What's the integral of #int (tanx)^2+(tanx)^4 dx#?

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Answer 1 · 2016-04-27T03:27:30

#tan^3x/3+C#

Factor a #tan^2x# inside the integral.

#=inttan^2x(1+tan^2x)dx#

We know #1+tan^2x=sec^2x# through the Pythagorean identity.

#=inttan^2xsec^2xdx#

We can now use substitution:

#u=tanx" "=>" "du=sec^2xdx#

This gives us:

#=intu^2du=u^3/3+C=tan^3x/3+C#