Đặt $x = u \Rightarrow dx = du $ và $\sin x \cos^2 x dx = dv \Rightarrow -\cos^2 x d(\cos x) =dv$
$\Rightarrow -\dfrac{1}{3}\cos^3 x = v$
$I = -x.\dfrac{1}{3}\cos^3 x +\dfrac{1}{3} \int \cos^3 x dx = -x.\dfrac{1}{3}\cos^3 x +\dfrac{1}{3} I_1$
$I_1= \int \cos^3 x dx =\int (1-\sin^2 x)\cos x dx =\int (1-\sin^2 x) d(\sin x) =\sin x -\dfrac{1}{3}\sin^3 x +C$
Vậy $I = -x.\dfrac{1}{3}\cos^3 x +\dfrac{1}{3}\sin x -\dfrac{1}{9}\sin^3 x +C$ tự thay cận