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Question Number 93907 by abdomathmax last updated on 16/May/20

1) calculate A_n =∫_0 ^(π/2) cos^n x dx 2) calculate ∫_(−∞) ^(+∞) (dx/((x^2 −x+1)^n )) n integr natural

$$\left.\mathrm{1}\right)\:{calculate}\:\:{A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{{n}} {x}\:{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{{n}} } \\ $$$${n}\:{integr}\:{natural} \\ $$

Commented by mathmax by abdo last updated on 17/May/20

1) A_(n+2) =∫_0 ^(π/2) cos^(n+2) x dx =∫_0 ^(π/2) (1−sin^2 x)cos^n x dx =A_n −∫_0 ^(π/2) sin^2 x cos^n x dx but by parts u =sinx and v^′ =sinx cos^n x ∫_0 ^(π/2) sin^2 x cos^n xdx =[−(1/(n+1))cos^(n+1) x sinx]_0 ^(π/2) +∫_0 ^(π/2) (1/(n+1))cos^(n+2) xdx ⇒ A_(n+2) =A_n −(1/(n+1)) A_(n+2 ) ⇒(1+(1/(n+1)))A_(n+2) =A_n ⇒ A_(n+2) =((n+1)/(n+2)) A_n ⇒ A_n =((n−1)/n) A_(n−2) A_(2k) =((2k−1)/(2k)) A_(2k−2) ⇒Π_(k=1) ^n (A_(2k) /A_(2k−2) ) =Π_(k=1) ^n ((2k−1)/(2k)) ⇒ (A_2 /A_0 )×(A_4 /A_2 )×....(A_(2n) /A_(2n−2) ) =((1.3.5.....(2n−1))/(2^n n!)) =((1.2.3.4...(2n−1)(2n))/(2^n n! 2^n n!)) =(((2n)!)/(2^(2n) (n!)^2 )) ⇒A_(2n) =(((2n)!)/(2^(2n) (n!)^2 )) A_0 ⇒A_(2n) =(((2n)!)/(2^(2n) (n!)^2 ))×(π/2) A_(2n+1) =((2n)/(2n+1)) A_(2n−1) ⇒Π_(k=1) ^n (A_(2k+1) /A_(2k−1) ) =Π_(k=1) ^n ((2k)/(2k+1)) ⇒ (A_3 /A_1 )×(A_5 /A_3 )×.....(A_(2n+1) /A_(2n−1) ) =((2^n n!)/(3.5.....(2n+1))) =((2^(2n) (n!)^2 )/(2.3.4.5....(2n)(2n+1))) =((2^(2n) (n!)^2 )/((2n+1)!)) ⇒A_(2n+1) =((2^(2n) (n!)^2 )/((2n+1)!))×A_1 but A_1 =1 ⇒ A_(2n+1) =((2^(2n) (n!)^2 )/((2n+1)!))

