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Question Number 138112 by 676597498 last updated on 10/Apr/21

$$provethat$$$${\int }_0^{\frac{\pi }{2}}\left({sin}^{2k}\left(x\right)dx\right)=\frac{\left(2k\right)!2^{-2k-1}}{{(k!)}^2}$$

Answered by Dwaipayan Shikari last updated on 10/Apr/21

$${\int }_0^{\frac{\pi }{2}}{sin}^{2a-1}x{cos}^{2b-1}xdx=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{2\mathrm{\Gamma }(a+b)}$$$$Matchingthis$$$${\int }_0^{\frac{\pi }{2}}{sin}^{2k}\left(x\right)dx={\int }_0^{\frac{\pi }{2}}{sin}^{2(k+1)-1}x{cos}^{2\left(\frac{1}{2}\right)-1}x=\frac{\mathrm{\Gamma }(k+1)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(k+\frac{3}{2})}$$

Answered by mathmax by abdo last updated on 10/Apr/21

$$\mathrm{let}{\mathrm{I}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}}\mathrm{xdx}\Rightarrow {\mathrm{I}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}-2}\mathrm{x}(1-{\cos }^2\mathrm{x})\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}-2}\mathrm{xdx}-{\int }_0^{\frac{\pi }{2}}{\cos }^2\mathrm{x}{\sin }^{\mathrm{n}-2}\mathrm{xdx}\mathrm{we}[\mathrm{have}$$$${\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}-2}\mathrm{xdx}={\mathrm{I}}_{\mathrm{n}-2}\mathrm{and}\mathrm{by}\mathrm{recurrence}$$$${\int }_0^{\frac{\pi }{2}}\mathrm{cosx}\left(\mathrm{cosx}{\sin }^{\mathrm{n}-2}\mathrm{x}\right)\mathrm{dx}={\left[\frac{{\sin }^{\mathrm{n}-1}\mathrm{x}}{\mathrm{n}-1}\mathrm{cosx}\right]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}\mathrm{sinx}\frac{{\sin }^{\mathrm{n}-1}\mathrm{x}}{\mathrm{n}-1}\mathrm{dx}$$$$=\frac{1}{\mathrm{n}-1}{\mathrm{I}}_{\mathrm{n}}\Rightarrow {\mathrm{I}}_{\mathrm{n}}={\mathrm{I}}_{\mathrm{n}-2}-\frac{1}{\mathrm{n}-1}{\mathrm{I}}_{\mathrm{n}}\Rightarrow (1+\frac{1}{\mathrm{n}-1}){\mathrm{I}}_{\mathrm{n}}={\mathrm{I}}_{\mathrm{n}-2}\Rightarrow$$$$\frac{\mathrm{n}}{\mathrm{n}-1}{\mathrm{I}}_{\mathrm{n}}={\mathrm{I}}_{\mathrm{n}-2}\Rightarrow {\mathrm{I}}_{\mathrm{n}}=\frac{\mathrm{n}-1}{\mathrm{n}}{\mathrm{I}}_{\mathrm{n}-2}\Rightarrow {\mathrm{I}}_{2\mathrm{n}}=\frac{2\mathrm{n}-1}{2\mathrm{n}}{\mathrm{I}}_{2\mathrm{n}-2}(\mathrm{n}⩾1)$$$$\Rightarrow \prod _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{I}}_{2\mathrm{k}}=\prod _{\mathrm{k}=1}^{\mathrm{n}}\frac{2\mathrm{k}-1}{2\mathrm{k}}{\mathrm{I}}_{2\mathrm{k}-2}\Rightarrow$$$${\mathrm{I}}_2.{\mathrm{I}}_4......{\mathrm{I}}_{2\mathrm{n}}=\frac{\prod _{\mathrm{k}=1}^{\mathrm{n}}(2\mathrm{k}-1)}{2^{\mathrm{n}}\mathrm{n}!}{\mathrm{I}}_0.{\mathrm{I}}_2.....{\mathrm{I}}_{2\mathrm{n}-2}\Rightarrow$$$${\mathrm{I}}_{2\mathrm{n}}=\frac{1.3.5....(2\mathrm{n}-1)}{2^{\mathrm{n}}.\mathrm{n}!}{\mathrm{I}}_0=\frac{\pi }{2}\frac{1.2.3.4.5.....(2\mathrm{n}-1)\left(2\mathrm{n}\right)}{2^{\mathrm{n}}\mathrm{n}!.2^{\mathrm{n}.}\mathrm{n}!}$$$$=\frac{\pi }{2}\times \frac{\left(2\mathrm{n}\right)!}{2^{2\mathrm{n}}{(\mathrm{n}!)}^2}\Rightarrow {\int }_0^{\frac{\pi }{2}}{\sin }^{2\mathrm{n}}\mathrm{xdx}=\frac{\pi \left(2\mathrm{n}\right)!}{2^{2\mathrm{n}+1}{(\mathrm{n}!)}^2}$$

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