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Question Number 125028 by ngahcedric last updated on 07/Dec/20

Answered by Dwaipayan Shikari last updated on 07/Dec/20

$${\int }_0^4{x}^n{(16-x^2)}^{\frac{1}{2}}dxx=4t$$$$=4{\int }_0^1{4}^n{t}^n{(16-16t^2)}^{\frac{1}{2}}dt$$$$=4^{n+2}{\int }_0^1{t}^n{(1-t^2)}^{\frac{1}{2}}dt=4^{n+\frac{3}{2}}{\int }_0^1{u}^{\frac{n}{2}-\frac{1}{2}}{(1-u)}^{\frac{1}{2}}du{t}^2=u$$$$=4^{n+\frac{3}{2}}\frac{\mathrm{\Gamma }(\frac{n}{2}+\frac{1}{2})\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(\frac{n}{2}+2)}=2^{2n+2}\sqrt{\pi }\frac{\mathrm{\Gamma }(\frac{n}{2}+\frac{1}{2})}{\mathrm{\Gamma }(\frac{n}{2}+2)}$$

Commented by ngahcedric last updated on 07/Dec/20

$${don}^{\prime }treallyunderstand$$$$pleasereadthequestionwellagain$$

Answered by mathmax by abdo last updated on 07/Dec/20

$${\mathrm{I}}_{\mathrm{n}}={\int }_0^4{\mathrm{x}}^{\mathrm{n}}\sqrt{16-{\mathrm{x}}^2}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=4\mathrm{sint}\Rightarrow$$$${\mathrm{I}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}{4}^{\mathrm{n}}{\sin }^{\mathrm{n}}\mathrm{t}\left(4\mathrm{cost}\right)4\mathrm{cost}\mathrm{dt}=4^{\mathrm{n}+2}{\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}}\mathrm{t}{\cos }^2\mathrm{t}\mathrm{dt}$$$$\mathrm{we}\mathrm{hsve}2{\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{p}-1}\mathrm{t}{\cos }^{2\mathrm{q}-1}\mathrm{t}\mathrm{dt}=\mathrm{B}(\mathrm{p},\mathrm{q})=\frac{\mathrm{\Gamma }\left(\mathrm{p}\right).\mathrm{\Gamma }\left(\mathrm{q}\right)}{\mathrm{\Gamma }(\mathrm{p}+\mathrm{q})}$$$$2\mathrm{p}-1=\mathrm{n}\Rightarrow \mathrm{p}=\frac{\mathrm{n}+1}{2}\mathrm{and}2\mathrm{q}-1=2\Rightarrow \mathrm{q}=\frac{3}{2}\Rightarrow$$$${\mathrm{I}}_{\mathrm{n}}=\frac{4^{\mathrm{n}+2}}{2}\mathrm{B}(\frac{\mathrm{n}+1}{2},\frac{3}{2})=\frac{4^{\mathrm{n}+2}}{2}\times \frac{\mathrm{\Gamma }\left(\frac{\mathrm{n}+1}{2}\right).\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(\frac{\mathrm{n}+1}{2}+\frac{3}{2})}$$$$=8.4^{\mathrm{n}}\times \frac{\mathrm{\Gamma }\left(\frac{\mathrm{n}+1}{2}\right)\times \frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{\mathrm{n}+4}{2}\right)}=4^{\mathrm{n}+1}\mathrm{\Gamma }\left(\frac{1}{2}\right)\times \frac{\mathrm{\Gamma }\left(\frac{\mathrm{n}+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{\mathrm{n}+4}{2}\right)}$$$$\mathrm{rest}\mathrm{to}\mathrm{prove}\mathrm{that}\frac{{\mathrm{I}}_{\mathrm{n}}}{{\mathrm{I}}_{\mathrm{n}-2}}=16\times \frac{\mathrm{n}-1}{\mathrm{n}+2}$$$$...\mathrm{be}\mathrm{continued}...$$

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