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0-xsin-4x-9-4x-2-dx-




Question Number 146621 by 777316 last updated on 14/Jul/21
∫_0 ^( ∞)  ((xsin(4x))/(9 + 4x^2 ))dx =
0xsin(4x)9+4x2dx=
Answered by mathmax by abdo last updated on 14/Jul/21
Ψ=∫_0 ^∞  ((xsin(4x))/(4x^2  +9)) ⇒Ψ=_(2x=3z)   ∫_0 ^∞   ((3zsin(4((3z)/2)))/(2.9(1+z^2 ))) (3/2)dz  =(1/4)∫_0 ^∞  ((zsin(6z))/(z^2  +1))dz =(1/8)∫_(−∞) ^(+∞)  ((zsin(6z))/(z^2  +1))dz =(1/8)Im(∫_(−∞) ^(+∞)  ((ze^(6iz) )/(z^2  )1))dz)  let ϕ(z)=((ze^(6iz) )/(z^2  +1)) ⇒ϕ(z)=((ze^(6iz) )/((z−i)(z+i)))  residus th.⇒∫_R ϕ(z)dz=2iπRes(ϕ,i)  Res(ϕ,i)=((ie^(−6) )/(2i)) ⇒∫_R ϕ(z)dz=2iπ.((ie^(−6) )/(2i))=((iπ)/e^6 ) ⇒  Ψ=(1/8)×(π/e^6 )=(π/(8e^6 ))
Ψ=0xsin(4x)4x2+9Ψ=2x=3z03zsin(43z2)2.9(1+z2)32dz=140zsin(6z)z2+1dz=18+zsin(6z)z2+1dz=18Im(+ze6izz2)1dz)letφ(z)=ze6izz2+1φ(z)=ze6iz(zi)(z+i)residusth.Rφ(z)dz=2iπRes(φ,i)Res(φ,i)=ie62iRφ(z)dz=2iπ.ie62i=iπe6Ψ=18×πe6=π8e6
Answered by KINMATICS last updated on 14/Jul/21
∫ ((xsin4x)/(9+4x^2 ))dx=∫sin4x.(x/((2x)^2 -(3i)^2 ))dx  ∫ ((xsin4x)/(9+4x^2 ))dx=∫sin4x.(1/4)((1/((2x+3i)))+(1/(2x-3i)))dx  ∫ ((xsin4x)/(9+4x^2 ))dx=(1/2)∫((sin4x)/((4x+6i)))dx+(1/2)∫((sin4x)/(4x-6i))dx  For ∫((sin4x)/(4x+6i))dx, let 4x+6i=u,4x=u-6i, dx=(du/4)  for ∫((sin4x)/(4x-6i))dx, let  4x-6i=y,  4x=y+6i, dx=(dy/4)  ⇒∫((sin4xdx)/(4x^2 +9))=(1/8)∫((sin(u-6i))/u)du+(1/8)∫((sin(y+6i))/y)dy    I=(1/8)∫((sin(u)cos(6i)-cos(u)sin(6i))/u)du+           (1/8)∫((sin(y)cos(6i)+cos(y)sin(6i))/y)dy    I=((cos(6i))/8)∫((sin(u))/u)du-((sin(6i))/8)∫((cos(u))/u)du+       ((cos(6i))/8)∫((sin(y))/y)dy+((sin(6i))/8)∫((cos(y))/y)dy     I= ((cos(6i))/8)Si(u)−((sin(6i))/8)Ci(u)+              ((cos(6i))/8)Si(y)+((sin(6i))/8)Ci(y)+C     I=((cos(6i))/8)Si(4x+6i)-((sin(6i))/8)Ci(4x+6i)                   ((cos(6i))/8)Si(4x-6i)+((sin(6i))/8)Ci(4x-6i)+C    By: Kin Tom  contact        www.hppt://kintom077∂gmail.com
xsin4x9+4x2dx=sin4x.x(2x)2(3i)2dxxsin4x9+4x2dx=sin4x.14(1(2x+3i)+12x3i)dxxsin4x9+4x2dx=12sin4x(4x+6i)dx+12sin4x4x6idxForsin4x4x+6idx,let4x+6i=u,4x=u6i,dx=du4forsin4x4x6idx,let4x6i=y,4x=y+6i,dx=dy4sin4xdx4x2+9=18sin(u6i)udu+18sin(y+6i)ydyI=18sin(u)cos(6i)cos(u)sin(6i)udu+18sin(y)cos(6i)+cos(y)sin(6i)ydyI=cos(6i)8sin(u)udusin(6i)8cos(u)udu+cos(6i)8sin(y)ydy+sin(6i)8cos(y)ydyI=cos(6i)8Si(u)sin(6i)8Ci(u)+cos(6i)8Si(y)+sin(6i)8Ci(y)+CI=cos(6i)8Si(4x+6i)sin(6i)8Ci(4x+6i)cos(6i)8Si(4x6i)+sin(6i)8Ci(4x6i)+CBy:KinTomcontactwww.hppt://kintom077gmail.com

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