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0-2pi-dx-1-ksinx-2-




Question Number 143399 by Ar Brandon last updated on 13/Jun/21
∫_0 ^(2π) (dx/((1−ksinx)^2 ))
02πdx(1ksinx)2
Answered by Ar Brandon last updated on 13/Jun/21
I=∫_0 ^(2π) (dx/((1−ksinx)^2 ))=2∫_(−(π/2)) ^(π/2) (dx/((1−ksinx)^2 )) t=tan(x/2)    =2∫_(−1) ^1 (1/((1−k((2t)/(1+t^2 )))^2 ))∙((2dt)/(1+t^2 ))=4∫_(−1) ^1 ((t^2 +1)/((t^2 −2kt+1)^2 ))dt    =4[((at+b)/(t^2 −2kt+1))+∫((ct+d)/(t^2 −2kt+1))dt]_(−1) ^1     =4∫[((a(t^2 −2kt+1)−(at+b)(2t−2k))/((t^2 +2kt+1)^2 ))+((ct+d)/(t^2 −2kt+1))]dt  c=0,a−2a+d=1⇒d=a+1,−2ka+2ka−2b−2kd=0⇒b=−kd  a+2bk+d=1⇒d+bk=1⇒d(1−k^2 )=1⇒d=(1/(1−k^2 )),b=(k/(k^2 −1))  ⇒a=(1/(1−k^2 ))−1=((1+k^2 −1)/(1−k^2 ))=(k^2 /(1−k^2 ))  I=4[(((k^2 /(1−k^2 ))t+(k/(k^2 −1)))/(t^2 −2kt+1))+(1/(1−k^2 ))∫(dt/(t^2 −2kt+1))]_(−1) ^1     =(4/(2−2k))((k^2 /(1−k^2 ))+(1/(k^2 −1)))−(4/(2+2k))((k/(k^2 −1))−(k^2 /(1−k^2 )))+(4/(1−k^2 ))∫(dt/(t^2 −2kt+1))  ∫(dt/(t^2 −2kt+1))=∫(dt/((t−k)^2 +1−k^2 ))=(1/( (√(1−k^2 ))))arctan(((t−k)/( (√(1−k^2 ))))), ∣k∣<1  ∫(dt/(t^2 −2kt+1))=∫(dt/((t−k)^2 −(k^2 −1)))=((−1)/( (√(k^2 −1))))argtanh(((t−k)/( (√(k^2 −1))))), ∣k∣>1  I=(2/(k−1))−(2/(k+1))((k/(k−1)))+((4(√(1−k^2 )))/( (1−k^2 )^2 ))[arctan((√((1−k)/(1+k))))+arctan(((k+1)/( (√(1−k^2 )))))], ∣k∣<1    =(2/(k−1))−(2/(k+1))((k/(k−1)))+((2(√(k^2 −1)))/((k^2 −1)^2 ))ln∣(((√(k^2 −1))−k+1)/( (√(k^2 −1))+k−1))∣, ∣k∣>1
I=02πdx(1ksinx)2=2π2π2dx(1ksinx)2t=tanx2=2111(1k2t1+t2)22dt1+t2=411t2+1(t22kt+1)2dt=4[at+bt22kt+1+ct+dt22kt+1dt]11=4[a(t22kt+1)(at+b)(2t2k)(t2+2kt+1)2+ct+dt22kt+1]dtc=0,a2a+d=1d=a+1,2ka+2ka2b2kd=0b=kda+2bk+d=1d+bk=1d(1k2)=1d=11k2,b=kk21a=11k21=1+k211k2=k21k2I=4[k21k2t+kk21t22kt+1+11k2dtt22kt+1]11=422k(k21k2+1k21)42+2k(kk21k21k2)+41k2dtt22kt+1dtt22kt+1=dt(tk)2+1k2=11k2arctan(tk1k2),k∣<1dtt22kt+1=dt(tk)2(k21)=1k21argtanh(tkk21),k∣>1I=2k12k+1(kk1)+41k2(1k2)2[arctan(1k1+k)+arctan(k+11k2)],k∣<1=2k12k+1(kk1)+2k21(k21)2lnk21k+1k21+k1,k∣>1
Commented by Ar Brandon last updated on 13/Jun/21
Please check !
