Đặt $x=\sin t$, $\frac{\pi}{2}<t<\frac{\pi}{2} $, suy ra:
$dx=\cos t.dt ; \frac{x^3dx}{\sqrt{1-x^2} }=\frac{\sin^3t.\cos tdt}{\cos t}=\sin^3t.dt=\frac{1}{4}(3\sin t-\sin3t)dt $
Khi đó:
$\int\limits f(x)dx=\frac{1}{4}\int\limits (3\sin t-\sin 3t)dt=-\frac{3}{4}\cos t+\frac{1}{12}\cos 3t+C $
$=\int\limits -\frac{3}{4}\cos t+\frac{1}{12}(4\cos^3t-3\cos t)+C=\frac{1}{3}\cos ^3t-\cos t+C $
$=(\frac{1}{3}\cos^2t-1)\cos t+C=[\frac{1}{3}(1-\sin^2t)-1]\cos t+C$
$=[\frac{1}{3}(1-x^2)-1]\sqrt{1-x^2}+C=-\frac{1}{3}(x^2+2)\sqrt{1-x^2}+C $