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Question Number 197891 by Erico last updated on 02/Oct/23

$$\mathrm{Soit}\mathrm{I}={\underset{0}{\int }}^1\sqrt{\mathrm{t}\sqrt{\mathrm{t}(1-\mathrm{t})}}\mathrm{dt}$$$$\mathrm{Comment}\mathrm{calculer}\mathrm{I}$$

Answered by Mathspace last updated on 03/Oct/23

$$I={\int }_0^1{t}^{\frac{1}{2}}.t^{\frac{1}{4}}.{(1-t)}^{\frac{1}{4}}dt$$$$={\int }_0^1{t}^{\frac{3}{4}}{(1-t)}^{\frac{1}{4}}dt$$$$=B(\frac{3}{4}+1,\frac{1}{4}+1)$$$$=B(\frac{7}{4},\frac{3}{4})=\frac{\mathrm{\Gamma }\left(\frac{7}{4}\right).\mathrm{\Gamma }\left(\frac{3}{4}\right)}{\mathrm{\Gamma }\left(\frac{5}{2}\right)}$$$$\mathrm{\Gamma }\left(\frac{7}{4}\right)=\mathrm{\Gamma }(\frac{3}{4}+1)=\frac{3}{4}\mathrm{\Gamma }\left(\frac{3}{4}\right)$$$$I=\frac{3}{4}{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right).\frac{1}{\mathrm{\Gamma }\left(\frac{5}{2}\right)}=\frac{3{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right)}{4\mathrm{\Gamma }\left(\frac{5}{2}\right)}$$

Answered by Frix last updated on 02/Oct/23

$$u=\frac{\sqrt[4]{1-t}}{\sqrt[4]{t}}\Rightarrow$$$$I=4\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{u^4}{{(u^4+1)}^3}du=\frac{3\sqrt{2}}{32}\pi$$

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