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Question Number 42798 by maxmathsup by imad last updated on 02/Sep/18
find I_n = ∫_0 ^1 x^n (√(1−x^2 ))dx
$${find}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 05/Sep/18
changement x=sint give I_n = ∫_0 ^(π/2) sin^n t cos^2 t dt =∫_0 ^(π/2) sin^n t(1−sin^2 t)dt = ∫_0 ^(π/2) sin^n t dt −∫_0 ^(π/2) sin^(n+2) dt =w_n −w_(n+2) let find w_n = ∫_0 ^(π/2) sin^n t dt we have w_n =∫_0 ^(π/2) sin^(n−2) t(1−cos^2 t)dt =w_(n−2) −∫_0 ^(π/2) cost( cost sin^(n−2) t)dt by parts u=cost and v^′ =cost sin^(n−2) t⇒ ∫_0 ^(π/2) cost (cost sin^(n−2) t)dt =[(1/(n−1)) cost sin^(n−1) t]_0 ^(π/2) −∫_0 ^(π/2) ((−sint)/(n−1)) sin^(n−1) tdt = ∫_0 ^(π/2) ((sin^n t)/(n−1)) dt =(1/(n−1)) w_n ⇒w_n =w_(n−2) −(1/(n−1)) w_n ⇒(1+(1/(n−1)))w_n =w_(n−2) ⇒ (n/(n−1))w_n =w_(n−2) ⇒w_n =((n−1)/n)w_(n−2) ⇒w_(2n) =((2n−1)/(2n))w_(2n−2) and w_(2n+1) =((2n)/(2n+1)) w_(2n−1) let find w_(2n) Π_(k=1) ^n w_(2k) =Π_(k=1) ^n ((2k−1)/(2k)) w_(2k−2) ⇒w_2 .w_4 .....w_(2n) =((1.3.5.....(2n−1))/((2)(4)....(2n)))w_0 w_2 ...w_(2n−2) ⇒ w_(2n) = ((1.3.5...(2n−1))/(2^n (n!))) w_0 = (((2n)!)/(2^(2n) (n!)^2 )) (π/2) let find w_(2n+1) I_(2n) =w_(2n) −w_(2n+2) = (((2n)!)/(2^(2n) (n!)^2 )) (π/2) − (((2n+4)!)/(2^(2n+4) {(n+2)!}^2 )) (π/2) let find w_(2n+1) we have w_(2n+1) =((2n)/(2n+1)) w_(2n−1) ⇒Π_(k=1) ^n w_(2k+1) =Π_(k=1) ^n ((2k)/(2k+1)) Π_(k=1) ^n w_(2k−1) ⇒ w_3 .w_5 .....w_(2n+1) =((2^n (n!))/(3.5....(2n+1))) w_1 .w_3 ....w_(2n−1) ⇒ w_(2n+1) = ((2^(2n) (n!)^2 )/((2n+1)!)) w_1 but w_1 =1 ⇒ I_(2n−1) =w_(2n−1) −w_(2n+1) =((2^(2n−2) {(n−1)!}^2 )/((2n−1)!)) −((2^(2n) (n!)^2 )/((2n+1)!)) .
