I-n-0-1-x-2n-1-1-x-2-dx-n-0-prove-that-n-0-2n-1-I-n-2-2nI-n-1- Tinku Tara June 4, 2023 Others 0 Comments FacebookTweetPin Question Number 158477 by LEKOUMA last updated on 04/Nov/21 In=∫01x2n+11+x2dx,n⩾0provethat∀n⩾0(2n+1)In=2−2nIn−1 Answered by Ar Brandon last updated on 04/Nov/21 In=∫01x2n+11+x2dx,x=tanϑ⇒dx=sec2ϑdϑ=∫0π4tan2n+1ϑsecϑ⋅sec2ϑdϑ=∫0π4tan2n+1ϑsecϑdϑ=∫0π4tan2nϑ⋅secϑtanϑdϑ,{u(ϑ)=tan2nϑv′(ϑ)=secϑtanϑ⇒{u′(ϑ)=2ntan2n−1ϑsec2ϑv(ϑ)=secϑ=[tan2nϑsecϑ]0π4−2n∫0π4tan2n−1sec3ϑdϑ=2−2n∫0π4tan2n−1secϑdϑ−2n∫0π4tan2n+1ϑsecϑdϑ=2−2nIn−1−2nIn⇒(2n+1)In=2−2nIn−1 Answered by Ar Brandon last updated on 04/Nov/21 In=∫01x2n+11+x2dx=12∫01x2n⋅2x1+x2dx{u(x)=x2nv′(x)=2x1+x2⇒{u′(x)=2nx2n−1v(x)=21+x2In=[x2n1+x2]01−2n∫01x2n−11+x2dx=2−2n∫01x2n−1(1+x2)1+x2dx=2−2n∫01x2n−11+x2dx−2n∫01x2n+11+x2dx=2−2nIn−1−2nIn⇒(2n+1)In=2−2nIn−1 Commented by mnjuly1970 last updated on 05/Nov/21 thanksalotyakhchider… Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 1-the-mean-of-14-9-22-14-22-18-Next Next post: Question-158482 Leave a Reply Cancel replyYour email address will not be published.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.