Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 140588 by mnjuly1970 last updated on 09/May/21

                          .......Advanced ....β˜…β˜…β˜…....Calculus.......          evaluation the value of :                    𝛗 :=∫_0 ^( (Ο€/2)) sin^2 (x).ln(sin(x))dx            solution::         ΞΎ (a):=∫_0 ^( (Ο€/2)) sin^(2+a) (x)dx =(1/2)Ξ² (((3+a)/2) ,(1/2))              :=(1/2)(((Ξ“(((3+a)/2))Ξ“((1/2)))/(Ξ“(2+(a/2))))).......βœ“             𝛗:= ΞΎ β€² (0) ..............βœ“             :=(1/2) (βˆšΟ€) (( Ξ“β€²(((3+a)/2)).Ξ“(2+(a/2))βˆ’Ξ“(((3+a)/2)).Ξ“β€²(2+(a/2)))/(Ξ“^2 (2+(a/2)))) ∣_(a=0)             :=(1/2)(βˆšΟ€)  ((Ξ“β€²((3/2))βˆ’Ξ“((3/2)).Ξ“β€²(2))/(( Ξ“^2 (2):=1 )))            :=(1/2)(βˆšΟ€) ((ψ((3/2))Ξ“((3/2))βˆ’Ξ“((3/2)).ψ(2))/1)            := ((βˆšΟ€)/4){ (2βˆ’Ξ³βˆ’2ln(2)βˆ’(1βˆ’Ξ³)}            :=((βˆšΟ€)/4)(1βˆ’ln(4))=(βˆšΟ€) ln(((e/4))^(1/4) )

.......Advanced....β˜…β˜…β˜…....Calculus.......evaluationthevalueof:Ο•:=∫0Ο€2sin2(x).ln(sin(x))dxsolution::ΞΎ(a):=∫0Ο€2sin2+a(x)dx=12Ξ²(3+a2,12):=12(Ξ“(3+a2)Ξ“(12)Ξ“(2+a2)).......βœ“Ο•:=ΞΎβ€²(0)..............βœ“:=12πΓ′(3+a2).Ξ“(2+a2)βˆ’Ξ“(3+a2).Ξ“β€²(2+a2)Ξ“2(2+a2)∣a=0:=12πΓ′(32)βˆ’Ξ“(32).Ξ“β€²(2)(Ξ“2(2):=1):=12Ο€Οˆ(32)Ξ“(32)βˆ’Ξ“(32).ψ(2)1:=Ο€4{(2βˆ’Ξ³βˆ’2ln(2)βˆ’(1βˆ’Ξ³)}:=Ο€4(1βˆ’ln(4))=Ο€ln(e44)

Answered by mathmax by abdo last updated on 10/May/21

Ξ¦=∫_0 ^(Ο€/2)  sin^2 x log(sinx)dx let f(a)=∫_0 ^(Ο€/2)  sin^a x dx β‡’  f(a)=∫_0 ^(Ο€/2)  e^(alog(sinx))  dx β‡’f^β€² (a)=∫_0 ^(Ο€/2)  log(sinx) sin^a x dx β‡’  f^β€² (2)=∫_0 ^(Ο€/2)  sin^2 x log(sinx)dx =Ξ¦  we have f(a)=∫_0 ^(Ο€/2)   sin^(2(((a+1)/2))βˆ’1)  cos^(2((1/2))βˆ’1) x dx  =(1/2)B(((a+1)/2),(1/2)) =(1/2).((Ξ“(((a+1)/2)).Ξ“((1/2)))/(Ξ“(((a+1)/(2 ))+(1/2))))=((βˆšΟ€)/2).((Ξ“(((a+1)/2)))/(Ξ“((a/2)+1)))  =((βˆšΟ€)/2)Γ—((Ξ“(((a+1)/2)))/((a/2)Ξ“((a/2)))) =((βˆšΟ€)/a) Γ—((Ξ“(((a+1)/2)))/(Ξ“((a/2)))) β‡’  f^β€² (a) =βˆ’((βˆšΟ€)/a^2 )Γ—((Ξ“(((a+1)/2)))/(Ξ“((a/2))))+((βˆšΟ€)/a)Γ—(((1/2)Ξ“^β€² (((a+1)/2)).Ξ“((a/2))βˆ’(1/2)Ξ“^β€² ((a/2))Ξ“(((a+1)/2)))/(Ξ“^2 ((a/2))))  β‡’ Ξ¦=f^β€² (2)=βˆ’((βˆšΟ€)/4)Γ—((Ξ“((3/2)))/1) +((βˆšΟ€)/4)Γ—((Ξ“^β€² ((3/2)).1βˆ’Ξ“^β€² (1).Ξ“((3/2)))/1^2 )  =βˆ’((βˆšΟ€)/4).Ξ“((3/2)) +((βˆšΟ€)/4)Γ—(Ξ“^β€² ((3/2))+Ξ³ .Ξ“((3/2)))  Ξ“^β€² (1)=∫_0 ^∞  e^(βˆ’t)  logt dt =βˆ’Ξ³

Ξ¦=∫0Ο€2sin2xlog(sinx)dxletf(a)=∫0Ο€2sinaxdxβ‡’f(a)=∫0Ο€2ealog(sinx)dxβ‡’fβ€²(a)=∫0Ο€2log(sinx)sinaxdxβ‡’fβ€²(2)=∫0Ο€2sin2xlog(sinx)dx=Ξ¦wehavef(a)=∫0Ο€2sin2(a+12)βˆ’1cos2(12)βˆ’1xdx=12B(a+12,12)=12.Ξ“(a+12).Ξ“(12)Ξ“(a+12+12)=Ο€2.Ξ“(a+12)Ξ“(a2+1)=Ο€2Γ—Ξ“(a+12)a2Ξ“(a2)=Ο€aΓ—Ξ“(a+12)Ξ“(a2)β‡’fβ€²(a)=βˆ’Ο€a2Γ—Ξ“(a+12)Ξ“(a2)+Ο€aΓ—12Ξ“β€²(a+12).Ξ“(a2)βˆ’12Ξ“β€²(a2)Ξ“(a+12)Ξ“2(a2)β‡’Ξ¦=fβ€²(2)=βˆ’Ο€4Γ—Ξ“(32)1+Ο€4Γ—Ξ“β€²(32).1βˆ’Ξ“β€²(1).Ξ“(32)12=βˆ’Ο€4.Ξ“(32)+Ο€4Γ—(Ξ“β€²(32)+Ξ³.Ξ“(32))Ξ“β€²(1)=∫0∞eβˆ’tlogtdt=βˆ’Ξ³

Terms of Service

Privacy Policy

Contact: info@tinkutara.com