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Question Number 40141 by maxmathsup by imad last updated on 16/Jul/18
find∫0π2dx3+sinx
Commented by maxmathsup by imad last updated on 19/Jul/18
letI=∫0π2dx3+sinxchangementtan(x2)=tgiveI=∫0113+2t1+t22dt1+t2=2∫01dt3+3t2+2t=2∫01dt3t2+2t+3but3t2+2t+3=3(t2+23t+1)=3(t2+23t+19+1−19)=3{(t+13)2+89}⇒I=23∫01dt(t+13)2+89=t+13=223u23∫1222189(1+u2)223du=42998∫1222du1+u2=22[arctanu]122⇒I=22{arctan(2)−arctan(12)}.
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