Evaluate-x-cos-2-x-dx- Tinku Tara June 3, 2023 Integration FacebookTweetPin Question Number 40 by user3 last updated on 25/Jan/15 Evaluate∫xcos2xdx. Answered by user3 last updated on 03/Nov/14 ∫xcos2xdx=∫x(1+cos2x2)dx=12∫xdx+12∫xcos2xdx=x24+12⋅[x⋅∫cos2xdx−∫{ddx(x)⋅∫cos2xdx}dx]=x24+12[xsin2x2−∫sin2x2dx]=x24+xsin2x4−14×(−cos2x)2+C=x24+xsin2x4+cos2x8+C Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-131110Next Next post: log-x-2-dx-