Find-0-pi-f-x-cosxdx-given-that-f-x-1-2-k-1-n-coskx- Tinku Tara June 3, 2023 Integration 0 Comments FacebookTweetPin Question Number 1780 by 112358 last updated on 23/Sep/15 Find∫0πf(x)cosxdxgiventhatf(x)=1+2∑nk=1coskx. Commented by Rasheed Soomro last updated on 23/Sep/15 ∫0πf(x)cosxdxf(x)=1+2∑nk=1coskx.−−−−−−−−−−−−−−−∫0π(1+2∑nk=1coskx)cosxdx∫0πcosxdx+2∑nk=1∫0π(coskx)(cosx)dx Answered by 123456 last updated on 24/Sep/15 f(x)=1+2∑nk=1coskxf(x)cosx=cosx+2∑nk=1coskxcosx∫π0f(x)cosxdx=∫π0cosx+2∑nk=1∫π0coskxcosxdx∫π0cosxdx=sinπ−sin0=0∫π0cos2xdx=12∫π0dx+12∫π0cos2xdx=12πcontinue Commented by 112358 last updated on 29/Sep/15 GraphicallyIfoundthevalueoftheintegraltobeapproximatelyπ.Considertheidentitycoskxcosx=12(cos(k+1)x+cos(k−1)x)Thentheintegralunderthesummationbecomes∫0π(cos(k+1)x+cos(k−1)x)dx=1k+1sin(k+1)x+1k−1sin(k−1)x∣0π=sin(k+1)πk+1+sin(k−1)πk−1Withk∈Z+andk⩾1,sin(k+1)π=0andsin(k−1)π=0.Butsin(k−1)πk−1isundefinedatk=1.∴∫0πf(x)cosxdx=π∑nk=1sin(k−1)π(k−1)πIfkwereacontinuousvariablewegetthatlimk→1sin(k−1)π(k−1)π=1sothat∫0πf(x)cosxdx=πbutIdon′tfeelcomfortablewiththisapproachsincekisadiscretequantity. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: The-n-positive-numbers-x-1-x-2-x-n-where-n-3-satisfy-x-1-1-1-x-2-x-2-1-1-x-3-x-n-1-1-1-x-n-and-x-n-1-1-x-1-Show-that-x-1-x-2-x-3-x-n-Next Next post: x-1-x-1-dx- Leave a Reply Cancel replyYour email address will not be published.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.