity tells us that #tan^2(x)=sec^2(x)-1#. Thus we can say that:

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

Now we can split up the integral:

#=inttan(x)sec^2(x)dx-inttan(x)dx#

For the first integral, let #u=tan(x)# so #du=sec^2(x)dx#.

#=intudu-inttan(x)dx#

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

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

Here, let #v=cos(x)# so #dv=-sin(x)dx#, so:

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

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

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

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