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Question Number 140588 by mnjuly1970 last updated on 09/May/21
.......Advanced....β β β ....Calculus.......evaluationthevalueof:Ο:=β«0Ο2sin2(x).ln(sin(x))dxsolution::ΞΎ(a):=β«0Ο2sin2+a(x)dx=12Ξ²(3+a2,12):=12(Ξ(3+a2)Ξ(12)Ξ(2+a2)).......βΟ:=ΞΎβ²(0)..............β:=12ΟΞβ²(3+a2).Ξ(2+a2)βΞ(3+a2).Ξβ²(2+a2)Ξ2(2+a2)β£a=0:=12ΟΞβ²(32)βΞ(32).Ξβ²(2)(Ξ2(2):=1):=12ΟΟ(32)Ξ(32)βΞ(32).Ο(2)1:=Ο4{(2βΞ³β2ln(2)β(1βΞ³)}:=Ο4(1βln(4))=Οln(e44)
Answered by mathmax by abdo last updated on 10/May/21
Ξ¦=β«0Ο2sin2xlog(sinx)dxletf(a)=β«0Ο2sinaxdxβf(a)=β«0Ο2ealog(sinx)dxβfβ²(a)=β«0Ο2log(sinx)sinaxdxβfβ²(2)=β«0Ο2sin2xlog(sinx)dx=Ξ¦wehavef(a)=β«0Ο2sin2(a+12)β1cos2(12)β1xdx=12B(a+12,12)=12.Ξ(a+12).Ξ(12)Ξ(a+12+12)=Ο2.Ξ(a+12)Ξ(a2+1)=Ο2ΓΞ(a+12)a2Ξ(a2)=ΟaΓΞ(a+12)Ξ(a2)βfβ²(a)=βΟa2ΓΞ(a+12)Ξ(a2)+ΟaΓ12Ξβ²(a+12).Ξ(a2)β12Ξβ²(a2)Ξ(a+12)Ξ2(a2)βΞ¦=fβ²(2)=βΟ4ΓΞ(32)1+Ο4ΓΞβ²(32).1βΞβ²(1).Ξ(32)12=βΟ4.Ξ(32)+Ο4Γ(Ξβ²(32)+Ξ³.Ξ(32))Ξβ²(1)=β«0βeβtlogtdt=βΞ³
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