Evaluating trig integral $\int \tan^3x \sec^2x dx$

Question

I am trying to evaluate this integral, and I was coming up short so I am looking at the solution from a website here.

I get lost on the u-substitution part. Where does $ \sec^2(\tan^{-1}(u)) $ come from in the denominator? Since we are using $u = \tan x, du = \sec^2x$, why isn't the denominator just $\sec^2x$?

Answer 1

$$\int \tan^3x\sec^2x\ dx$$ Take $u=\tan x$, then $du=\sec^2x\ dx$ and we get $$\int u^3\ du=\dfrac{u^4}{4}+C$$ Now substitute back $u=\tan x$ and we finally get $$\dfrac14\tan^4x+C$$