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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
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