Question: The random variables X and Y.

DELETED... AS IT IS OFF TOPIC.

Answer: 
When we substitute $u = 1- x^2 \implies du = -2x \, dx$, we get, $$I = \int x^3\sqrt{1-x^2} \, dx = \int \frac{1}{2} (-x^2)(-2x)\sqrt{1-x^2} \, dx$$ $$=\int \frac{1}{2}((1-x^2)-1)\sqrt{1-x^2}(-2x) \, dx$$ $$=\int \frac{1}{2}(u-1) \sqrt{u} \, du$$ $$= \frac12[\int u^{\frac32}\, du - \int \sqrt{u}\, du]$$
$$\frac{du}{dx}=-2x$$
as you say, and
$$x^3\sqrt{1-x^2}=-\frac{x^2}2\sqrt{1-x^2}\frac{du}{dx}
=-\frac{1-u}2\sqrt u\frac{du}{dx}=\frac{u^{3/2}-u^{1/2}}{2}\frac{du}{dx}.$$
This means that
$$\int x^3\sqrt{1-x^2}\,dx=\int \frac{u^{3/2}-u^{1/2}}{2}\,du.$$