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I want to integrate

$$\int x \sqrt{1-x} dx $$

I set

$ U = 1 - x $

However when I continue on I'm stuck here -

$\int x(u)^{1/2} -du $

I can't remove $x$ ,

How should I change my $U$ ?

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    $\begingroup$ You would remove $x$ by having $x=1-u$. $\endgroup$
    – Dave
    Commented Jul 17, 2017 at 12:56

4 Answers 4

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letting $1-x=u$ so $x=1-u$ and $-dx=du$ so our integral is $\int(u-1)\sqrt{u}du$ which can be easily solved.

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letting \begin{equation} \sqrt{1-x}=U\\ U^{2}=1-x \longrightarrow 2U du=-dx\\ x= 1- U^{2}\\ \int (1-U^{2})U(-2U)du=\\ \int (-2U^{2}+2U^{4})du=\\ \frac{-2}{3}U^{3}+\frac{2}{5}U^{5}=\\ \frac{-2}{3}(\sqrt{1-x})^{3}+\frac{2}{5}(\sqrt{1-x})^{5}\\ \end{equation}

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$$\int x\sqrt{1-x}dx=\int\left((x-1)\sqrt{1-x}+\sqrt{1-x}\right)dx=$$ $$=\int\left(-\sqrt{(1-x)^3}+\sqrt{1-x}\right)dx=-\frac{(1-x)^{\frac{5}{2}}}{-\frac{5}{2}}+\frac{(1-x)^{\frac{3}{2}}}{-\frac{3}{2}}+C=$$ $$=\frac{2}{5}\sqrt{(1-x)^5}-\frac{2}{3}\sqrt{(1-x)^3}+C.$$

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If you let $u = 1-x$, then $du = -dx$, making $\int x\sqrt{1-x}\,dx$ become $\int (u-1)\,\sqrt u\, du$ = $\int u^{3/2}du$ - $\int u^{1/2} du$.

The rest is just homework.

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