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Question Number 40139 by maxmathsup by imad last updated on 16/Jul/18
let  I_n = ∫_0 ^1   x^n (√(1−x)) dx  1) calculate I_0   and I_1   2) prove that  ∀n∈ N^★     (3+2n) I_n =2n I_(n−1)   3) find I_n   interms of n
$${let}\:\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{{n}} \sqrt{\mathrm{1}−{x}}\:{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{I}_{\mathrm{0}} \:\:{and}\:{I}_{\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\:\forall{n}\in\:{N}^{\bigstar} \:\:\:\:\left(\mathrm{3}+\mathrm{2}{n}\right)\:{I}_{{n}} =\mathrm{2}{n}\:{I}_{{n}−\mathrm{1}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{I}_{{n}} \:\:{interms}\:{of}\:{n} \\ $$
Commented by prof Abdo imad last updated on 20/Jul/18
1) I_0 =∫_0 ^1 (√(1−x))dx  =_(x=sin^2 t)   ∫_0 ^(π/2)   cost 2sint cost?dt  =2 ∫_0 ^(π/2)  sint cos^2 t dt =2[−(1/3) cos^3 t]_0 ^(π/2)   =−(2/3){0−1}=(2/3)  I_1 = ∫_0 ^1  x(√(1−x))dx =_((√(1−x))=t)     −∫_0 ^1 (1−t^2 )t(−2tdt)  =2 ∫_0 ^1  t^2 (1−t^2 )dt =2 ∫_0 ^1  (t^2  −t^4 )dt  =2 [(t^3 /3) −(t^5 /5)]_0 ^1  =2{(1/3) −(1/5)}=2(2/(15)) =(4/(15))
$$\left.\mathrm{1}\right)\:{I}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}}{dx}\:\:=_{{x}={sin}^{\mathrm{2}} {t}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cost}\:\mathrm{2}{sint}\:{cost}?{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sint}\:{cos}^{\mathrm{2}} {t}\:{dt}\:=\mathrm{2}\left[−\frac{\mathrm{1}}{\mathrm{3}}\:{cos}^{\mathrm{3}} {t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\left\{\mathrm{0}−\mathrm{1}\right\}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${I}_{\mathrm{1}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\sqrt{\mathrm{1}−{x}}{dx}\:=_{\sqrt{\mathrm{1}−{x}}={t}} \:\:\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right){t}\left(−\mathrm{2}{tdt}\right) \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left({t}^{\mathrm{2}} \:−{t}^{\mathrm{4}} \right){dt} \\ $$$$=\mathrm{2}\:\left[\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\:−\frac{{t}^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{5}}\right\}=\mathrm{2}\frac{\mathrm{2}}{\mathrm{15}}\:=\frac{\mathrm{4}}{\mathrm{15}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 24/Jul/18
2) changement (√(1−x))=t give 1−x=t^2   I_n = ∫_0 ^1   (1−t^2 )^n t  2tdt  =  ∫_0 ^1 2t^2 (1−t^2 )^n  dt  =_(t=sinα)   ∫_0 ^(π/2)  2sin^2 α cos^(2n) α cosα dα  = ∫_0 ^(π/2)  2sin^2 αcos^(2n) α dα  by parts  u^′  =sinα cos^(2n) α  and v=2sinα  I_n = [−(1/(2n+1)) cos^(2n+1) α 2sinα]_0 ^(π/2)  −∫_0 ^(π/2)  −(1/(2n+1))cos^(2n+1) α .2cosα dα  =(2/(2n+1)) ∫_0 ^(π/2)   cos^(2n+2) α dα = (2/(2n+1)) ∫_0 ^(π/2)  (1 −sin^2 α)cos^(2n) α dα  = (2/(2n+1)) ∫_0 ^(π/2)  cos^(2n) α dα −(1/(2n+1)) ∫_0 ^(π/2)  2sin^2 α cos^(2n) α dα  =(2/(2n+1)) ∫_0 ^(π/2) (1−sin^2 α)cos^(2(n−1)) α dα −(1/(2n+1)) I_n  ⇒  (1+(1/(2n+1))) I_n = (2/(2n+1)) ∫_0 ^(π/2)  cos^(2(n−1)) α dα −(1/(2n+1)) I_(n−1)   ⇒  ((2n+2)/(2n+1)) I_n = (2/(2n+1)) ∫_0 ^(π/2)  cos^(2(n−1)) α dα −(1/(2n+1)) I_(n−1)    ...be continued...
