cos-3-x-sinx-dx- Tinku Tara June 3, 2023 Integration 0 Comments FacebookTweetPin Question Number 5262 by Kasih last updated on 03/May/16 ∫cos3xsinxdx Answered by Yozzii last updated on 03/May/16 J=∫cos3xsinxdx=∫cosxcos2xsinxdxJ=∫cosx(1−sin2x)sinxdx.Letu=sinx⇒du=cosxdx.∴J=∫(1−u2)u1/2duJ=∫(u1/2−u5/2)duJ=2u3/23−2u7/27+KJ=2sin3/2x3−2sin7/2x7+K Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: x-2014-sin-x-2015-2015-Next Next post: calculate-0-t-a-1-t-t-2-dt-study-first-the-convergence-a-real- Leave a Reply Cancel replyYour email address will not be published.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.