Evaluate-0-pi-2-xcos-x-dx- Tinku Tara June 3, 2023 Integration FacebookTweetPin Question Number 51 by surabhi last updated on 25/Jan/15 Evaluate∫0π/2xcosxdx Answered by surabhi last updated on 04/Nov/14 ∫0π/2xcosxdx=[xsinx]0π/2−∫0π/21⋅sinxdx=π2+[cosx]0π/2=π2−1 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Evaluate-x-1-1-x-x-2-dx-Next Next post: Question-65589