The indefinite integral $\int x^2\sqrt{1-x}\,dx$

I'm trying get the integral $\int x^2\sqrt{1-x}\,dx$ 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 $du = -\frac{1}{2\sqrt{1-x}}$

and then it becomes $-2 \int u^2(1-u^2)^2\,du$

I don't understand how it becomes like that?

-

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.

-
As I just commented elsewhere, from $u=\sqrt{1-x}$ you can get $u^2=1-x$ and then differentiate both sides to get $2u\,du=-dx$. It's a bit simpler that way. –  Michael Hardy Mar 25 '13 at 17:33
I had just finished adding my comment! From the point of view of minimizing the probability of mechanical error, it is a lot simpler. –  André Nicolas Mar 25 '13 at 17:36

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 $-2\int u^2(1-u^2)^2 \mathrm{d}u$.

-
Here's a slightly simpler way: Once you have $x=1-u^2$, differentiate both sides to get $dx=-2u\,du$. ${}\qquad{}$ –  Michael Hardy Mar 25 '13 at 17:30