What is the Integral of tan^2 x sec^4 x dx?

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##tan^3x/3 + tan^5x/5 + C## where C = regular of integration

Using the Pythagorean unity, ##sec^2x = 1 + tan^2x## we keep ##int tan^2 x sec^4x dx = int tan^2x sec^2 x sec^2x dx = inttan^2x(1+tan^2x)sec^2xdx## Now let ##tanx = u, ##so ##(du)/dx = sec^2x or du = sec^2x dx## Now integrating after a while deference to u, we keep ##intu^2(1+u^2)du == int (u^2 + u^4) du = u^3/3 + u^5/5 + C ##

where C = regular of integration.

Resubstituting u = tanx, we keep the decisive integration vindication as

##tan^3x/3 +tan^5x/5 + C##

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