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?
 A: We carry out the details. In a comment at the end, we show a somewhat simpler way. 
Let $u=\sqrt{1-x}$. Then $\dfrac{du}{dx}=-\dfrac{1}{2\sqrt{1-x}}$.
Thus $du=-\dfrac{1}{2\sqrt{1-x}}\,dx$. You left out the $dx$, which may be part of the reason you are puzzled.
So $dx=-2\sqrt{1-x} \,du=-2u\,du$.
Also, since $u^2=1-x$, we have $x=1-u^2$, and therefore $x^2=(1-u^2)^2$.
Expressing everything in terms of $u$, we get
$$\int (1-u^2)^2 (u)(-2u)\,du.$$
Note that by everything, we include $dx$. 
Now expand the $(1-u^2)^2$, multiply through by $2u^2$, and integrate term by term. 
Remark: I would prefer to do the same substitution in the form $u^2=1-x$. Then $2u\,du=-dx$, no unpleasant square roots, less algebra. Try it, you will like it.
A: 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 $-2\int u^2(1-u^2)^2 \mathrm{d}u$.
