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Question Number 125028 by ngahcedric last updated on 07/Dec/20

Answered by Dwaipayan Shikari last updated on 07/Dec/20

∫_0 ^4 x^n (16−x^2 )^(1/2) dx x=4t =4∫_0 ^1 4^n t^n (16−16t^2 )^(1/2) dt =4^(n+2) ∫_0 ^1 t^n (1−t^2 )^(1/2) dt =4^(n+(3/2)) ∫_0 ^1 u^((n/2)−(1/2)) (1−u)^(1/2) du t^2 =u =4^(n+(3/2)) ((Γ((n/2)+(1/2))Γ((3/2)))/(Γ((n/2)+2))) =2^(2n+2) (√π)((Γ((n/2)+(1/2)))/(Γ((n/2)+2)))

$$\int_{\mathrm{0}} ^{\mathrm{4}} {x}^{{n}} \left(\mathrm{16}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{4}{t} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{4}^{{n}} {t}^{{n}} \left(\mathrm{16}−\mathrm{16}{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dt} \\ $$$$=\mathrm{4}^{{n}+\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dt}\:=\mathrm{4}^{{n}+\frac{\mathrm{3}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{du}\:\:\:\:\:\:{t}^{\mathrm{2}} ={u} \\ $$$$=\mathrm{4}^{{n}+\frac{\mathrm{3}}{\mathrm{2}}} \frac{\Gamma\left(\frac{{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{n}}{\mathrm{2}}+\mathrm{2}\right)}\:=\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \sqrt{\pi}\frac{\Gamma\left(\frac{{n}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{n}}{\mathrm{2}}+\mathrm{2}\right)} \\ $$

Commented by ngahcedric last updated on 07/Dec/20

don′t really understand please read the question well again

$${don}'{t}\:{really}\:{understand} \\ $$$${please}\:{read}\:{the}\:{question}\:{well}\:{again} \\ $$

Answered by mathmax by abdo last updated on 07/Dec/20

I_n =∫_0 ^(4 ) x^n (√(16−x^2 ))dx we do the changement x=4sint ⇒ I_n =∫_0 ^(π/2) 4^n sin^n t(4cost)4cost dt =4^(n+2) ∫_0 ^(π/2) sin^n t cos^2 t dt we hsve 2 ∫_0 ^(π/2) sin^(2p−1) t cos^(2q−1) t dt =B(p,q)=((Γ(p).Γ(q))/(Γ(p+q))) 2p−1=n ⇒p=((n+1)/2) and 2q−1=2 ⇒q=(3/2) ⇒ I_n =(4^(n+2) /2)B(((n+1)/2),(3/2)) =(4^(n+2) /2)×((Γ(((n+1)/2)).Γ((3/2)))/(Γ(((n+1)/2)+(3/2)))) =8.4^n ×((Γ(((n+1)/2))×(1/2)Γ((1/2)))/(Γ(((n+4)/2)))) =4^(n+1) Γ((1/2))×((Γ(((n+1)/2)))/(Γ(((n+4)/2)))) rest to prove that (I_n /I_(n−2) )=16×((n−1)/(n+2)) ...be continued...

$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{4}\:} \mathrm{x}^{\mathrm{n}} \sqrt{\mathrm{16}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\mathrm{4sint}\:\Rightarrow \\ $$$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{4}^{\mathrm{n}} \:\mathrm{sin}^{\mathrm{n}} \mathrm{t}\left(\mathrm{4cost}\right)\mathrm{4cost}\:\mathrm{dt}\:=\mathrm{4}^{\mathrm{n}+\mathrm{2}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{n}} \mathrm{t}\:\mathrm{cos}^{\mathrm{2}} \mathrm{t}\:\mathrm{dt} \\ $$$$\mathrm{we}\:\mathrm{hsve}\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{sin}^{\mathrm{2p}−\mathrm{1}} \mathrm{t}\:\mathrm{cos}^{\mathrm{2q}−\mathrm{1}} \mathrm{t}\:\mathrm{dt}\:=\mathrm{B}\left(\mathrm{p},\mathrm{q}\right)=\frac{\Gamma\left(\mathrm{p}\right).\Gamma\left(\mathrm{q}\right)}{\Gamma\left(\mathrm{p}+\mathrm{q}\right)} \\ $$$$\mathrm{2p}−\mathrm{1}=\mathrm{n}\:\Rightarrow\mathrm{p}=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\:\:\mathrm{and}\:\mathrm{2q}−\mathrm{1}=\mathrm{2}\:\Rightarrow\mathrm{q}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{I}_{\mathrm{n}} =\frac{\mathrm{4}^{\mathrm{n}+\mathrm{2}} }{\mathrm{2}}\mathrm{B}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\frac{\mathrm{4}^{\mathrm{n}+\mathrm{2}} }{\mathrm{2}}×\frac{\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right).\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$=\mathrm{8}.\mathrm{4}^{\mathrm{n}} \:×\frac{\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)×\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{n}+\mathrm{4}}{\mathrm{2}}\right)}\:=\mathrm{4}^{\mathrm{n}+\mathrm{1}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)×\frac{\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{n}+\mathrm{4}}{\mathrm{2}}\right)} \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:\:\frac{\mathrm{I}_{\mathrm{n}} }{\mathrm{I}_{\mathrm{n}−\mathrm{2}} }=\mathrm{16}×\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}+\mathrm{2}} \\ $$$$...\mathrm{be}\:\mathrm{continued}... \\ $$

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