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Question Number 42501 by maxmathsup by imad last updated on 26/Aug/18
calculateI=∫02πcos(4x)cosx+sinxandJ=∫02πsin(4x)cosx+sinxdx
Commented by maxmathsup by imad last updated on 27/Aug/18
wehaveI=∫0πcos(4x)cosx+sinxdx+∫π2πcos(4x)cosx+sinxbutchangementx=π+tgive∫π2πcos(4x)cosx+sinxdx=∫0πcos(4t)−cost−sintdt⇒I=0thesamemethodshowthatJ=0
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