pi-2-pi-2-dx-1-e-sin-x- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 89946 by jagoll last updated on 20/Apr/20 ∫π2−π2dx1+esinx Answered by john santu last updated on 20/Apr/20 I=∫π2−π2(dx1+esinx)replacexby−xI=∫−π2π2(−dx1+e−sinx)=∫π2−π2(dx1+e−sinx)I+I=∫π2−π21+esinx1+esinxdx2I=∫π2−π21dx=πI=π2 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-24411Next Next post: log-2-sin-x-5pi-12-log-2-sin-x-pi-12-1- Leave a Reply Cancel replyYour email address will not be published.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.