$${changement}\:{x}={sint}\:{give}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{{n}} {t}\:{cos}^{\mathrm{2}} {t}\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{sin}^{{n}} {t}\left(\mathrm{1}−{sin}^{\mathrm{2}} {t}\right){dt}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} {t}\:{dt}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{sin}^{{n}+\mathrm{2}} {dt}\:={w}_{{n}} −{w}_{{n}+\mathrm{2}} \\ $$$${let}\:{find}\:{w}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{{n}} {t}\:{dt}\:\:{we}\:{have}\:{w}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{{n}−\mathrm{2}} {t}\left(\mathrm{1}−{cos}^{\mathrm{2}} {t}\right){dt} \\ $$$$={w}_{{n}−\mathrm{2}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cost}\left(\:{cost}\:{sin}^{{n}−\mathrm{2}} {t}\right){dt}\:\:{by}\:{parts}\:{u}={cost}\:{and}\:{v}^{'} ={cost}\:{sin}^{{n}−\mathrm{2}} {t}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cost}\:\left({cost}\:{sin}^{{n}−\mathrm{2}} {t}\right){dt}\:=\left[\frac{\mathrm{1}}{{n}−\mathrm{1}}\:{cost}\:{sin}^{{n}−\mathrm{1}} {t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{−{sint}}{{n}−\mathrm{1}}\:{sin}^{{n}−\mathrm{1}} {tdt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sin}^{{n}} {t}}{{n}−\mathrm{1}}\:{dt}\:=\frac{\mathrm{1}}{{n}−\mathrm{1}}\:{w}_{{n}} \:\Rightarrow{w}_{{n}} ={w}_{{n}−\mathrm{2}} −\frac{\mathrm{1}}{{n}−\mathrm{1}}\:{w}_{{n}} \:\Rightarrow\left(\mathrm{1}+\frac{\mathrm{1}}{{n}−\mathrm{1}}\right){w}_{{n}} ={w}_{{n}−\mathrm{2}} \:\Rightarrow \\ $$$$\frac{{n}}{{n}−\mathrm{1}}{w}_{{n}} \:={w}_{{n}−\mathrm{2}} \:\Rightarrow{w}_{{n}} =\frac{{n}−\mathrm{1}}{{n}}{w}_{{n}−\mathrm{2}} \Rightarrow{w}_{\mathrm{2}{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}{w}_{\mathrm{2}{n}−\mathrm{2}} \:{and} \\ $$$${w}_{\mathrm{2}{n}+\mathrm{1}} \:=\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}\:{w}_{\mathrm{2}{n}−\mathrm{1}} \:\:{let}\:{find}\:{w}_{\mathrm{2}{n}} \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \:{w}_{\mathrm{2}{k}} \:=\prod_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}{k}}\:{w}_{\mathrm{2}{k}−\mathrm{2}} \:\Rightarrow{w}_{\mathrm{2}} .{w}_{\mathrm{4}} …..{w}_{\mathrm{2}{n}} =\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}…..\left(\mathrm{2}{n}−\mathrm{1}\right)}{\left(\mathrm{2}\right)\left(\mathrm{4}\right)….\left(\mathrm{2}{n}\right)}{w}_{\mathrm{0}} {w}_{\mathrm{2}} …{w}_{\mathrm{2}{n}−\mathrm{2}} \Rightarrow \\ $$$${w}_{\mathrm{2}{n}} =\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}…\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}^{{n}} \left({n}!\right)}\:{w}_{\mathrm{0}} =\:\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\:\frac{\pi}{\mathrm{2}}\:\:\:{let}\:{find}\:{w}_{\mathrm{2}{n}+\mathrm{1}} \\ $$$${I}_{\mathrm{2}{n}} ={w}_{\mathrm{2}{n}} −{w}_{\mathrm{2}{n}+\mathrm{2}} =\:\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\:\frac{\pi}{\mathrm{2}}\:−\:\frac{\left(\mathrm{2}{n}+\mathrm{4}\right)!}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{4}} \:\left\{\left({n}+\mathrm{2}\right)!\right\}^{\mathrm{2}} }\:\frac{\pi}{\mathrm{2}}\:\:{let}\:{find}\:{w}_{\mathrm{2}{n}+\mathrm{1}} \\ $$$${we}\:{have}\:{w}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}\:{w}_{\mathrm{2}{n}−\mathrm{1}} \:\Rightarrow\prod_{{k}=\mathrm{1}} ^{{n}} \:{w}_{\mathrm{2}{k}+\mathrm{1}} \:=\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{k}}{\mathrm{2}{k}+\mathrm{1}}\:\prod_{{k}=\mathrm{1}} ^{{n}} {w}_{\mathrm{2}{k}−\mathrm{1}} \:\Rightarrow \\ $$$${w}_{\mathrm{3}} .{w}_{\mathrm{5}} …..