$\sin^2 x \tan x = \sin^2 x \dfrac{\sin x }{\cos x} = (1-\cos^2 x) \dfrac{\sin x}{\cos x}$
đặt $\cos x = t \Rightarrow -\sin x dx = dt \Rightarrow \sin x dx = -dt$
$I = -\int \dfrac{1-t^2}{t} dt = \int t dt - \int \dfrac{1}{t}dt = \dfrac{1}{2}t^2 - \ln | t| + C$