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Question Number 198074 by sonukgindia last updated on 09/Oct/23
Answered by a.lgnaoui last updated on 10/Oct/23
x^(20) +2x^(10) +1=(x^(10) +1)^2 x^8 =(x^(10) /x^2 )⇒ 50(x^(10) /(x^2 (x^(10) +1)^2 ))=((50)/x)×(x^9 /((x^(10) +1)^2 )) t=x^(10) 10lnx=lnt ⇒x=((ln t)/(10)) x=e^(lnt/10) dt=10x^9 dx dx=(dt/x^9 ) =(dt/c^(9lnt/10) ) 50∫(t^(4/5) /((t+1)^2 e^(lnt/10) ))dt; ( x^2 =^5 (√t) ) ...............
$$\mathrm{x}^{\mathrm{20}} +\mathrm{2x}^{\mathrm{10}} +\mathrm{1}=\left(\mathrm{x}^{\mathrm{10}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{8}} =\frac{\mathrm{x}^{\mathrm{10}} }{\mathrm{x}^{\mathrm{2}} }\Rightarrow \\ $$$$\:\mathrm{50}\frac{\mathrm{x}^{\mathrm{10}} }{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{10}} +\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{50}}{\mathrm{x}}×\frac{\mathrm{x}^{\mathrm{9}} }{\left(\mathrm{x}^{\mathrm{10}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{t}=\mathrm{x}^{\mathrm{10}} \:\:\mathrm{10}\boldsymbol{\mathrm{lnx}}=\boldsymbol{\mathrm{lnt}}\:\:\Rightarrow\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{ln}}\:\boldsymbol{\mathrm{t}}}{\mathrm{10}}\:\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{lnt}}/\mathrm{10}} \\ $$$$\:\:\mathrm{dt}=\mathrm{10x}^{\mathrm{9}} \mathrm{dx}\:\:\:\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{x}^{\mathrm{9}} }\:=\frac{\boldsymbol{\mathrm{dt}}}{\boldsymbol{\mathrm{c}}^{\mathrm{9}\boldsymbol{\mathrm{lnt}}/\mathrm{10}} }\: \\ $$$$ \\ $$$$\:\mathrm{50}\int\frac{\boldsymbol{\mathrm{t}}^{\mathrm{4}/\mathrm{5}} }{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} \boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{lnt}}/\mathrm{10}} }\boldsymbol{\mathrm{dt}};\:\:\:\:\left(\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} =^{\mathrm{5}} \sqrt{\boldsymbol{\mathrm{t}}}\:\right) \\ $$$$…………… \\ $$$$\:\: \\ $$
Answered by witcher3 last updated on 10/Oct/23
x→(1/x) I=∫_0 ^∞ ((50x^(10) )/(x^(20) +2x^(10) +1))dx=∫_0 ^∞ ((50x^(10) )/((x^(10) +1)^2 ))dx =50∫_0 ^∞ (1/(1+x^(10) ))−(1/((x^(10) +1)^2 ))dx x^(10) =t =5∫_0 ^∞ (t^(−(1/9)) /((1+t)))−(t^(−(1/9)) /((t+1)^2 ))dt =5β((8/9),(1/9))−5β((8/9),1+(1/9)) =((5π)/(sin((π/9))))−5,((Γ((8/9))Γ((1/9)))/(9Γ(2)))=((5π)/(sin((π/9)))).(8/9) I=((40π)/(sin((π/9)))) sin((π/9)) can be expressed sin(3x)=2sin(x)(1−sin^2 (x))+(1−2sin^2 (x))sin(x) =3sin(x)−4sin^3 (x) sin(x) root of −4Y^3 +Y=((√3)/2) can be solved cardan formula
$$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{50x}^{\mathrm{10}} }{\mathrm{x}^{\mathrm{20}} +\mathrm{2x}^{\mathrm{10}} +\mathrm{1}}\mathrm{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{50x}^{\mathrm{10}} }{\left(\mathrm{x}^{\mathrm{10}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\mathrm{50}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{10}} }−\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{10}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$${x}^{\mathrm{10}} ={t} \\ $$$$=\mathrm{5}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{9}}} }{\left(\mathrm{1}+\mathrm{t}\right)}−\frac{\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{9}}} }{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\mathrm{5}\beta\left(\frac{\mathrm{8}}{\mathrm{9}},\frac{\mathrm{1}}{\mathrm{9}}\right)−\mathrm{5}\beta\left(\frac{\mathrm{8}}{\mathrm{9}},\mathrm{1}+\frac{\mathrm{1}}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{5}\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{9}}\right)}−\mathrm{5},\frac{\Gamma\left(\frac{\mathrm{8}}{\mathrm{9}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{9}}\right)}{\mathrm{9}\Gamma\left(\mathrm{2}\right)}=\frac{\mathrm{5}\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{9}}\right)}.\frac{\mathrm{8}}{\mathrm{9}} \\ $$$$\mathrm{I}=\frac{\mathrm{40}\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{9}}\right)} \\ $$$$\mathrm{sin}\left(\frac{\pi}{\mathrm{9}}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{expressed} \\ $$$$\mathrm{sin}\left(\mathrm{3x}\right)=\mathrm{2sin}\left(\mathrm{x}\right)\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)+\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)\mathrm{sin}\left(\mathrm{x}\right) \\ $$$$=\mathrm{3sin}\left(\mathrm{x}\right)−\mathrm{4sin}^{\mathrm{3}} \left(\mathrm{x}\right) \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\:\mathrm{root}\:\mathrm{of} \\ $$$$−\mathrm{4Y}^{\mathrm{3}} +\mathrm{Y}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{cardan}\:\mathrm{formula} \\ $$
Commented by mr W last updated on 10/Oct/23
but the question is I=∫_0 ^∞ ((50x^(10 8) )/(x^(20) +2x^(10) +1))dx
$${but}\:{the}\:{question}\:{is} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{50x}^{\cancel{\mathrm{10}}\:\mathrm{8}} }{\mathrm{x}^{\mathrm{20}} +\mathrm{2x}^{\mathrm{10}} +\mathrm{1}}\mathrm{dx} \\ $$
Commented by witcher3 last updated on 10/Oct/23
x→(1/t) =∫_0 ^∞ (((50)/t^8 )/((1/t^(20) )+(2/t^(10) )+1)).(dt/( t^2 ))=∫_0 ^∞ ((50t^(10) )/(t^(20) +2t^(10) +1))
$$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{t}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{50}}{\mathrm{t}^{\mathrm{8}} }}{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{20}} }+\frac{\mathrm{2}}{\mathrm{t}^{\mathrm{10}} }+\mathrm{1}}.\frac{\mathrm{dt}}{\:\mathrm{t}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{50t}^{\mathrm{10}} }{\mathrm{t}^{\mathrm{20}} +\mathrm{2t}^{\mathrm{10}} +\mathrm{1}} \\ $$
Commented by mr W last updated on 10/Oct/23
i see. thanks sir!
