The indefinite integral $\int x^2\sqrt{1-x}\,\mathrm dx$ I'm trying get the integral $\int x^2\sqrt{1-x}\,\mathrm dx$ but I don't know how to proceed. I know I have to use substitution, but that's it.
I tried to get some help with the wolfram alpha step-by-step integral calculation, but I don't quite get how it gets there
It subtitutes $u=\sqrt{1-x}$ and $\mathrm du = -\frac{1}{2\sqrt{1-x}}$
and then it becomes $-2 \int u^2(1-u^2)^2\,\mathrm du$
I don't understand how it becomes like that?

If you use change of variable $u=1-x$ and develop, you'll have to integrate a sum of monomials.
With your method, you have $u=\sqrt{1-x}$, thus $x=1-u^2$, thus the factor $(1-u^2)^2$.
Then $\mathrm{d}u=-\frac{\mathrm{d}x}{2\sqrt{1-x}}=-\frac{\mathrm{d}x}{2u}$, hence $\mathrm{d}x=-2 u \mathrm{d}u$. With the additional factor $\sqrt{1-x}=u$, you get $-2\int u^2(1-u^2)^2 \mathrm{d}u$.
