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Question Number 174838 by mnjuly1970 last updated on 12/Aug/22

       calculate     Ω = ∫_0 ^( (π/2)) (( x^( 2) )/(( sin(x)+cos(x))^( 2) ))dx=?

calculateΩ=0π2x2(sin(x)+cos(x))2dx=?

Answered by princeDera last updated on 12/Aug/22

           calculate     Ω = ∫_0 ^( (π/2)) (( x^( 2) )/(( sin(x)+cos(x))^( 2) ))dx=?  Ω = ∫_0 ^(π/2) (x^2 /(1 + sin (2x)))dx  sin (2x) = ((1 − e^(−4ix)  )/(2ie^(−2ix) ))  Ω = 2i∫_0 ^(π/2) ((x^2 e^(−2ix) )/(2ie^(−2ix)  + 1 − e^(−4ix) ))dx  let z = e^(−2ix)   dz =−2ie^(−2ix) dx ⇒ −(dz/(2iz))=dx and x = −(1/(2i))ln (z)  Ω = −(1/(4i))∫_C ((ln^2 (z) )/(2iz+1−z^2 ))dz where the contour C is in the first quadrant    Ω = (1/(4i))∫_C ((ln^2 (z))/((z−i)^2 ))dz   double poles appear at z=i and they are in the contour    Ω = (1/(4i))×2πi[(d/dz)ln^2 (z)]_(z=i)  = (π/2){((2ln (i))/i)}  ln (i) = i(π/2)  Ω = (π/2)×((2iπ)/(2i)) = (π^2 /2)    Ω = (π^2 /2)

calculateΩ=0π2x2(sin(x)+cos(x))2dx=?Ω=0π2x21+sin(2x)dxsin(2x)=1e4ix2ie2ixΩ=2i0π2x2e2ix2ie2ix+1e4ixdxletz=e2ixdz=2ie2ixdxdz2iz=dxandx=12iln(z)Ω=14iCln2(z)2iz+1z2dzwherethecontourCisinthefirstquadrantΩ=14iCln2(z)(zi)2dzdoublepolesappearatz=iandtheyareinthecontourΩ=14i×2πi[ddzln2(z)]z=i=π2{2ln(i)i}ln(i)=iπ2Ω=π2×2iπ2i=π22Ω=π22

Commented by Tawa11 last updated on 12/Aug/22

Great sir

Greatsir

Commented by Frix last updated on 12/Aug/22

something went wrong...

somethingwentwrong...

Commented by Frix last updated on 12/Aug/22

after substituting  z=e^(−2ix)   I get  Ω=−(1/4)∫(((z^2 +2iz−1)ln^2  z)/((z^2 +1)^2 ))dz  I cannot solve this but your result is  Ω≈4.935  and approximating I get  Ω=∫_0 ^(π/2) (x^2 /((sin x +cos x)^2 ))dx≈0.862

aftersubstitutingz=e2ixIgetΩ=14(z2+2iz1)ln2z(z2+1)2dzIcannotsolvethisbutyourresultisΩ4.935andapproximatingIgetΩ=π/20x2(sinx+cosx)2dx0.862

Commented by princeDera last updated on 12/Aug/22

  how did u get that first integral

howdidugetthatfirstintegral

Commented by Frix last updated on 13/Aug/22

∫_0 ^(π/2) (x^2 /(1+sin 2x))dx=  with sin 2x =((e^(4ix) −1)/(2ie^(2ix) )) (=((1−e^(4ix) )/(2ie^(−2ix) )))  =2∫_0 ^(π/2) ((x^2 e^(2ix) (2e^(2ix) +(e^(4ix) −1)i))/((e^(4ix) +1)^2 ))dx=  with z=e^(2ix)  ⇔ x=−i((ln z)/2) ⇔ dx=−i(dz/(2z))  =−(1/4)∫_1 ^(−1) (((z^2 −2iz−1)ln^2  z)/((z^2 +1)^2 ))dz

π/20x21+sin2xdx=withsin2x=e4ix12ie2ix(=1e4ix2ie2ix)=2π/20x2e2ix(2e2ix+(e4ix1)i)(e4ix+1)2dx=withz=e2ixx=ilnz2dx=idz2z=1411(z22iz1)ln2z(z2+1)2dz

Answered by mnjuly1970 last updated on 13/Aug/22

  Ω= (1/2) ∫_0 ^( (π/2)) (( x^( 2) )/(sin^( 2) (x+(π/4))))dx      =^(i.b.p)  (1/2) {[−cot(x+(π/4))x^( 2) ]_0 ^( (π/2)) +2∫_0 ^( (π/2)) xcot(x+(π/4))dx    = (π^( 2) /8) +∫_0 ^( (π/2)) x.cot(x+(π/4))dx    =^(i.b.p)  (π^( 2) /8) + {[xlnsin(x+(π/4))]_0 ^(π/2) −∫_0 ^( (π/2)) lnsin(x+(π/4))dx}    =(π^( 2) /8) +(π/2) ln(((√2)/2)) −∫_(−(π/4)) ^( (π/4)) lncos(x)dx     = (π^( 2) /8) −(π/4) ln(2)−2∫_0 ^( (π/4)) lncos(x)dx    =_(ln(cos(x))^( ∗∗) ) ^(fourier series) (π^( 2) /8) −(π/4)ln(2)−2{(G/2)−(π/4)ln(2)}         = (π^( 2) /8) +(π/4) ln(2)−G     ∗∗ :  ln(cos(x))=^? −ln(2)−Σ_(n=1) ^∞ (((−1)^(n−1) cos(2nx)))/n)

Ω=120π2x2sin2(x+π4)dx=i.b.p12{[cot(x+π4)x2]0π2+20π2xcot(x+π4)dx=π28+0π2x.cot(x+π4)dx=i.b.pπ28+{[xlnsin(x+π4)]0π20π2lnsin(x+π4)dx}=π28+π2ln(22)π4π4lncos(x)dx=π28π4ln(2)20π4lncos(x)dx=fourierseriesln(cos(x))π28π4ln(2)2{G2π4ln(2)}=π28+π4ln(2)G:ln(cos(x))=?ln(2)n=1(1)n1cos(2nx))n

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