I would proceed by thus , let $y = [\sec (x)]^2 $
then
$dy = 2 \cdot \sec(x) \cdot \sec(x) \cdot \tan(x) \cdot dx = 2 \cdot ( \sec (x))^2 \cdot \tan(x) \cdot dx $
so, $ 2 \tan^2(x) \sec^2 (x) dx = \sec(x) \cdot \tan(x) \cdot dy = y(y-1)^\frac{1}{2} \cdot dy $
since $\sec(x) = y^{\frac{1}{2}}$ and by considering positive square roots only $\tan y = ( \sec^2(x) - 1)^{1/2} = (y - 1)^{1/2}$. Thus the substitution $y = \sec^2 x$ yields $ 2 \int \tan^2 (x) \sec^3(x) dx = \int (y(y - 1) )^{1/2} dy $ and this later form can be reduced to the standard form $\int(z^2 - a^2)^{1/2} dz$ since $ y(y-1)=(y-(1/2))^2 - (1/2)^2 . $ What are the other ways to integrate this expression, except for the substitution
$\tan^2(x)^2=\sec (x)^2 -1$ which gives $ \sec^5(x) - \sec^3 (x) $ in the integrand which I quite don't like.