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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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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