Đặt $x=\sin t \Rightarrow dx =\cos t dt$
$I=\int \sqrt{(1-\sin^2 t)^3} .\cos t dt =\int \cos^4 t dt =\dfrac{1}{4} \int (1+\cos 2t)^2 dt$
$=\dfrac{1}{4}\int \bigg ( \cos^2 2t +2\cos 2t +1 \bigg ) dt =\dfrac{1}{4} \int \dfrac{1+\cos 4t}{2} dt +\dfrac{1}{2}\int \cos 2t dt +\dfrac{1}{4}\int dt$
Trong đó $\int \cos at dt =\dfrac{1}{a} \sin at +C$ Ráp vào nhé