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Question Number 107835 by PNL last updated on 12/Aug/20

$$computeIn={\int }_0^1{x}^n\sqrt{1-x}dx$$

Answered by hgrocks last updated on 12/Aug/20

$${\mathrm{I}}_{\mathrm{n}}=\underset{0}{\stackrel{1}{\int }}{\mathrm{x}}^{\mathrm{n}}{(1-\mathrm{x})}^{\frac{1}{2}}\mathrm{dx}$$$$=\beta (\mathrm{n}+1,\frac{3}{2})$$

Answered by mathmax by abdo last updated on 12/Aug/20

$${\mathrm{I}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}\sqrt{1-\mathrm{x}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}={\sin }^2\mathrm{t}\Rightarrow$$$${\mathrm{I}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{n}}\mathrm{t}\mathrm{cost}\left(2\mathrm{sint}\right)\mathrm{cost}\mathrm{dt}$$$$=2{\int }_0^{\frac{\pi }{2}}{\cos }^2\mathrm{t}{\sin }^{2\mathrm{n}+1}\mathrm{t}\mathrm{dt}\mathrm{we}\mathrm{know}$$$$2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{p}-1}\mathrm{x}{\sin }^{2\mathrm{q}-1}\mathrm{x}\mathrm{dx}=\mathrm{B}(\mathrm{p},\mathrm{q})=\frac{\mathrm{\Gamma }\left(\mathrm{p}\right).\mathrm{\Gamma }\left(\mathrm{q}\right)}{\mathrm{\Gamma }(\mathrm{p}+\mathrm{q})}$$$$2\mathrm{p}-1=2\Rightarrow \mathrm{p}=3\mathrm{and}2\mathrm{q}-1=2\mathrm{n}+1\Rightarrow 2\mathrm{q}=2\mathrm{n}+2\Rightarrow \mathrm{q}=\mathrm{n}+1\Rightarrow$$$$2{\int }_0^{\frac{\pi }{2}}{\cos }^2\mathrm{t}{\sin }^{2\mathrm{n}+1}\mathrm{t}\mathrm{dt}=2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\left(\frac{3}{2}\right)-1}{\sin }^{2(\mathrm{n}+1)-1}\mathrm{t}\mathrm{dt}$$$$=\mathrm{B}(\frac{3}{2},\mathrm{n}+1)=\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)\mathrm{\Gamma }(\mathrm{n}+1)}{\mathrm{\Gamma }(\frac{3}{2}+\mathrm{n}+1)}=\frac{\mathrm{\Gamma }(\frac{1}{2}+1)\mathrm{n}!}{(\mathrm{n}+\frac{3}{2})\mathrm{\Gamma }(\mathrm{n}+\frac{3}{2})}$$$$=\frac{1}{2}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{n}!}{(\mathrm{n}+\frac{3}{2})(\mathrm{n}+\frac{1}{2})\mathrm{\Gamma }(\mathrm{n}+\frac{1}{2})}$$

Commented by PNL last updated on 13/Aug/20

Commented by PNL last updated on 13/Aug/20

$${it}^{\prime }salittledifficulttounderstand$$

Commented by mathmax by abdo last updated on 13/Aug/20

$$\mathrm{i}\mathrm{have}\mathrm{used}\mathrm{\Gamma }(\mathrm{x}+1)=\mathrm{x}\mathrm{\Gamma }\left(\mathrm{x}\right)$$

Commented by PNL last updated on 13/Aug/20

$$whatistheexpressionof\mathrm{\Gamma }?$$

Commented by Aziztisffola last updated on 13/Aug/20

$$\mathrm{\Gamma }\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=(\mathrm{x}-1)!$$

Commented by PNL last updated on 13/Aug/20

$$oktanks.Godblessyou$$

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