C-0-pi-2-cos-2-x-4sin-2-x-cos-2-x-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 178988 by cortano1 last updated on 23/Oct/22 C=∫π/20cos2x4sin2x+cos2xdx=? Answered by qaz last updated on 23/Oct/22 C=∫0π/2cos2x4sin2x+cos2xdx=∫0π/2sec2xdx(4tan2x+1)(1+tan2x)=∫0∞dx(4x2+1)(1+x2)=13∫0∞(41+4x2−11+x2)dx=23arctan2x−13arctanx∣0∞=π6 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-113452Next Next post: Question-178994 Leave a Reply Cancel replyYour email address will not be published.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.