Question Number 93907 by abdomathmax last updated on 16/May/20
$$\left.\mathrm{1}\right)\:{calculate}\:\:{A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{{n}} {x}\:{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{{n}} } \\ $$$${n}\:{integr}\:{natural} \\ $$
Commented by mathmax by abdo last updated on 17/May/20
$$\left.\mathrm{1}\right)\:{A}_{{n}+\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{{n}+\mathrm{2}} {x}\:{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right){cos}^{{n}} {x}\:{dx} \\ $$$$={A}_{{n}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} {x}\:{cos}^{{n}} {x}\:{dx}\:{but}\:{by}\:{parts}\:{u}\:={sinx}\:{and}\:{v}^{'} \:={sinx}\:{cos}^{{n}} {x} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} {x}\:{cos}^{{n}} {xdx}\:=\left[−\frac{\mathrm{1}}{{n}+\mathrm{1}}{cos}^{{n}+\mathrm{1}} {x}\:{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}{cos}^{{n}+\mathrm{2}} \:{xdx}\:\Rightarrow \\ $$$${A}_{{n}+\mathrm{2}} ={A}_{{n}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{A}_{{n}+\mathrm{2}\:} \:\Rightarrow\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){A}_{{n}+\mathrm{2}} ={A}_{{n}} \:\Rightarrow \\ $$$${A}_{{n}+\mathrm{2}} =\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}\:{A}_{{n}} \:\Rightarrow\:{A}_{{n}} =\frac{{n}−\mathrm{1}}{{n}}\:{A}_{{n}−\mathrm{2}} \\ $$$${A}_{\mathrm{2}{k}} =\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}{k}}\:{A}_{\mathrm{2}{k}−\mathrm{2}} \:\:\Rightarrow\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{{A}_{\mathrm{2}{k}} }{{A}_{\mathrm{2}{k}−\mathrm{2}} }\:=\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}{k}}\:\Rightarrow \\ $$$$\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{0}} }×\frac{{A}_{\mathrm{4}} }{{A}_{\mathrm{2}} }×….\frac{{A}_{\mathrm{2}{n}} }{{A}_{\mathrm{2}{n}−\mathrm{2}} }\:=\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}…..\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}^{{n}} \:{n}!}\:=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}…\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)}{\mathrm{2}^{{n}} {n}!\:\mathrm{2}^{{n}} {n}!} \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\:\Rightarrow{A}_{\mathrm{2}{n}} =\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\:{A}_{\mathrm{0}} \:\Rightarrow{A}_{\mathrm{2}{n}} =\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }×\frac{\pi}{\mathrm{2}} \\ $$$${A}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}\:{A}_{\mathrm{2}{n}−\mathrm{1}} \:\Rightarrow\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{{A}_{\mathrm{2}{k}+\mathrm{1}} }{{A}_{\mathrm{2}{k}−\mathrm{1}} }\:=\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}}{\mathrm{2}{k}+\mathrm{1}}\:\Rightarrow \\ $$$$\frac{{A}_{\mathrm{3}} }{{A}_{\mathrm{1}} }×\frac{{A}_{\mathrm{5}} }{{A}_{\mathrm{3}} }×…..\frac{{A}_{\mathrm{2}{n}+\mathrm{1}} }{{A}_{\mathrm{2}{n}−\mathrm{1}} }\:=\frac{\mathrm{2}^{{n}} {n}!}{\mathrm{3}.\mathrm{5}…..\left(\mathrm{2}{n}+\mathrm{1}\right)}\:=\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}….\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow{A}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}×{A}_{\mathrm{1}} \:\:{but}\:{A}_{\mathrm{1}} =\mathrm{1}\:\Rightarrow \\ $$$${A}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 17/May/20
$$\left.\mathrm{2}\right)\:{let}\:{I}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{{n}} }\:\Rightarrow{I}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left\{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right\}^{{n}} } \\ $$$$=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \:\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:=\:\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} } \\ $$$$=_{{t}={tan}\theta} \:\:\:\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{{n}} } \\ $$$$=\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{{n}−\mathrm{1}} }\:=\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}−\mathrm{2}} \theta\:{d}\theta \\ $$$${but}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{n}} \theta\:{d}\theta\:={A}_{\mathrm{2}{n}} =\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }×\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${I}_{{n}} =\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} ×\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }×\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 17/May/20
$$\mathrm{Amazing}. \\ $$$$\mathrm{Please}\:\mathrm{I}'\mathrm{ll}\:\mathrm{like}\:\mathrm{to}\:\mathrm{know}\:\mathrm{is}\:\mathrm{A}_{\mathrm{2n}} \:\mathrm{same}\:\mathrm{as} \\ $$$$\mathrm{A}_{\mathrm{2n}} =\frac{\left(\mathrm{2n}−\mathrm{1}\right)!}{\left(\mathrm{2n}\right)!}×\frac{\pi}{\mathrm{2}}\:\:\:\:? \\ $$
Commented by mathmax by abdo last updated on 17/May/20
$${sir}\:{verify}\:{for}\:{n}=\mathrm{1},\mathrm{2},\mathrm{3}\:\:{and}\:{prove}\:{it}\:{by}\:{recurrence}..{if}\:{itis}\:{correct}… \\ $$
Answered by Ar Brandon last updated on 16/May/20
$$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{3}}{\mathrm{4}}\centerdot\frac{\mathrm{5}}{\mathrm{6}}\centerdot\centerdot\centerdot\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}\centerdot\frac{\pi}{\mathrm{2}}\:\mathrm{if}\:\mathrm{n}\:\mathrm{is}\:\mathrm{even} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{5}}\centerdot\frac{\mathrm{6}}{\mathrm{7}}\centerdot\centerdot\centerdot\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}\:\:\mathrm{if}\:\mathrm{n}\:\mathrm{is}\:\mathrm{odd} \\ $$
Answered by Ar Brandon last updated on 16/May/20
$$\mathrm{2}\backslash\mathrm{B}=\int_{−\infty} ^{+\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}} }=\int_{−\infty} ^{+\infty} \frac{\mathrm{dx}}{\left[\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \right]^{\mathrm{n}} } \\ $$$$\mathrm{Consider}\:\mathrm{J}_{\mathrm{n}} =\int_{−\infty} ^{+\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{n}} }\Rightarrow\mathrm{J}_{\mathrm{n}} =\frac{\mathrm{2n}−\mathrm{3}}{\mathrm{2u}^{\mathrm{2}} \left(\mathrm{n}−\mathrm{1}\right)}\mathrm{J}_{\mathrm{n}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{u}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow\mathrm{B}_{\mathrm{n}} =\frac{\mathrm{4n}−\mathrm{6}}{\mathrm{3}\left(\mathrm{n}−\mathrm{1}\right)}\mathrm{B}_{\mathrm{n}−\mathrm{1}} \\ $$