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Question Number 127774 by Bird last updated on 02/Jan/21

$$calculate{u}_n={\int }_0^1{x}^n\sqrt{1-x^4}dx$$

Answered by Dwaipayan Shikari last updated on 02/Jan/21

$${\int }_0^1{x}^n\sqrt{1-x^4}{x}^4=u$$$$=\frac{1}{4}{\int }_0^1{u}^{\frac{n}{4}-\frac{1}{3}}{(1-u)}^{\frac{1}{2}}du=\frac{1}{4}.\frac{\mathrm{\Gamma }(n+\frac{2}{3})\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(n+\frac{13}{6})}$$

Answered by mathmax by abdo last updated on 02/Jan/21

$${\mathrm{u}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}\sqrt{1-{\mathrm{x}}^4}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}={\mathrm{t}}^{\frac{1}{4}}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}={\int }_0^1{\mathrm{t}}^{\frac{\mathrm{n}}{4}}{(1-\mathrm{t})}^{\frac{1}{2}}\frac{1}{4}{\mathrm{t}}^{\frac{1}{4}-1}\mathrm{dt}=\frac{1}{4}{\int }_0^1{\mathrm{t}}^{\frac{\mathrm{n}}{4}-\frac{3}{4}}{(1-\mathrm{t})}^{\frac{1}{2}}\mathrm{dt}$$$$=\frac{1}{4}{\int }_0^1{\mathrm{t}}^{\frac{\mathrm{n}+1}{4}-1}{(1-\mathrm{t})}^{\frac{3}{2}-1}\mathrm{dt}=\frac{1}{4}\mathrm{B}(\frac{\mathrm{n}+1}{4},\frac{3}{2})$$$$=\frac{1}{4}\times \frac{\mathrm{\Gamma }\left(\frac{\mathrm{n}+1}{4}\right)\times \mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(\frac{\mathrm{n}+1}{4}+\frac{3}{2})}=\frac{1}{4}\mathrm{\Gamma }\left(\frac{3}{2}\right)\frac{\mathrm{\Gamma }\left(\frac{\mathrm{n}+1}{4}\right)}{\mathrm{\Gamma }\left(\frac{\mathrm{n}+7}{4}\right)}$$$$\mathrm{\Gamma }\left(\frac{3}{2}\right)=\mathrm{\Gamma }(\frac{1}{2}+1)=\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{\sqrt{\pi }}{2}\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\frac{\sqrt{\pi }}{8}\times \frac{\mathrm{\Gamma }\left(\frac{\mathrm{n}+1}{4}\right)}{\mathrm{\Gamma }\left(\frac{\mathrm{n}+7}{4}\right)}$$

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