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Question Number 117221 by bemath last updated on 10/Oct/20

∫_(−∞) ^(     ∞)  ((cos (3x))/(x^2 +4)) dx =?

cos(3x)x2+4dx=?

Commented by bemath last updated on 10/Oct/20

thank you sirs

thankyousirs

Answered by AbduraufKodiriy last updated on 10/Oct/20

Answered by bobhans last updated on 10/Oct/20

Note ∫ _(−∞) ^∞ ((1/(x^2 +4))) dx =  (π/2)  ∫ _(−∞) ^∞ ((sin x)/x) dx =  (π/2)  I(t)=  2∫ _0 ^∞  ((cos (tx))/(x^2 +4)) dx ; I(0)=  (π/2)  I′(t) =  −2∫ _0 ^∞  ((xsin (tx))/(x^2 +4))dx =  −2∫ _0 ^∞  ((x^2  sin (tx))/(x(x^2 +4)))  I′(t) =  −2∫_0 ^( ∞)  (((x^2 +4)sin (tx))/(x(x^2 +4)))dx+8∫ _0 ^∞  ((sin (tx))/(x(x^2 +4)))dx  = −2∫_0 ^∞  ((sin u)/u) du+8∫ _0 ^∞  ((sin (tx))/(x(x^2 +4))) dx  =  −π +8∫ _0 ^∞  ((sin (tx))/(x(x^2 +4))) dx  I′′(t)=  8 ∫ _0 ^∞  ((cos (tx))/(x^2 +4)) dx =  4I(t)  I(t) =  C_1 e^(2t)  + C_2 e^(−2t)   ⇒C_1 +C_2 =  (π/2) ; 2C_1 −2C_2 =  0  C_1 =  0 ∧ C_2 =  (π/2)  I(t)=  (π/2)e^(−2t)  ⇔ I(3)=  (π/(2 e^6 ))

Note(1x2+4)dx=π2sinxxdx=π2I(t)=20cos(tx)x2+4dx;I(0)=π2I(t)=20xsin(tx)x2+4dx=20x2sin(tx)x(x2+4)I(t)=20(x2+4)sin(tx)x(x2+4)dx+80sin(tx)x(x2+4)dx=20sinuudu+80sin(tx)x(x2+4)dx=π+80sin(tx)x(x2+4)dxI(t)=80cos(tx)x2+4dx=4I(t)I(t)=C1e2t+C2e2tC1+C2=π2;2C12C2=0C1=0C2=π2I(t)=π2e2tI(3)=π2e6

Answered by Bird last updated on 11/Oct/20

I =∫_(−∞) ^(+∞)  ((cos(3x))/(x^(2 ) +4))dx ⇒  I =_(x=2t)   ∫_(−∞) ^(+∞)  ((cos(6t))/(4(t^2 +1)))(2dt)  =(1/2)∫_(−∞) ^(+∞)  ((cos(6t))/(t^2 +1))dt   =(1/2)Re(∫_(−∞) ^(+∞)  (e^(6it) /(t^2  +1))dt) but  ∫_(−∞) ^(+∞ )  (e^(6iz) /(z^2  +1))dz?=2iπ Res(f,i)  =2iπ.(e^(6i(i)) /(2i)) =π e^(−6)  =(π/e^6 ) ⇒  I =(π/(2e^6 ))

I=+cos(3x)x2+4dxI=x=2t+cos(6t)4(t2+1)(2dt)=12+cos(6t)t2+1dt=12Re(+e6itt2+1dt)but+e6izz2+1dz?=2iπRes(f,i)=2iπ.e6i(i)2i=πe6=πe6I=π2e6

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