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
$$\mathrm{2020}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2019}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 30/Jul/20
![let A_n =∫_0 ^(π/2) cos^(2n+1) (x)dx ⇒ A_n =∫_0 ^(π/2) cos^(2n+2−1) (x)dx =∫_0 ^(π/2) cos^(2n−1) x(1−sin^2 x)dx =∫_0 ^(π/2) cos^(2n−1) x−∫_0 ^(π/2) sin^2 x cos^(2n−1) (x)dx =A_(n−1) +∫_0 ^(π/2) sinx(−sinx)cos^(2n−1) (x)dx by parts u=sinx and v^′ =−sinx cos^(2n−1) (x) ⇒ ∫_0 ^(π/2) sinx(−sinx cos^(2n−1) x)dx =[((sinx)/(2n)) cos^(2n) x]_0 ^(π/2) −∫_0 ^(π/2) cosx((cos^(2n) (x))/(2n))dx =−(1/(2n)) ∫_0 ^(π/2) cos^(2n+1) (x)dx =−(1/(2n)) A_n ⇒A_n =A_(n−1) −(1/(2n)) A_n ⇒ (1+(1/(2n)))A_n =A_(n−1 ) ⇒((2n+1)/(2n)) A_n =A_(n−1) ⇒A_n =((2n)/(2n+1)) A_(n−1) ⇒Π_(k=1) ^n A_k =Π_(k=1) ^n ((2k)/(2k+1)) ×Π_(k=1) ^n A_(n−1) ⇒ A_1 .A_2 ....A_n =Π_(k=1) ^n ((2k)/(2k+1))×A_0 .A_1 ...A_(n−1) ⇒ A_n =((2.4.6....(2n))/(3.5.7...(2n+1))) =((2^n n!2.4.....(2n))/(2.3.4.5.....(2n)(2n+1))) =((2^(2n) (n!)^2 )/((2n+1)!)) ⇒2020∫_0 ^(π/2) cos^(2019) x dx =2020 A_(1009) =2020×((2^(2018) (1009!)^2 )/((2019)!))](https://www.tinkutara.com/question/Q105561.png)
$$\mathrm{let}\:\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2n}+\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx}\:\:\Rightarrow\:\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2n}+\mathrm{2}−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2n}−\mathrm{1}} \mathrm{x}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2n}−\mathrm{1}} \mathrm{x}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{cos}^{\mathrm{2n}−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=\mathrm{A}_{\mathrm{n}−\mathrm{1}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sinx}\left(−\mathrm{sinx}\right)\mathrm{cos}^{\mathrm{2n}−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx}\:\mathrm{by}\:\mathrm{parts}\:\mathrm{u}=\mathrm{sinx}\:\mathrm{and} \\ $$$$\mathrm{v}^{'} \:=−\mathrm{sinx}\:\mathrm{cos}^{\mathrm{2n}−\mathrm{1}} \left(\mathrm{x}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sinx}\left(−\mathrm{sinx}\:\mathrm{cos}^{\mathrm{2n}−\mathrm{1}} \mathrm{x}\right)\mathrm{dx}\:=\left[\frac{\mathrm{sinx}}{\mathrm{2n}}\:\mathrm{cos}^{\mathrm{2n}} \mathrm{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cosx}\frac{\mathrm{cos}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\mathrm{2n}}\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2n}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2n}+\mathrm{1}} \left(\mathrm{x}\right)\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2n}}\:\mathrm{A}_{\mathrm{n}} \:\Rightarrow\mathrm{A}_{\mathrm{n}} =\mathrm{A}_{\mathrm{n}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2n}}\:\mathrm{A}_{\mathrm{n}} \:\Rightarrow \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}}\right)\mathrm{A}_{\mathrm{n}} =\mathrm{A}_{\mathrm{n}−\mathrm{1}\:} \:\:\Rightarrow\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{2n}}\:\mathrm{A}_{\mathrm{n}} =\mathrm{A}_{\mathrm{n}−\mathrm{1}} \:\Rightarrow\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{2n}}{\mathrm{2n}+\mathrm{1}}\:\mathrm{A}_{\mathrm{n}−\mathrm{1}} \\ $$$$\Rightarrow\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{A}_{\mathrm{k}} \:=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{2k}}{\mathrm{2k}+\mathrm{1}}\:×\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{A}_{\mathrm{n}−\mathrm{1}} \:\Rightarrow \\ $$$$\mathrm{A}_{\mathrm{1}} .\mathrm{A}_{\mathrm{2}} ….\mathrm{A}_{\mathrm{n}} =\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{2k}}{\mathrm{2k}+\mathrm{1}}×\mathrm{A}_{\mathrm{0}} .\mathrm{A}_{\mathrm{1}} …\mathrm{A}_{\mathrm{n}−\mathrm{1}} \:\:\Rightarrow \\ $$$$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}….\left(\mathrm{2n}\right)}{\mathrm{3}.\mathrm{5}.\mathrm{7}…\left(\mathrm{2n}+\mathrm{1}\right)}\:=\frac{\mathrm{2}^{\mathrm{n}} \mathrm{n}!\mathrm{2}.\mathrm{4}…..\left(\mathrm{2n}\right)}{\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}…..\left(\mathrm{2n}\right)\left(\mathrm{2n}+\mathrm{1}\right)}\:=\frac{\mathrm{2}^{\mathrm{2n}} \left(\mathrm{n}!\right)^{\mathrm{2}} }{\left(\mathrm{2n}+\mathrm{1}\right)!} \\ $$$$\Rightarrow\mathrm{2020}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\mathrm{cos}^{\mathrm{2019}} \mathrm{x}\:\mathrm{dx}\:=\mathrm{2020}\:\mathrm{A}_{\mathrm{1009}} \:=\mathrm{2020}×\frac{\mathrm{2}^{\mathrm{2018}} \left(\mathrm{1009}!\right)^{\mathrm{2}} }{\left(\mathrm{2019}\right)!} \\ $$
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
$$\mathrm{Question} \\ $$$$\mathrm{please}\:\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\:\:\:\mathrm{2020}\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2019}} \mathrm{x}\:\:\mathrm{dx}\:\:\:\:\:\:\mathrm{at}\:\mathrm{first}. \\ $$
Commented by 1549442205PVT last updated on 30/Jul/20
$$\boldsymbol{\mathrm{Great}}\:\boldsymbol{\mathrm{Sir}}.\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{way}}\:\boldsymbol{\mathrm{instead}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{Beta}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{Gamma}} \\ $$
Commented by mathmax by abdo last updated on 31/Jul/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$