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Question Number 159309 by LEKOUMA last updated on 15/Nov/21

$$Resolve{I}_n={\int }_{-1}^1{(1-x^2)}^{n}dx$$

Answered by mathmax by abdo last updated on 15/Nov/21

$${\mathrm{I}}_{\mathrm{n}}=2{\int }_0^1{(1-{\mathrm{x}}^2)}^{\mathrm{n}}\mathrm{dx}{=}_{\mathrm{x}=\mathrm{sint}}2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}}\mathrm{tcost}\mathrm{dt}$$$$=2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}+1}\mathrm{t}\mathrm{dt}\mathrm{we}\mathrm{know}\mathrm{that}2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{p}-1}\mathrm{t}{\sin }^{2\mathrm{q}-1}\mathrm{t}=\mathrm{B}(\mathrm{p},\mathrm{a})$$$$=\frac{\mathrm{\Gamma }\left(\mathrm{p}\right).\mathrm{\Gamma }\left(\mathrm{q}\right)}{\mathrm{\Gamma }(\mathrm{p}+\mathrm{q})}$$$$2\mathrm{p}-1=2\mathrm{n}+1\Rightarrow \mathrm{p}=\mathrm{n}+1\mathrm{and}2\mathrm{q}-1=0\Rightarrow \mathrm{q}=\frac{1}{2}\Rightarrow$$$$2{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}+1}\mathrm{t}\mathrm{dt}=2{\int }_0^{\frac{\pi }{2}}{\cos }^{2(\mathrm{n}+1)-1}\mathrm{t}{\sin }^{2\left(\frac{1}{2}\right)-1}\mathrm{tdt}$$$$=\mathrm{B}(\mathrm{n}+1,\frac{1}{2})=\frac{\mathrm{\Gamma }(\mathrm{n}+1).\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\mathrm{n}+1+\frac{1}{2})}=\frac{\sqrt{\pi }\mathrm{\Gamma }(\mathrm{n}+1)}{\mathrm{\Gamma }(\mathrm{n}+\frac{3}{2})}$$

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