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Question-196983

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Question Number 196983 by universe last updated on 05/Sep/23
Commented by Frix last updated on 06/Sep/23
I can′t solve the first 2 but wolframalpha can (C is the Catalan constant) 1. I_1 =∫_0 ^π (x^3 /(...))dx=((π(−1440C+1144+15π(−41+20π+24ln 2)))/(3780)) 2. I_2 =∫_0 ^π (x^2 /(...))dx=((−1440C+1144+15π(−41+30π+24ln 2))/(5670)) 3. I_3 =∫_0 ^π (x/(...))dx=((5π)/(63)) 4. I_4 =∫_0 ^π (1/(...))dx=((10)/(63))
$$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{first}\:\mathrm{2}\:\mathrm{but}\:\mathrm{wolframalpha}\:\mathrm{can} \\ $$$$\left(\mathrm{C}\:\mathrm{is}\:\mathrm{the}\:\mathrm{Catalan}\:\mathrm{constant}\right) \\ $$$$\mathrm{1}.\:{I}_{\mathrm{1}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}^{\mathrm{3}} }{…}{dx}=\frac{\pi\left(−\mathrm{1440C}+\mathrm{1144}+\mathrm{15}\pi\left(−\mathrm{41}+\mathrm{20}\pi+\mathrm{24ln}\:\mathrm{2}\right)\right)}{\mathrm{3780}} \\ $$$$\mathrm{2}.\:{I}_{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}^{\mathrm{2}} }{…}{dx}=\frac{−\mathrm{1440C}+\mathrm{1144}+\mathrm{15}\pi\left(−\mathrm{41}+\mathrm{30}\pi+\mathrm{24ln}\:\mathrm{2}\right)}{\mathrm{5670}} \\ $$$$\mathrm{3}.\:{I}_{\mathrm{3}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}}{…}{dx}=\frac{\mathrm{5}\pi}{\mathrm{63}} \\ $$$$\mathrm{4}.\:{I}_{\mathrm{4}} =\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{\mathrm{1}}{…}{dx}=\frac{\mathrm{10}}{\mathrm{63}} \\ $$
Answered by Blackpanther last updated on 05/Sep/23
7
$$\mathrm{7} \\ $$
Commented by Frix last updated on 05/Sep/23
I also get ≈7 by approximating the integral
$$\mathrm{I}\:\mathrm{also}\:\mathrm{get}\:\approx\mathrm{7}\:\mathrm{by}\:\mathrm{approximating}\:\mathrm{the}\:\mathrm{integral} \\ $$
Commented by universe last updated on 05/Sep/23
send your solution sir
$$\:{send}\:{your}\:{solution}\:{sir} \\ $$
Commented by mokys last updated on 05/Sep/23
can you solve by steps
$${can}\:{you}\:{solve}\:{by}\:{steps} \\ $$
Answered by witcher3 last updated on 06/Sep/23
∫_0 ^π ((4x^3 −6πx^2 +2x+1)/((sin(x)+1)^5 ))sin(x)dx=A A=∫_0 ^π ((sin(x))/((sin(x)+1)^5 ))(4(π−x)^3 −6π(π−x)^2 +2(π−x)+1))dx 2A=∫_0 ^π ((sim(x)(−2π^3 +2π+2))/((sin(x)+1)^5 )) A=(−π^3 +π+1)∫_0 ^π ((sin(x))/((sin(x)+1)^5 )) ∫_0 ^π ((sin(x))/((sin(x)+1)^5 ))dx=B ∫_0 ^π ((2tg((x/2))(1+tg^2 ((x/2)))^4 )/((1+tg((x/2)))^(10) ))dx =∫_0 ^∞ ((4x(1+x^2 )^3 )/((1+x)^(10) ))dx=4∫_0 ^∞ (x/((1+x)^(10) ))+12∫_0 ^∞ (x^3 /((1+x)^(10) ))+12∫_0 ^∞ (x^5 /((1+x)^(10) )) +4∫_0 ^∞ (x^7 /((1+x)^(10) )) “β(x,y)=∫_0 ^∞ (t^(x−1) /((1+t)^(x+y) ))” B=4β(2,8)+12β(4,6)+12β(6,4)+4β(8,2) =8((Γ(2)Γ(8))/(Γ(10)))+24((Γ(4)Γ(6))/(Γ(10))) =(1/9)+((24.3.2)/(9.8.7.6))=(3/(63))+(1/9)=((10)/(63)) A=(−π^3 +π^2 +1)B=((10)/(63))(−π^3 +π^2 +1) ((a+3)/(9a))=((10)/(63)),⇒a=7