$$\left.\mathrm{1}\right)\:{A}_{{n}+\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{{n}+\mathrm{2}} {x}\:{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right){cos}^{{n}} {x}\:{dx} \\ $$$$={A}_{{n}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} {x}\:{cos}^{{n}} {x}\:{dx}\:{but}\:{by}\:{parts}\:{u}\:={sinx}\:{and}\:{v}^{'} \:={sinx}\:{cos}^{{n}} {x} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} {x}\:{cos}^{{n}} {xdx}\:=\left[−\frac{\mathrm{1}}{{n}+\mathrm{1}}{cos}^{{n}+\mathrm{1}} {x}\:{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}{cos}^{{n}+\mathrm{2}} \:{xdx}\:\Rightarrow \\ $$$${A}_{{n}+\mathrm{2}} ={A}_{{n}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{A}_{{n}+\mathrm{2}\:} \:\Rightarrow\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){A}_{{n}+\mathrm{2}} ={A}_{{n}} \:\Rightarrow \\ $$$${A}_{{n}+\mathrm{2}} =\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}\:{A}_{{n}} \:\Rightarrow\:{A}_{{n}} =\frac{{n}−\mathrm{1}}{{n}}\:{A}_{{n}−\mathrm{2}} \\ $$$${A}_{\mathrm{2}{k}} =\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}{k}}\:{A}_{\mathrm{2}{k}−\mathrm{2}} \:\:\Rightarrow\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{{A}_{\mathrm{2}{k}} }{{A}_{\mathrm{2}{k}−\mathrm{2}} }\:=\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}{k}}\:\Rightarrow \\ $$$$\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{0}} }×\frac{{A}_{\mathrm{4}} }{{A}_{\mathrm{2}} }×....\frac{{A}_{\mathrm{2}{n}} }{{A}_{\mathrm{2}{n}−\mathrm{2}} }\:=\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.....\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}^{{n}} \:{n}!}\:=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}...\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)}{\mathrm{2}^{{n}} {n}!\:\mathrm{2}^{{n}} {n}!} \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\:\Rightarrow{A}_{\mathrm{2}{n}} =\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\:{A}_{\mathrm{0}} \:\Rightarrow{A}_{\mathrm{2}{n}} =\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }×\frac{\pi}{\mathrm{2}} \\ $$$${A}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}\:{A}_{\mathrm{2}{n}−\mathrm{1}} \:\Rightarrow\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{{A}_{\mathrm{2}{k}+\mathrm{1}} }{{A}_{\mathrm{2}{k}−\mathrm{1}} }\:=\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}}{\mathrm{2}{k}+\mathrm{1}}\:\Rightarrow \\ $$$$\frac{{A}_{\mathrm{3}} }{{A}_{\mathrm{1}} }×\frac{{A}_{\mathrm{5}} }{{A}_{\mathrm{3}} }×.....\frac{{A}_{\mathrm{2}{n}+\mathrm{1}} }{{A}_{\mathrm{2}{n}−\mathrm{1}} }\:=\frac{\mathrm{2}^{{n}} {n}!}{\mathrm{3}.\mathrm{5}.....\left(\mathrm{2}{n}+\mathrm{1}\right)}\:=\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}....\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\Rightarrow{A}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}×{A}_{\mathrm{1}} \:\:{but}\:{A}_{\mathrm{1}} =\mathrm{1}\:\Rightarrow \\ $$$${A}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 17/May/20

2) let I_n =∫_(−∞) ^(+∞) (dx/((x^2 −x+1)^n )) ⇒I_n =∫_(−∞) ^(+∞) (dx/({(x−(1/2))^2 +(3/4)}^n )) =_(x−(1/2)=((√3)/2)t) ((4/3))^n ∫_(−∞) ^(+∞) (1/((t^2 +1)^n ))×((√3)/2)dt =((√3)/2)((4/3))^n ∫_(−∞) ^(+∞) (dt/((t^2 +1)^n )) = (√3)((4/3))^n ∫_0 ^∞ (dt/((1+t^2 )^n )) =_(t=tanθ) (√3)((4/3))^n ∫_0 ^(π/2) (((1+tan^2 θ)dθ)/((1+tan^2 θ)^n )) =(√3)((4/3))^n ∫_0 ^(π/2) (dθ/((1+tan^2 θ)^(n−1) )) =(√3)((4/3))^n ∫_0 ^(π/2) cos^(2n−2) θ dθ but ∫_0 ^(π/2) cos^(2n) θ dθ =A_(2n) =(((2n)!)/(2^(2n) (n!)^2 ))×(π/2) ⇒ I_n =(√3)((4/3))^n ×(((2n−2)!)/(2^(2n−2) ((n−1)!)^2 ))×(π/2)

$$\left.\mathrm{2}\right)\:{let}\:{I}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{{n}} }\:\Rightarrow{I}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left\{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right\}^{{n}} } \\ $$$$=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \:\:\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:=\:\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} } \\ $$$$=_{{t}={tan}\theta} \:\:\:\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{{n}} } \\ $$$$=\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{{n}−\mathrm{1}} }\:=\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}−\mathrm{2}} \theta\:{d}\theta \\ $$$${but}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{n}} \theta\:{d}\theta\:={A}_{\mathrm{2}{n}} =\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }×\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${I}_{{n}} =\sqrt{\mathrm{3}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{{n}} ×\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }×\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 17/May/20

