calculate-0-4pi-sinx-3-cosx-2-dx- Tinku Tara June 4, 2023 Relation and Functions 0 Comments FacebookTweetPin Question Number 144214 by Mathspace last updated on 23/Jun/21 calculate∫04πsinx(3+cosx)2dx Answered by bemath last updated on 23/Jun/21 =−∫4π0d(3+cosx)(3+cosx)2=−[−13+cosx]04π=[13+cos4π]−[13+cos0]=0 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-144209Next Next post: find-0-e-3x-x-2-x-1-dx- Leave a Reply Cancel replyYour email address will not be published.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.