calculate-0-pi-2-cosx-sinx-cos-8-x-sin-8-x-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 57235 by maxmathsup by imad last updated on 31/Mar/19 calculate∫0π2cosx−sinxcos8x+sin8xdx Commented by maxmathsup by imad last updated on 01/Apr/19 changementx=π2−tgiveI=−∫0π2cos(π2−t)−sin(π2−t)cos8(π2−t)+sin8(π2(t)(−dt)=∫0π2sint−costsin8t+cos8tdt=−I⇒2I=0⇒I=0. Answered by tanmay.chaudhury50@gmail.com last updated on 01/Apr/19 I=∫0π2cosx−sinxcos8x+sin8xdx=∫0π2sinx−cosxsin8x+cos8xdx[∫0af(x)dx=∫0af(a−x)dx]2I=0→I=0 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: find-the-value-of-0-x-4-1-x-2-x-4-2-dx-Next Next post: clalculate-A-n-0-1-t-2n-1-t-n-dt-with-n-integr-natural- Leave a Reply Cancel replyYour email address will not be published.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.