Pleasecheck!
Answered by mathmax by abdo last updated on 13/Jun/21
f(t)=∫_0 ^(2π)  (dx/(t−ksinx)) ⇒f^′ (t)=−∫_0 ^(2π)   (dx/((t−ksinx)^2 )) ⇒  ∫_0 ^(2π)  (dx/((1−ksinx)^2 ))=−f^′ (1)  changement e^(ix) =t give  f(t)=∫_(∣z∣=1)      (dz/(iz(t−k((z−z^(−1) )/(2i))))) =∫_(∣z∣=1)     ((2idz)/(iz(2it−kz+kz^(−1) )))  =∫  ((2dz)/(2itz−kz^2 +k)) =∫_(∣z∣=1)    ((−2dz)/(kz^2 −2itz −k))  let Φ(z)=((−2)/(kz^2 −2itz −k))   poles of Φ?  Δ^′  =−t^2 +k^2   case 1  −t^2 +k^2 <0 ⇒∣k∣<∣t∣ ⇒z_1 =((it+i(√(k^2 −t^2 )))/k)  z_2 =((it−i(√(k^2 −t^2 )))/k)  ∣z_1 ∣=(1/(∣k∣))(√(t^2 +k^2 −t^2 ))=1  also ∣z_2 ∣=1 and Φ(z)=((−2)/(k(z−z_1 )(z−z_2 )))  ∫_R Φ(z)dz =2iπ{Res(Φ,z_1 )+Res(Φ,z_2 )}  Res(Φ,z_1 )=((−2)/(k(z_1 −z_2 )))  Res(Φ,z_2 ) =((−2)/(k(z_2 −z_1 ))) ⇒Σ Res=0=∫_R Φ(z)dz ⇒f(t)=0 ⇒f^′ (t)=0  case 2  −t^2 +k^2 >0 ⇒∣k∣>∣t∣ ⇒z_1 =((it+(√(k^2 −t^2 )))/k)  z_2 =((it−(√(k^2 −t^2 )))/k)  ∣z_1 ∣ =(1/(∣k∣))(√(t^2 +k^2 −t^2 ))=1  and ∣z_2 ∣=1  ∫_(∣z∣=1)   Φ(z)dz =2iπ{Res(Φ,z_1 )+Res(Φ,z_2 )}  but we get Res(Φ,z_1 )=−Res(Φ,z_2 ) ⇒Σ Res=0⇒∫_(∣z∣) Φ(z)dz=0 ⇒  f(t)=0 ⇒f^′ (t)=0
f(t)=02πdxtksinxf(t)=02πdx(tksinx)202πdx(1ksinx)2=f(1)changementeix=tgivef(t)=z∣=1dziz(tkzz12i)=z∣=12idziz(2itkz+kz1)=2dz2itzkz2+k=z∣=12dzkz22itzkletΦ(z)=2kz22itzkpolesofΦ?Δ=t2+k2case1t2+k2<0⇒∣k∣<∣tz1=it+ik2t2kz2=itik2t2kz1∣=1kt2+k2t2=1alsoz2∣=1andΦ(z)=2k(zz1)(zz2)RΦ(z)dz=2iπ{Res(Φ,z1)+Res(Φ,z2)}Res(Φ,z1)=2k(z1z2)Res(Φ,z2)=2k(z2z1)ΣRes=0=RΦ(z)dzf(t)=0f(t)=0case2t2+k2>0⇒∣k∣>∣tz1=it+k2t2kz2=itk2t2kz1=1kt2+k2t2=1andz2∣=1z∣=1Φ(z)dz=2iπ{Res(Φ,z1)+Res(Φ,z2)}butwegetRes(Φ,z1)=Res(Φ,z2)ΣRes=0zΦ(z)dz=0f(t)=0f(t)=0
Commented by Ar Brandon last updated on 13/Jun/21
Hum... The^� ore^� me de Re^� sidus.  Thanks Sir ! Grateful !
HumTheor´eme`deResidus´.ThanksSir!Grateful!
Commented by mathmax by abdo last updated on 13/Jun/21
you are welcome
youarewelcome

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