calculate-0-pi-ln-x-2-2xsin-1-d- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 62145 by maxmathsup by imad last updated on 16/Jun/19 calculate∫0πln(x2−2xsinθ+1)dθ Commented by maxmathsup by imad last updated on 17/Jun/19 letf(x)=∫0πln(x2−2xsinθ+1)dθ⇒f′(x)=∫0π2x−2sinθx2−2xsinθ+1dθ⇒f′(x)2=∫0πx−sinθx2−2xsinθ+1dθ=θ=t2∫02πx−sin(t2)x2−2xsin(t2)+1dt2⇒f′(x)=z=eit2∫∣z∣=1x−z−z−12ix2−2xz−z−12i2dziz=∫∣z∣=12ix−z+z−1−z{2ix2−2xz+2xz−1}(2i)dz=−∫∣z∣=1−4x−2iz+2iz−12ix2z−2xz2+2xdz=∫4x+2iz−2iz−1−2xz2+2ix2z+2xdz=1x∫∣z∣=14xz+2iz2−2iz(−2xz2+2ix2z+2x}dz(x≠0)=−1x∫∣z∣=1iz2+2xz−iz{z2−ixz−1}dzletw(z)=iz2+2xz−iz{z2−ixz−1}polesofw?z2−ixz−1=0→Δ′=(−ix)2+1=1−x2case1x>1⇒Δ′=(ix2−1)2⇒z1=ix+ix2−1z2=ix−ix2−1andw(z)=iz2+2xz−iz(z−z1)(z−z2)∣z1∣−1=∣x+x2−1∣=x+x2−1−1=x−1+x2−1>0(z1isoutofcircle)∣z2∣−1=∣x−x2−1∣−1=x−1−x2−1(x−1)2−(x2−1)=x2−2x+1−x2+1=2−2x=2(1−x)<0⇒∣z2∣<1∫∣z∣=1w(z)dz=2iπ{Res(w,o)+Res(w,z2)}Res(w,0)=limz→0zw(z)=−iz1z2=−i−1=iRes(w,z2)=limz→z2(z−z2)w(z)=iz22+2xz2−iz2(z2−z1)=iz22+2xz2−i(ix−ix2−1)2+1=i(ix−ix2−1)2+2x(ix−x2−1)−i1−(x+x2−1)2=−i(x+x2−1)2+2ix2−2xx2−1−i1−(x2+2xx2−1+x2−1)=….becontinued…m Answered by perlman last updated on 16/Jun/19 letA(θ)=∫0πln(x2−2xcos(θ)+1)dθθ∈IR∖{2kπ,k∈IZ}A(θ)=[xln(x2−2cos(θ)x+1)]0π−∫2x2−2xcos(θ)x2−2cos(θ)x+1dx=πln(π2−2cos(θ)π+1)−2∫0πx2−xcos(θ)x2−2cos(θ)x+1dxx2−xcos(θ)x2−2cos(θ)x+1=1+xcos(θ)−1x2−2cos(θ)x+1A(θ)=πln(π2−2cos(θ)π+1)−2∫0π1+xcos(θ)−1x2−2cos(θ)x+1dxthe2ndintegral=−2π−2∫0πxcos(θ)−1x2−2cos(θ)x+1dx=−2π−∫0πcos(θ)[2x−2cos(θ)]−2+2cos2(θ)x2−2cos(θ)x+1dx=−2π−cos(θ)∫0π2x−2cos(θ)x2−2cos(θ)x+1dx+2∫sin2(θ)x2−2cos(θ)x+1dx=−2π−cos(θ)[ln[x2−2cos(θ)x+1⌉0π+2sin2θ∫0π1(x−cos(θ))2+sin2(θ)dx=2π−cos(θ)ln(π2−2cos(θ)π+1)+2∫0π1(xsin(θ)−cot(θ))2+1dxifsin≠0=−2π−cos(θ)ln(π2−2cos(θ)π+1)+2sin(θ){tan−1{xsinθ−cot(θ)}0π}=−2π−cos(θ)ln{π2−2cos(θ)π+1}+2sin(θ){tan−1{πsin(θ)−cot(θ)}−tan−1{cot(−θ)}=−2π−cos(θ)ln{π2−2cos(θ)π+1}+2sin(θ){arctan{πsin(θ)−cot(θ)}−arctan(tg(π2+θ))A(θ)=−2π+(π−cos(θ))ln(π2−2cos(θ)x+1)+2sin(θ)arctan(π−cos(θ)sin(θ))−2sin(θ)arctan(tan(π2+θ))arctan(tg(a))=aifa∈]−π2.π2[ifa∈]−π2+kπ.π2+kπ[arctantg(a)=a−kπ Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 6-38-0-2-Next Next post: calculate-x-1-x-2-3-dx- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![let A(θ)=∫_0 ^π ln(x^2 −2xcos(θ)+1)dθ θ∈IR\{2kπ,k∈IZ} A(θ)=[xln(x^2 −2cos(θ)x+1)]_0 ^π −∫((2x^2 −2xcos(θ))/(x^2 −2cos(θ)x+1))dx =πln(π^2 −2cos(θ)π+1)−2∫_0 ^π ((x^2 −xcos(θ))/(x^2 −2cos(θ)x+1))dx ((x^2 −xcos(θ))/(x^2 −2cos(θ)x+1))=1+((xcos(θ)−1)/(x^2 −2cos(θ)x+1)) A(θ)=πln(π^2 −2cos(θ)π+1)−2∫_0 ^π 1+((xcos(θ)−1)/(x^2 −2cos(θ)x+1))dx the 2 nd integral =−2π −2∫_0 ^π ((xcos(θ)−1)/(x^2 −2cos(θ)x+1))dx=−2π−∫_0 ^π ((cos(θ)[2x−2cos(θ)]−2+2cos^2 (θ))/(x^2 −2cos(θ)x+1))dx =−2π−cos(θ)∫_0 ^π ((2x−2cos(θ))/(x^2 −2cos(θ)x+1))dx+2∫((sin^2 (θ))/(x^2 −2cos(θ)x+1))dx =−2π−cos(θ)[ln[x^2 −2cos(θ)x+1⌉_0 ^π +2sin^2 θ∫_0 ^π (1/((x−cos(θ))^2 +sin^2 (θ)))dx =2π−cos(θ)ln(π2−2cos(θ)π+1)+2∫_0 ^π (1/(((x/(sin(θ)))−cot (θ))^2 +1))dx if sin≠0 =−2π−cos(θ)ln(π^2 −2cos(θ)π+1)+2sin(θ){tan^(−1) {(x/(sinθ))−cot (θ)}_0 ^π } =−2π−cos(θ)ln{π^2 −2cos(θ)π+1}+2sin(θ){tan^(−1) {(π/(sin(θ)))−cot (θ)}−tan^(−1) {cot (−θ)} =−2π−cos(θ)ln{π^2 −2cos(θ)π+1}+2sin(θ){arctan{(π/(sin(θ)))−cot(θ)}−arctan(tg((π/2)+θ)) A(θ)=−2π+(π−cos(θ))ln(π^2 −2cos(θ)x+1)+2sin(θ)arctan(((π−cos(θ))/(sin(θ))))−2sin(θ)arctan(tan((π/2)+θ)) arctan(tg(a))=a if a∈]((−π)/2).(π/2)[ if a∈]−(π/2)+kπ.(π/2)+kπ[ arctantg(a)=a−kπ](https://www.tinkutara.com/question/Q62163.png)