$${i}\:{see}.\:{thanks}\:{sir}! \\ $$
Commented by witcher3 last updated on 10/Oct/23
y′re welcom sir
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom}\:\mathrm{sir} \\ $$
Answered by Frix last updated on 10/Oct/23
I=[((5x^9 )/(x^(10) +1))]_0 ^∞ +5∫_0 ^∞ (x^8 /(x^(10) +1))dx= =5∫_0 ^∞ (x^8 /(x^(10) +1))dx I get I=((1+(√5))/2)π
$${I}=\left[\frac{\mathrm{5}{x}^{\mathrm{9}} }{{x}^{\mathrm{10}} +\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} +\mathrm{5}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{x}^{\mathrm{8}} }{{x}^{\mathrm{10}} +\mathrm{1}}{dx}= \\ $$$$=\mathrm{5}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{x}^{\mathrm{8}} }{{x}^{\mathrm{10}} +\mathrm{1}}{dx} \\ $$$$\mathrm{I}\:\mathrm{get}\:{I}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\pi \\ $$
Answered by Mathspace last updated on 10/Oct/23
I=50∫_0 ^∞ (x^8 /((1+x^(10) )^2 ))dx (x=t^(1/(10)) ) =50∫_0 ^∞ (t^(4/5) /((1+t)^2 )).(1/(10))t^((1/(10))−1) dt =5∫_0 ^∞ (t^((9/(10))−1) /((1+t)^2 ))dt B(x,y)=∫_0 ^∞ (t^(x−1) /((1+t)^(x+y) ))dt we take x=(9/(10))and x+y=2 ⇒ y=2−x=2−(9/(10))=((11)/(10)) ⇒ I=5B((9/(10)),((11)/(10))) =5((Γ((9/(10)))Γ(((11)/(10))))/(Γ((9/(10))+((11)/(10))))) =5((Γ((9/(10))).Γ(1+(1/(10))))/(Γ(2))) =(5/(10))Γ((9/(10))).Γ((1/(10))) =(1/2)Γ((1/(10)))Γ(1−(1/(10))) (1/2)×(π/(sin((π/(10)))))⇒I=(π/(2sin((π/(10)))))
$${I}=\mathrm{50}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{8}} }{\left(\mathrm{1}+{x}^{\mathrm{10}} \right)^{\mathrm{2}} }{dx}\:\:\:\left({x}={t}^{\frac{\mathrm{1}}{\mathrm{10}}} \right) \\ $$$$=\mathrm{50}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\frac{\mathrm{4}}{\mathrm{5}}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{10}}{t}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} {dt} \\ $$$$=\mathrm{5}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{9}}{\mathrm{10}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\ $$$${B}\left({x},{y}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{x}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{{x}+{y}} }{dt} \\ $$$${we}\:{take}\:{x}=\frac{\mathrm{9}}{\mathrm{10}}{and}\:{x}+{y}=\mathrm{2}\:\Rightarrow \\ $$$${y}=\mathrm{2}−{x}=\mathrm{2}−\frac{\mathrm{9}}{\mathrm{10}}=\frac{\mathrm{11}}{\mathrm{10}}\:\Rightarrow \\ $$$${I}=\mathrm{5}{B}\left(\frac{\mathrm{9}}{\mathrm{10}},\frac{\mathrm{11}}{\mathrm{10}}\right)\:=\mathrm{5}\frac{\Gamma\left(\frac{\mathrm{9}}{\mathrm{10}}\right)\Gamma\left(\frac{\mathrm{11}}{\mathrm{10}}\right)}{\Gamma\left(\frac{\mathrm{9}}{\mathrm{10}}+\frac{\mathrm{11}}{\mathrm{10}}\right)} \\ $$$$=\mathrm{5}\frac{\Gamma\left(\frac{\mathrm{9}}{\mathrm{10}}\right).\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{10}}\right)}{\Gamma\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{5}}{\mathrm{10}}\Gamma\left(\frac{\mathrm{9}}{\mathrm{10}}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{10}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{10}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{10}}\right)}\Rightarrow{I}=\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{10}}\right)} \\ $$
Commented by Frix last updated on 10/Oct/23
(π/(2sin (π/(10))))=(π/(2×((−1+(√5))/4)))=((1+(√5))/2)π
$$\frac{\pi}{\mathrm{2sin}\:\frac{\pi}{\mathrm{10}}}=\frac{\pi}{\mathrm{2}×\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\pi \\ $$

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