<s>[INST] 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.
 [/INST] What about by integration by parts? It looks pretty simple:
$$u=x^2\;,\;\;u'=2x\\v'=x\sqrt{1-x^2}\;,\;\;v=-\frac13(1-x^2)^{3/2}$$
and thus
$$\int x^2\cdot x\sqrt{1-x^2}\,dx=-\frac13x^2(1-x^2)^{3/2}+\frac23\int x(1-x^2)^{3/2}dx=$$
$$-\frac13x^2(1-x^2)^{3/2}-\frac2{15}(1-x^2)^{5/2}+C$$
Note: We used above the following:
$$\int x(1-x^2)^k\,dx=-\frac12\int (-2x\,dx)(1-x^2)^k=-\frac12\frac{(1-x^2)^{k+1}}{k+1}\;\ldots$$
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