Amazing. Please I′ll like to know is A_(2n) same as A_(2n) =(((2n−1)!)/((2n)!))×(π/2) ?

$$\mathrm{Amazing}. \\ $$$$\mathrm{Please}\:\mathrm{I}'\mathrm{ll}\:\mathrm{like}\:\mathrm{to}\:\mathrm{know}\:\mathrm{is}\:\mathrm{A}_{\mathrm{2n}} \:\mathrm{same}\:\mathrm{as} \\ $$$$\mathrm{A}_{\mathrm{2n}} =\frac{\left(\mathrm{2n}−\mathrm{1}\right)!}{\left(\mathrm{2n}\right)!}×\frac{\pi}{\mathrm{2}}\:\:\:\:? \\ $$

Commented by mathmax by abdo last updated on 17/May/20

sir verify for n=1,2,3 and prove it by recurrence..if itis correct...

$${sir}\:{verify}\:{for}\:{n}=\mathrm{1},\mathrm{2},\mathrm{3}\:\:{and}\:{prove}\:{it}\:{by}\:{recurrence}..{if}\:{itis}\:{correct}... \\ $$

Answered by Ar Brandon last updated on 16/May/20

A_n =(1/2)∙(3/4)∙(5/6)∙∙∙((n−1)/n)∙(π/2) if n is even =(2/3)∙(4/5)∙(6/7)∙∙∙((n−1)/n) if n is odd

$$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{3}}{\mathrm{4}}\centerdot\frac{\mathrm{5}}{\mathrm{6}}\centerdot\centerdot\centerdot\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}\centerdot\frac{\pi}{\mathrm{2}}\:\mathrm{if}\:\mathrm{n}\:\mathrm{is}\:\mathrm{even} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{5}}\centerdot\frac{\mathrm{6}}{\mathrm{7}}\centerdot\centerdot\centerdot\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}\:\:\mathrm{if}\:\mathrm{n}\:\mathrm{is}\:\mathrm{odd} \\ $$

Answered by Ar Brandon last updated on 16/May/20

2\B=∫_(−∞) ^(+∞) (dx/((x^2 −x+1)^n ))=∫_(−∞) ^(+∞) (dx/([(x−(1/2))^2 +(((√3)/2))^2 ]^n )) Consider J_n =∫_(−∞) ^(+∞) (dx/((x^2 +u^2 )^n ))⇒J_n =((2n−3)/(2u^2 (n−1)))J_(n−1) ⇒u=((√3)/2)⇒B_n =((4n−6)/(3(n−1)))B_(n−1)

$$\mathrm{2}\backslash\mathrm{B}=\int_{−\infty} ^{+\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}} }=\int_{−\infty} ^{+\infty} \frac{\mathrm{dx}}{\left[\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \right]^{\mathrm{n}} } \\ $$$$\mathrm{Consider}\:\mathrm{J}_{\mathrm{n}} =\int_{−\infty} ^{+\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{n}} }\Rightarrow\mathrm{J}_{\mathrm{n}} =\frac{\mathrm{2n}−\mathrm{3}}{\mathrm{2u}^{\mathrm{2}} \left(\mathrm{n}−\mathrm{1}\right)}\mathrm{J}_{\mathrm{n}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{u}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow\mathrm{B}_{\mathrm{n}} =\frac{\mathrm{4n}−\mathrm{6}}{\mathrm{3}\left(\mathrm{n}−\mathrm{1}\right)}\mathrm{B}_{\mathrm{n}−\mathrm{1}} \\ $$

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