calculate-0-arctanx-1-x-2-2-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 145636 by mathmax by abdo last updated on 06/Jul/21 calculate∫0∞arctanx(1+x2)2dx Answered by Dwaipayan Shikari last updated on 06/Jul/21 ∫0∞tan−1(x)(1+x2)2dxx=tanθ=∫0π2θsec4θsec2θdθ=∫0π2θcos2θdθ=12∫0π2θ+θcos2θdθ=π216−14∫0π2sin2θdθ=π216−14 Answered by mathmax by abdo last updated on 07/Jul/21 Ψ=∫0∞arctanx(1+x2)2dxcha7gementx=tantgiveΨ=∫0π2t(1+tan2t)2(1+tan2t)dt=∫0π2t1+tan2tdt=∫0π2tcos2tdt=∫0π2t(1+cos(2t)2)dt=12∫0π2tdt+12∫0π2tcos(2t)dt∫0π2tdt=[t22]0π2=π28∫0π2tcos(2t)dt=[t2sin(2t)]0π2−∫0π2sin(2t)2dt=[14cos(2t)]0π2=14(−2)=−12⇒Ψ=π216−14 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: What-is-the-last-2-digits-of-2-613-Next Next post: Find-the-arc-lenght-of-the-function-y-2-x-3-a-where-a-is-a-constant-for-0-x-7a-3- Leave a Reply Cancel replyYour email address will not be published.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.