$$\left.\mathrm{2}\right)\:{changement}\:\sqrt{\mathrm{1}−{x}}={t}\:{give}\:\mathrm{1}−{x}={t}^{\mathrm{2}} \\ $$$${I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {t}\:\:\mathrm{2}{tdt}\:\:=\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} \:{dt} \\ $$$$=_{{t}={sin}\alpha} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{2}{sin}^{\mathrm{2}} \alpha\:{cos}^{\mathrm{2}{n}} \alpha\:{cos}\alpha\:{d}\alpha \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{2}{sin}^{\mathrm{2}} \alpha{cos}^{\mathrm{2}{n}} \alpha\:{d}\alpha\:\:{by}\:{parts}\:\:{u}^{'} \:={sin}\alpha\:{cos}^{\mathrm{2}{n}} \alpha\:\:{and}\:{v}=\mathrm{2}{sin}\alpha \\ $$$${I}_{{n}} =\:\left[−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{cos}^{\mathrm{2}{n}+\mathrm{1}} \alpha\:\mathrm{2}{sin}\alpha\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{cos}^{\mathrm{2}{n}+\mathrm{1}} \alpha\:.\mathrm{2}{cos}\alpha\:{d}\alpha \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}^{\mathrm{2}{n}+\mathrm{2}} \alpha\:{d}\alpha\:=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}\:−{sin}^{\mathrm{2}} \alpha\right){cos}^{\mathrm{2}{n}} \alpha\:{d}\alpha \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{n}} \alpha\:{d}\alpha\:−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{2}{sin}^{\mathrm{2}} \alpha\:{cos}^{\mathrm{2}{n}} \alpha\:{d}\alpha \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−{sin}^{\mathrm{2}} \alpha\right){cos}^{\mathrm{2}\left({n}−\mathrm{1}\right)} \alpha\:{d}\alpha\:−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{I}_{{n}} \:\Rightarrow \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)\:{I}_{{n}} =\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}\left({n}−\mathrm{1}\right)} \alpha\:{d}\alpha\:−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{I}_{{n}−\mathrm{1}} \:\:\Rightarrow \\ $$$$\frac{\mathrm{2}{n}+\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:{I}_{{n}} =\:\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}\left({n}−\mathrm{1}\right)} \alpha\:{d}\alpha\:−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{I}_{{n}−\mathrm{1}} \:\:\:…{be}\:{continued}… \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 24/Jul/18
3) we have I_n =((2n)/(2n+3)) I_(n−1)     with n≥1 ⇒  Π_(k=1) ^n  I_k =Π_(k=1) ^n   ((2k)/(2k+3)) Π_(k=1) ^n  I_(k−1)   ⇒  I_1 .I_2 .I_3 ...I_n =Π_(k=1) ^n  ((2k)/(2k+3)) I_0 .I_1 .I_2 ....I_(n−1)  ⇒  I_n = Π_(k=1) ^n  ((2k)/(2k+3)) . I_0 = (2/3) Π_(k=1) ^n   ((2k)/(2k+3))
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{I}_{{n}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{3}}\:{I}_{{n}−\mathrm{1}} \:\:\:\:{with}\:{n}\geqslant\mathrm{1}\:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \:{I}_{{k}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{2}{k}}{\mathrm{2}{k}+\mathrm{3}}\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{I}_{{k}−\mathrm{1}} \:\:\Rightarrow \\ $$$${I}_{\mathrm{1}} .{I}_{\mathrm{2}} .{I}_{\mathrm{3}} …{I}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}}{\mathrm{2}{k}+\mathrm{3}}\:{I}_{\mathrm{0}} .{I}_{\mathrm{1}} .{I}_{\mathrm{2}} ….{I}_{{n}−\mathrm{1}} \:\Rightarrow \\ $$$${I}_{{n}} =\:\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}}{\mathrm{2}{k}+\mathrm{3}}\:.\:{I}_{\mathrm{0}} =\:\frac{\mathrm{2}}{\mathrm{3}}\:\prod_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{2}{k}}{\mathrm{2}{k}+\mathrm{3}} \\ $$

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