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Question Number 107835 by PNL last updated on 12/Aug/20
compute In=∫_0 ^1 x^n (√(1−x)) dx
$${compute}\:{In}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \sqrt{\mathrm{1}−{x}}\:{dx} \\ $$
Answered by hgrocks last updated on 12/Aug/20
I_n = ∫_0 ^1 x^n (1−x)^(1/2) dx = β(n+1,(3/2))
$$\mathrm{I}_{\mathrm{n}} \:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{x}^{\mathrm{n}} \left(\mathrm{1}−\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{dx} \\ $$$$\:\:\:\:\:=\:\beta\left(\mathrm{n}+\mathrm{1},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$
Answered by mathmax by abdo last updated on 12/Aug/20
I_n =∫_0 ^1 x^n (√(1−x))dx we do the changement x =sin^2 t ⇒ I_n =∫_0 ^(π/2) sin^(2n) t cost (2sint)cost dt =2∫_0 ^(π/2) cos^2 t sin^(2n+1) t dt we know 2∫_0 ^(π/2) cos^(2p−1) x sin^(2q−1) x dx =B(p,q) =((Γ(p).Γ(q))/(Γ(p+q))) 2p−1 =2 ⇒p=3 and 2q−1 =2n+1 ⇒2q=2n+2 ⇒q=n+1 ⇒ 2 ∫_0 ^(π/2) cos^2 t sin^(2n+1) t dt =2 ∫_0 ^(π/2) cos^(2((3/2))−1) sin^(2(n+1)−1) t dt =B((3/2),n+1) =((Γ((3/2))Γ(n+1))/(Γ((3/2)+n+1))) =((Γ((1/2)+1)n!)/((n+(3/2))Γ(n+(3/2)))) =(1/2)((Γ((1/2))n!)/((n+(3/2))(n+(1/2))Γ(n+(1/2))))
$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{n}} \sqrt{\mathrm{1}−\mathrm{x}}\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}\:=\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2n}} \mathrm{t}\:\mathrm{cost}\:\left(\mathrm{2sint}\right)\mathrm{cost}\:\mathrm{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\mathrm{cos}^{\mathrm{2}} \mathrm{t}\:\:\mathrm{sin}^{\mathrm{2n}+\mathrm{1}} \mathrm{t}\:\mathrm{dt}\:\:\mathrm{we}\:\mathrm{know} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2p}−\mathrm{1}} \mathrm{x}\:\mathrm{sin}^{\mathrm{2q}−\mathrm{1}} \mathrm{x}\:\mathrm{dx}\:=\mathrm{B}\left(\mathrm{p},\mathrm{q}\right)\:=\frac{\Gamma\left(\mathrm{p}\right).\Gamma\left(\mathrm{q}\right)}{\Gamma\left(\mathrm{p}+\mathrm{q}\right)} \\ $$$$\mathrm{2p}−\mathrm{1}\:=\mathrm{2}\:\Rightarrow\mathrm{p}=\mathrm{3}\:\mathrm{and}\:\mathrm{2q}−\mathrm{1}\:=\mathrm{2n}+\mathrm{1}\:\Rightarrow\mathrm{2q}=\mathrm{2n}+\mathrm{2}\:\Rightarrow\mathrm{q}=\mathrm{n}+\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2}} \mathrm{t}\:\mathrm{sin}^{\mathrm{2n}+\mathrm{1}} \mathrm{t}\:\mathrm{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{1}} \:\mathrm{sin}^{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)−\mathrm{1}} \mathrm{t}\:\mathrm{dt} \\ $$$$=\mathrm{B}\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{n}+\mathrm{1}\right)\:=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{n}+\mathrm{1}\right)}\:=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\mathrm{n}!}{\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{n}!}{\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$
Commented by PNL last updated on 13/Aug/20
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Commented by PNL last updated on 13/Aug/20
it′s a little difficult to understand
$${it}'{s}\:{a}\:{little}\:{difficult}\:{to}\:{understand} \\ $$
Commented by mathmax by abdo last updated on 13/Aug/20
i have used Γ(x+1) =xΓ(x)
$$\mathrm{i}\:\mathrm{have}\:\mathrm{used}\:\Gamma\left(\mathrm{x}+\mathrm{1}\right)\:=\mathrm{x}\Gamma\left(\mathrm{x}\right) \\ $$
Commented by PNL last updated on 13/Aug/20
what is the expression of Γ?
$${what}\:{is}\:{the}\:{expression}\:{of}\:\Gamma? \\ $$
Commented by Aziztisffola last updated on 13/Aug/20
Γ(x)=∫_0 ^( ∞) t^(x−1) e^(−t) dt=(x−1)!
$$\Gamma\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\:\infty} \mathrm{t}^{\mathrm{x}−\mathrm{1}} \mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}=\left(\mathrm{x}−\mathrm{1}\right)! \\ $$
Commented by PNL last updated on 13/Aug/20
ok tanks. God bless you
$${ok}\:{tanks}.\:{God}\:{bless}\:{you} \\ $$

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