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, $$I = \int x^3\sqrt{1-x^2} \, dx = \int \frac{1}{2} (-x^2)(-2x)\sqrt{1-x^2} \, dx$$ $$=\int \frac{1}{2}((1-x^2)-1)\sqrt{1-x^2}(-2x) \, dx$$ $$=\int \frac{1}{2}(u-1) \sqrt{u} \, du$$ $$= \frac12[\int u^{\frac32}\, du - \int \sqrt{u}\, du]$$
Alternatively:
$$\int x^3\sqrt{1-x^2} \, dx=\frac12\int x^2\sqrt{1-x^2} \, dx^2 \stackrel{x^2=t}=\frac12\int t\sqrt{1-t} \, dt\stackrel{\text{by parts}}=$$
$$\frac12\cdot t\cdot \left(-\frac23(1-t)^{3/2}\right)-\frac12 \int \left(-\frac23(1-t)^{3/2}\right)dt=$$
$$\frac13t(1-t)^{3/2}-\frac{2}{15}(1-t)^{5/2}+C=$$
$$\frac13x^2(1-x^2)^{3/2}-\frac{2}{15}(1-x^2)^{5/2}+C.$$