0-pi-2-cos-2x-ln-sin-x-dx-pi-4-solution-1-0-pi-2-2cos-2-x-1-ln-sin-x-dx-2-0-pi-2-cos-2-x-ln-sin-x-dx-0-pi-2- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 153555 by mnjuly1970 last updated on 08/Sep/21 Ω:=∫0π2cos(2x).ln(sin(x))dx=?−π4solution(1)Ω:=∫0π2(2cos2(x)−1)ln(sin(x))dx:=2∫0π2cos2(x).ln(sin(x))dx−∫0π2ln(sin(x))dxweknowthat:∫0π2ln(sin(x))dx=derivedearlier−π2ln(2)∫0π2cos2(x).ln(sin(x))dx=previousposts−π4ln(2)−π8∴Ω:=−π2ln(2)−π4+π2ln(2)◂Ω=−π4▸m.n Answered by Ar Brandon last updated on 08/Sep/21 Ω=∫0π2cos(2x)ln(sinx)dx{u(x)=ln(sinx)v′(x)=cos(2x)⇒{u′(x)=cosxsinxv(x)=sin2x2Ω=[ln(sinx)sin2x2−∫cos2xdx]0π2=−∫0π2cos2xdx=−12∫0π2(1+cos2x)dx=−12[x+sin2x2]0π2=−π4 Commented by mnjuly1970 last updated on 08/Sep/21 verynicesirbrandoasalways.. Commented by Ar Brandon last updated on 08/Sep/21 Thanks Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-22481Next Next post: Question-88021 Leave a Reply Cancel replyYour email address will not be published.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.