$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{4x}^{\mathrm{3}} −\mathrm{6}\pi\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}}{\left(\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{1}\right)^{\mathrm{5}} }\mathrm{sin}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{A} \\ $$$$\left.\mathrm{A}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\left(\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{1}\right)^{\mathrm{5}} }\left(\mathrm{4}\left(\pi−\mathrm{x}\right)^{\mathrm{3}} −\mathrm{6}\pi\left(\pi−\mathrm{x}\right)^{\mathrm{2}} +\mathrm{2}\left(\pi−\mathrm{x}\right)+\mathrm{1}\right)\right)\mathrm{dx} \\ $$$$\mathrm{2A}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sim}\left(\mathrm{x}\right)\left(−\mathrm{2}\pi^{\mathrm{3}} +\mathrm{2}\pi+\mathrm{2}\right)}{\left(\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{1}\right)^{\mathrm{5}} } \\ $$$$\mathrm{A}=\left(−\pi^{\mathrm{3}} +\pi+\mathrm{1}\right)\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\left(\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{1}\right)^{\mathrm{5}} } \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\left(\mathrm{sin}\left(\mathrm{x}\right)+\mathrm{1}\right)^{\mathrm{5}} }\mathrm{dx}=\mathrm{B} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)^{\mathrm{4}} }{\left(\mathrm{1}+\mathrm{tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)^{\mathrm{10}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{10}} }\mathrm{dx}=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{10}} }+\mathrm{12}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{10}} }+\mathrm{12}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{5}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{10}} } \\ $$$$+\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{7}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{10}} } \\ $$$$“\beta\left(\mathrm{x},\mathrm{y}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{x}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{x}+\mathrm{y}} }'' \\ $$$$\mathrm{B}=\mathrm{4}\beta\left(\mathrm{2},\mathrm{8}\right)+\mathrm{12}\beta\left(\mathrm{4},\mathrm{6}\right)+\mathrm{12}\beta\left(\mathrm{6},\mathrm{4}\right)+\mathrm{4}\beta\left(\mathrm{8},\mathrm{2}\right) \\ $$$$=\mathrm{8}\frac{\Gamma\left(\mathrm{2}\right)\Gamma\left(\mathrm{8}\right)}{\Gamma\left(\mathrm{10}\right)}+\mathrm{24}\frac{\Gamma\left(\mathrm{4}\right)\Gamma\left(\mathrm{6}\right)}{\Gamma\left(\mathrm{10}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{24}.\mathrm{3}.\mathrm{2}}{\mathrm{9}.\mathrm{8}.\mathrm{7}.\mathrm{6}}=\frac{\mathrm{3}}{\mathrm{63}}+\frac{\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{10}}{\mathrm{63}} \\ $$$$\mathrm{A}=\left(−\pi^{\mathrm{3}} +\pi^{\mathrm{2}} +\mathrm{1}\right)\mathrm{B}=\frac{\mathrm{10}}{\mathrm{63}}\left(−\pi^{\mathrm{3}} +\pi^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\frac{\mathrm{a}+\mathrm{3}}{\mathrm{9a}}=\frac{\mathrm{10}}{\mathrm{63}},\Rightarrow\mathrm{a}=\mathrm{7} \\ $$$$ \\ $$$$ \\ $$
Commented by universe last updated on 06/Sep/23
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by lain_math last updated on 17/Sep/23
a=7 :)
$${a}=\mathrm{7} \\ $$$$\left.:\right) \\ $$

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