$\int \sin^2 x \cos^4 x \cos x dx =\int \sin^2 x (1-\sin^2 x)^2 d(\sin x) =\int t^2(1-t^2)^2 dt$
$=\int t^2 (1+t^4 -2t^2)dt =\int (t^6-2t^4 +t^2)dt =\dfrac{1}{7}t^7 -\dfrac{2}{5}t^5+\dfrac{1}{3}t^3 +C$
$=\dfrac{1}{7}\sin^7 x -\dfrac{2}{5}\sin^5 x+\dfrac{1}{3}\sin^3 x +C$
Tự thế cận nhé