find-pi-4-pi-4-xsinx-cos-2-x-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 53462 by maxmathsup by imad last updated on 22/Jan/19 find∫−π4π4xsinxcos2xdx Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19 ∫xtanxsecxdxx∫tanxsecxdx−∫[dxdx∫tanxsecxdx]dxxsecx−∫secxdxxsecx−ln(secx+tanx)+csoI=∣xsecx−ln(secx+tanx)∣−π4π4=[{π4×2−ln(2+1)}−{−π4×2−ln(2−1)}]=[2×π4×2+ln(2−12+1)]=π2+ln(2−12+1) Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-118997Next Next post: 1-let-0-lt-lt-pi-2-and-A-0-pi-2-dx-x-2-2sin-x-1-calculate-A-2-calculate-0-pi-2-dx-x-2-2-x-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Leave a Reply Cancel replyYour email address will not be published.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.