<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] You can avoid all the trig by making a much simpler substitution:$$u^2=1-x^2$$so: $$x^2=1-u^2$$ $$ u= \sqrt{1-x^2}$$ $$2u du=-2xdx$$
Rewriting the integral, factoring out one $x$:$$\int x^3 \sqrt{1-x^2} dx=\int x^2 \sqrt{1-x^2} xdx=-\int (1-u^2) u^2 du$$Multiply out the integrand, integrate with the power formula term by term and substitute back for $x$...
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