How to find these integrals?

How to find the following two integrals? $$\int_{0}^{1}{\sqrt{{{x}^{3}}-{{x}^{4}}}dx}$$ and $$\int_{0}^{1}{x\sqrt{{{x}^{3}}-{{x}^{4}}}dx}$$

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$$I:=\int_0^1 x\sqrt {x^3-x^4}\,dx=\int_0^1 x^2\sqrt{x-x^2}\,dx=\int_0^1x^2\sqrt{\frac{1}{4}-\left(\frac{1}{2}-x\right)^2}\,dx=$$

$$=\frac{1}{2}\int_0^1x^2\sqrt{1-\left(1-2x\right)^2}\,dx$$

Substitute now

$$\sin u=1-2x\Longrightarrow \cos u\,du=-2\,dx\,,\,x=0\Longrightarrow u=\frac{\pi}{2}\;\;,\;x=1\Longrightarrow u=\frac{3\pi}{2}\,\,,\,\text{thus:}$$

$$I=-\frac{1}{4}\int_{\pi/2}^{3\pi/2}\left(\frac{1-\sin u}{2}\right)^2\sqrt{1-\sin^2u}\cos u\,du=$$

$$-\frac{1}{16}\int_{\pi/2}^{3\pi/2}\left(\cos^2u-2\sin u\cos^2u+\cos^2u\sin^2u\right)du=\ldots\,\,etc.$$

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Shouldn't it read $x = 0 \implies u = \frac{\pi}{2}$ and $x = 1 \implies u = \frac{3\pi}{2}$? –  Joe Dec 9 '12 at 1:38
Thanks Joe: big mistake there! –  DonAntonio Dec 9 '12 at 2:35
No problem, +1. You may want to include a justification for only choosing the principal branch for $\cos u$ in your last step now since the bounds have been changed and $\cos u$ is negative in quadrants $2$ and $3$. –  Joe Dec 9 '12 at 2:47
Ah, you're right. Overlooked that fact. Long day of studying, sorry. –  Joe Dec 9 '12 at 2:51
Thank you very much! It helped me a lot! –  Sleepingip Dec 9 '12 at 17:39

One semi-mechanical but unpleasant way to attack both is to bring an $x$ "out." We end up having to integrate $x\sqrt{x-x^2}$ and $x^2\sqrt{x-x^2}$.

Note that $x-x^2=\frac{1}{4}-(x-\frac{1}{2})^2$. This may suggest a substitution. Best would be $x-\frac{1}{2}=\frac{1}{2}u$. After the substitutions, we end up with integrals of familiar shape.

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@DonAntonio: Thanks. I often have trouble with minus signs. Fixed. –  André Nicolas Dec 9 '12 at 1:09
@Graphth: Yes, I am so familiar with minus sign trouble that I assumed that I had done it again. –  André Nicolas Dec 9 '12 at 1:12
Seems to be a nice hint, +1. –  Joe Dec 9 '12 at 2:50