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, $$I = \int x^3\sqrt{1-x^2} \, dx = \int \frac{1}{2} (-x^2)(-2x)\sqrt{1-x^2} \, dx$$ $$=\int \frac{1}{2}((1-x^2)-1)\sqrt{1-x^2}(-2x) \, dx$$ $$=\int \frac{1}{2}(u-1) \sqrt{u} \, du$$ $$= \frac12[\int u^{\frac32}\, du - \int \sqrt{u}\, du]$$
If $u=1-x^2$ then $du=-2xdx,$ which gives
$$\int x^3\sqrt{1-x^2}dx=-\frac{1}{2}\int x(1-u)\sqrt{u}\frac{du}{x}=-\frac{1}{2}\int\left(\sqrt{u}-u^{\frac{3}{2}}\right)du=...$$