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Question Number 32040 by abdo imad last updated on 18/Mar/18
letgivef(x)=∫0π2dt1+xtant1)findasimpleformoff(x)2)calculate∫0π2tant(1+xtant)2dt3)givethevalueof∫0π2tant(1+3tant)2dt.
Commented by abdo imad last updated on 21/Mar/18
ch.tant=ugivef(x)=∫0∞11+xu2du1+u2=∫0∞du(1+u2)(1+xu2)letdecomposeF(u)=1(1+u2)(1+xu2)=au+b1+u2+cu+d1+xu2F(−u)=F(u)⇒a=c=0⇒F(u)=b1+u2+d1+xu2F(0)=1=b+dF(1)=12(1+x)=b2+d1+x=b2+2d2(1+x)⇒11+x=b+2d1+x⇒1=(1+x)b+2d⇒1=(1+x)b+2(1−b)⇒1=2+(x−1)b⇒b=11−xd=1−b=1−11−x=−x1−x⇒F(u)=1(1−x)(1+u2)−x(1−x)(1+xu2)f(x)=11−x∫0∞du1+u2−x1−x∫0∞du1+xu2butwehave∫0∞du1+u2=[arctanu]0∞=π2andch.xu=t(wesupposex>0)∫0∞du1+xu2=∫0∞11+t2dtx=π2x⇒f(x)=π2(1−x)−πx2x(1−x)=π2(1−x)(1−x)=π21−x(1+x)(1−x)=π2(1+x).2)wehavef(x)=∫0π2dt1+xtant⇒f′(x)=−∫0π2tant(1+xtant)2dt⇒∫0π2tant(1+xtant)2dt=−f′(x)butf′(x)=−π2(1+x)′(1+x)2=−π212x(1+x)2⇒∫0π2tant(1+xtant)2dt=π4x(1+x)23)lettakex=3weget∫0π2tant(1+3tant)2dt=π443(1+43)2.
x>0.
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