{w}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}^{{n}} \left({n}!\right)}{\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{w}_{\mathrm{1}} .{w}_{\mathrm{3}} ….{w}_{\mathrm{2}{n}−\mathrm{1}} \Rightarrow \\ $$$${w}_{\mathrm{2}{n}+\mathrm{1}} =\:\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:{w}_{\mathrm{1}} \:\:\:{but}\:{w}_{\mathrm{1}} =\mathrm{1}\:\Rightarrow\:{I}_{\mathrm{2}{n}−\mathrm{1}} ={w}_{\mathrm{2}{n}−\mathrm{1}} \:−{w}_{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} \left\{\left({n}−\mathrm{1}\right)!\right\}^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)!}\:−\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:. \\ $$
Commented by maxmathsup by imad last updated on 05/Sep/18
another way changement (√(1−x^2 ))=t give x^2 =1−t^2 ⇒dx =−dt ⇒ I_n = ∫_0 ^1 (1−t^2 )^(n/2) dt ⇒ I_(2n) = ∫_0 ^1 (1−t^2 )^n dt =∫_0 ^1 Σ_(k=0) ^n C_n ^k (−1)^k t^(2k) dt =Σ_(k=0) ^n (−1)^k C_n ^k (1/(2k+1)) =Σ_(k=0) ^n (((−1)^k )/(2k+1)) C_n ^k I_(2n+1) = ∫_0 ^1 (1−t^2 )^(n+(1/2)) dt = ∫_0 ^1 (1−t^2 )^n (√(1−t^2 ))dt = ∫_0 ^1 (√(1−t^2 ))(Σ_(k=0) ^n C_n ^k (−1)^k t^(2k) )dt = Σ_(k=0) ^n (−1)^k C_n ^k ∫_0 ^1 t^(2k) (√(1−t^2 ))dt then we calculate w_k =∫_0 ^1 t^(2k) (√(1−t^2 ))dt by recurrence....
$${another}\:{way}\:{changement}\:\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }={t}\:{give}\:{x}^{\mathrm{2}} \:=\mathrm{1}−{t}^{\mathrm{2}} \:\Rightarrow{dx}\:=−{dt}\:\Rightarrow \\ $$$${I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{{n}}{\mathrm{2}}} \:\:{dt}\:\:\Rightarrow\:{I}_{\mathrm{2}{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} \:{t}^{\mathrm{2}{k}} {dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:\:\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\:{C}_{{n}} ^{{k}} \\ $$$${I}_{\mathrm{2}{n}+\mathrm{1}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} {dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} \:{t}^{\mathrm{2}{k}} \right){dt} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{k}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:\:{then}\:{we}\:{calculate}\:{w}_{{k}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{k}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt} \\ $$$${by}\:{recurrence}…. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18
x=sinα dx=cosα dα ∫_0 ^(Π/2) sin^n α cos^2 α dα using gamma beta function ∫_0 ^(Π/2) sin^(2p−1) αcos^(2q−1) ∝ dα =((⌈p ⌈q)/(2⌈(p+q))) 2p−1=n p=((n+1)/2) 2q−1=2 q=(3/2) ∫_0 ^(Π/2) sin^(2×((n+1)/2)−1) α×cos^(2×(3/2)−1) α dα =((⌈(((n+1)/2))×⌈((3/2)))/(2⌈(((n+4)/2)))) pls check
$${x}={sin}\alpha\:\:{dx}={cos}\alpha\:{d}\alpha \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{{n}} \alpha\:{cos}^{\mathrm{2}} \alpha\:{d}\alpha \\ $$$${using}\:{gamma}\:{beta}\:{function} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \alpha{cos}^{\mathrm{2}{q}−\mathrm{1}} \propto\:{d}\alpha \\ $$$$=\frac{\lceil{p}\:\lceil{q}}{\mathrm{2}\lceil\left({p}+{q}\right)} \\ $$$$\mathrm{2}{p}−\mathrm{1}={n}\:\:{p}=\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\:\:\mathrm{2}{q}−\mathrm{1}=\mathrm{2}\:\:\:{q}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\mathrm{2}×\frac{{n}+\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \alpha×{cos}^{\mathrm{2}×\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} \alpha\:{d}\alpha \\ $$$$=\frac{\lceil\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)×\lceil\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}\lceil\left(\frac{{n}+\mathrm{4}}{\mathrm{2}}\right)} \\ $$$${pls}\:{check} \\ $$

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