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Question Number 127774 by Bird last updated on 02/Jan/21

calculate u_n =∫_0 ^1 x^n (√(1−x^4 ))dx

$${calculate}\:\:{u}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Jan/21

∫_0 ^1 x^n (√(1−x^4 )) x^4 =u =(1/4)∫_0 ^1 u^((n/4)−(1/3)) (1−u)^(1/2) du=(1/4).((Γ(n+(2/3))Γ((3/2)))/(Γ(n+((13)/6))))

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }\:\:\:\:\:{x}^{\mathrm{4}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{{n}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {du}=\frac{\mathrm{1}}{\mathrm{4}}.\frac{\Gamma\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left({n}+\frac{\mathrm{13}}{\mathrm{6}}\right)} \\ $$

Answered by mathmax by abdo last updated on 02/Jan/21

u_n =∫_0 ^1 x^n (√(1−x^4 ))dx we do the changement x=t^(1/4) ⇒ u_n =∫_0 ^1 t^(n/4) (1−t)^(1/2) (1/4)t^((1/4)−1) dt =(1/4)∫_0 ^1 t^((n/4)−(3/4)) (1−t)^(1/2) dt =(1/4)∫_0 ^1 t^(((n+1)/4)−1) (1−t)^((3/2)−1) dt =(1/4)B(((n+1)/4),(3/2)) =(1/4)×((Γ(((n+1)/4))×Γ((3/2)))/(Γ(((n+1)/4)+(3/2)))) =(1/4)Γ((3/2))((Γ(((n+1)/4)))/(Γ(((n+7)/4)))) Γ((3/2))=Γ((1/2)+1)=(1/2)Γ((1/2))=((√π)/2) ⇒u_n =((√π)/8)×((Γ(((n+1)/4)))/(Γ(((n+7)/4))))

$$\mathrm{u}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{n}} \sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow \\ $$$$\mathrm{u}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\mathrm{t}^{\frac{\mathrm{n}}{\mathrm{4}}} \left(\mathrm{1}−\mathrm{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\frac{\mathrm{n}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}} \:\left(\mathrm{1}−\mathrm{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\frac{\mathrm{n}+\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−\mathrm{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{B}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{4}}\right)×\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\frac{\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{n}+\mathrm{7}}{\mathrm{4}}\right)} \\ $$$$\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow\mathrm{u}_{\mathrm{n}} =\frac{\sqrt{\pi}}{\mathrm{8}}×\frac{\Gamma\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{n}+\mathrm{7}}{\mathrm{4}}\right)} \\ $$

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