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Question Number 105534 by I want to learn more last updated on 29/Jul/20

Commented by I want to learn more last updated on 29/Jul/20

$$\mathrm{the}\mathrm{bracket}\mathrm{is}\mathrm{a}\mathrm{floor}\mathrm{function}\mathrm{sir}.\mathrm{error}\mathrm{in}\mathrm{bracket}.$$

Commented by Ar Brandon last updated on 29/Jul/20

OK ��

Answered by Ar Brandon last updated on 29/Jul/20

$$2020{\int }_0^{\frac{\pi }{2}}{\cos }^{2019}\left(\mathrm{x}\right)\mathrm{dx}$$

Commented by mathmax by abdo last updated on 30/Jul/20

$$\mathrm{let}{\mathrm{A}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}+1}\left(\mathrm{x}\right)\mathrm{dx}\Rightarrow {\mathrm{A}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}+2-1}\left(\mathrm{x}\right)\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}-1}\mathrm{x}(1-{\sin }^2\mathrm{x})\mathrm{dx}={\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}-1}\mathrm{x}-{\int }_0^{\frac{\pi }{2}}{\sin }^2\mathrm{x}{\cos }^{2\mathrm{n}-1}\left(\mathrm{x}\right)\mathrm{dx}$$$$={\mathrm{A}}_{\mathrm{n}-1}+{\int }_0^{\frac{\pi }{2}}\mathrm{sinx}(-\mathrm{sinx}){\cos }^{2\mathrm{n}-1}\left(\mathrm{x}\right)\mathrm{dx}\mathrm{by}\mathrm{parts}\mathrm{u}=\mathrm{sinx}\mathrm{and}$$$${\mathrm{v}}^{{}^{\prime }}=-\mathrm{sinx}{\cos }^{2\mathrm{n}-1}\left(\mathrm{x}\right)\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\mathrm{sinx}(-\mathrm{sinx}{\cos }^{2\mathrm{n}-1}\mathrm{x})\mathrm{dx}={\left[\frac{\mathrm{sinx}}{2\mathrm{n}}{\cos }^{2\mathrm{n}}\mathrm{x}\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\mathrm{cosx}\frac{{\cos }^{2\mathrm{n}}\left(\mathrm{x}\right)}{2\mathrm{n}}\mathrm{dx}$$$$=-\frac{1}{2\mathrm{n}}{\int }_0^{\frac{\pi }{2}}{\cos }^{2\mathrm{n}+1}\left(\mathrm{x}\right)\mathrm{dx}=-\frac{1}{2\mathrm{n}}{\mathrm{A}}_{\mathrm{n}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}={\mathrm{A}}_{\mathrm{n}-1}-\frac{1}{2\mathrm{n}}{\mathrm{A}}_{\mathrm{n}}\Rightarrow$$$$(1+\frac{1}{2\mathrm{n}}){\mathrm{A}}_{\mathrm{n}}={\mathrm{A}}_{\mathrm{n}-1}\Rightarrow \frac{2\mathrm{n}+1}{2\mathrm{n}}{\mathrm{A}}_{\mathrm{n}}={\mathrm{A}}_{\mathrm{n}-1}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{2\mathrm{n}}{2\mathrm{n}+1}{\mathrm{A}}_{\mathrm{n}-1}$$$$\Rightarrow \prod _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{A}}_{\mathrm{k}}=\prod _{\mathrm{k}=1}^{\mathrm{n}}\frac{2\mathrm{k}}{2\mathrm{k}+1}\times \prod _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{A}}_{\mathrm{n}-1}\Rightarrow$$$${\mathrm{A}}_1.{\mathrm{A}}_2....{\mathrm{A}}_{\mathrm{n}}=\prod _{\mathrm{k}=1}^{\mathrm{n}}\frac{2\mathrm{k}}{2\mathrm{k}+1}\times {\mathrm{A}}_0.{\mathrm{A}}_1...{\mathrm{A}}_{\mathrm{n}-1}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=\frac{2.4.6....\left(2\mathrm{n}\right)}{3.5.7...(2\mathrm{n}+1)}=\frac{2^{\mathrm{n}}\mathrm{n}!2.4.....\left(2\mathrm{n}\right)}{2.3.4.5.....\left(2\mathrm{n}\right)(2\mathrm{n}+1)}=\frac{2^{2\mathrm{n}}{(\mathrm{n}!)}^2}{(2\mathrm{n}+1)!}$$$$\Rightarrow 2020{\int }_0^{\frac{\pi }{2}}{\cos }^{2019}\mathrm{x}\mathrm{dx}=2020{\mathrm{A}}_{1009}=2020\times \frac{2^{2018}{(1009!)}^2}{\left(2019\right)!}$$

Commented by I want to learn more last updated on 30/Jul/20

$$\mathrm{Wow},\mathrm{thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

Commented by I want to learn more last updated on 30/Jul/20

$$\mathrm{Question}$$$$\mathrm{please}\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{get}2020{\int }_0^{\frac{\pi }{2}}{\cos }^{2019}\mathrm{x}\mathrm{dx}\mathrm{at}\mathrm{first}.$$

Commented by 1549442205PVT last updated on 30/Jul/20

$$\mathbf{Great}\mathbf{Sir}.\mathbf{A}\mathbf{way}\mathbf{instead}\mathbf{of}\mathbf{using}\mathbf{the}$$$$\mathbf{function}\mathbf{Beta}\mathbf{and}\mathbf{Gamma}$$

Commented by mathmax by abdo last updated on 31/Jul/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}$$

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