$I=\int_{-1}^0\dfrac{\sqrt{2-x}.\sqrt{2+x}}{\sqrt{(2+x)^2}}dx=\int_{-1}^0 \sqrt{\dfrac{2-x}{2+x}}dx$
Đặt $x=2\cos 2t \Rightarrow dx =-4\sin 2t dt$
$I=-4\int \sqrt{\dfrac{2-2\cos 2t}{2+2\cos 2t}}.\sin 2t dt=-4\int \sqrt{\dfrac{1-\cos 2t}{1+\cos 2t}} .\sin 2t dt$
$=-4\int \sqrt{\dfrac{2\sin^2 t}{2\cos^2 t}}.\sin 2t dt=-4\int 2\sin^2 t dt$ tự hạ bậc mà làm nốt