0-2pi-dx-1-sinxcosx-4piln2-3- Tinku Tara February 16, 2025 Coordinate Geometry 0 Comments FacebookTweetPin Question Number 216695 by sniper237 last updated on 16/Feb/25 ∫02πdx1+sinxcosx=?4πln23 Answered by Ghisom last updated on 16/Feb/25 ∫2π0dx1+sinxcosx=8∫3π/4π/4dx2+sin2x==8∫π/20dx2+cos2x=[t=tanx]=8∫∞0dtt2+3=833[arctan3t3]0∞==4π33 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Find-Sin-5x-2-Sin-x-2-dx-Next Next post: lim-x-3-x-1-3-x-0- Leave a Reply Cancel replyYour email address will not be published.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.