calculate-pi-4-pi-3-cosx-sinx-2-sin-2x-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 52197 by maxmathsup by imad last updated on 04/Jan/19 calculate∫π4π3cosx−sinx2+sin(2x)dx Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19 2+sin2x1+sin2x+cos2x+2sinxcosx1+(sinx+cosx)2∫π4π3d(sinx+cosx)1+(sinx+cosx)2∣tan−1(sinx+cosx)∣π4π3tan−1(32+12)−tan−1(12+12)tan−1(1+32)−tan−1(2)tan−1(1+32−21+1+32)=tan−1(1+3−2222+1+32)=tan−1(1+3−222(2+1+3)tan−1(1+3−222+2+6) Commented by Abdo msup. last updated on 05/Jan/19 thankyousirTanmay. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Solution-d-2-y-dx-2-3-dy-dx-4y-x-2-Next Next post: Question-52198 Leave a Reply Cancel replyYour email address will not be published.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.