Calcul-pi-2-0-t-tan-t-dt- Tinku Tara August 12, 2023 Logarithms 0 Comments FacebookTweetPin Question Number 195895 by Erico last updated on 12/Aug/23 Calcul∫0π2ttan(t)dt Answered by witcher3 last updated on 13/Aug/23 ∫0∞tan−1(t).t1+t2=∫0∞t1+t2.∫01tdx1+(xt)2.dx=∫01∫0∞t32(1+t2)(1+x2t2)dtdx∫0∞t32(1+t2)(1+x2t2)dt=A∫−∞0t32(1+t2)(1+x2t2)=∫0∞−it32(1+t2)(1+x2t2)dt=A∫Ct32(1+t2)(1+x2t2)=(i)32(2i)(1−x2).+(i)32x2x32(x2−1).2ix=i122(1−x2)(1−1x)=i32πx(1+x)(1+x)=−πei3π4x(1+x)(1+x).∫01π2(1+t)(1+t2).=∫01π2[1+t2+(1−t)(1+t)2(1+t)(1+t2)=π2{∫01dt1+t+∫01dt1+t2−∫01tdt1+t2}=π2{ln(2)+π4−12ln(2))}=π2[ln(2)2+π4]=∫0π2xtan(x)dx Commented by Rodier97 last updated on 17/Aug/23 Hi…..i′mhavingtroubleunderstandinghowyoudidthesecondline…i′mstilltryingtonoticethattan−1(t)=∫0111+(xt)2dxbutwheredoes″tdx″comefrominyourterm∫01tdx1+(xt)2dx???please! Commented by witcher3 last updated on 20/Aug/23 ∫01tdx1+(xt)2=a;xt=y⇒dy=tdx⇒a=∫0tdy1+y2=tan−1(t) Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: dy-dx-1-y-2-1-x-2-0-Next Next post: Question-195892 Leave a Reply Cancel replyYour email address will not be published.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.