Trig substitution $\int x^3 \sqrt{1-x^2} dx$ $$\int x^3 \sqrt{1-x^2} dx$$
$x = \sin \theta $
$dx = \cos \theta d \theta$ 
$$\int \sin^3 \theta d \theta$$
$$\int (1 - \cos^2 \theta) \sin \theta  d \theta$$ 
$u = \cos  \theta$
$du = -\sin\theta d \theta$
$$-\int u^2 du$$
$$\frac{-u^3}{3} $$
$$\frac{\cos^3 \theta}{3}$$
With the triangle trick I get:
$$\frac{-\sqrt{1-x^2}^3}{3}$$
This is wrong but I am not sure where I went wrong.

Let $x=\sin{\theta}$, then $dx = \cos{\theta} \, d\theta$; the integral becomes
$$\int d\theta \, \sin^3{\theta} \, \cos^2{\theta} = \int d\theta \, \sin^3{\theta} -\int d\theta \, \sin^5{\theta} $$
$$\int d\theta \, \sin^3{\theta} = \int d\theta \, \sin{\theta} (1-\cos^2{\theta}) = -\int d(\cos{\theta}) (1-\cos^2{\theta})= -\cos{\theta} + \frac13 \cos^3{\theta}+C$$
Similarly
$$\int d\theta \, \sin^5{\theta} = -\int d(\cos{\theta}) (1-\cos^2{\theta})^2 = -\cos{\theta} + \frac{2}{3} \cos^3{\theta}-\frac15 \cos^5{\theta}+C'$$
Subtracting the two, I get
$$\int d\theta \, \sin^3{\theta} \, \cos^2{\theta} = -\frac13 \cos^3{\theta}+\frac15 \cos^5{\theta}+C$$
Then use $x=\sin{\theta}$ and get
$$\int dx \, x^3 \, \sqrt{1-x^2} = \frac{1}{15} (3 x^4-x^2-2) \sqrt{1-x^2}+C$$
EDIT
I see that the answer can be simplified further to
$$-\frac{1}{15} (1-x^2)^{3/2} (3 x^2+2) + C$$
