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Question Number 32040 by abdo imad last updated on 18/Mar/18
let give f(x) =∫_0 ^(π/2)      (dt/(1+x tant))    1) find a simple form of f(x)  2) calculate  ∫_0 ^(π/2)   ((tant)/((1+xtant)^2 ))dt  3)give the value of  ∫_0 ^(π/2)     ((tant)/((1+(√3) tant)^2 )) dt .
letgivef(x)=0π2dt1+xtant1)findasimpleformoff(x)2)calculate0π2tant(1+xtant)2dt3)givethevalueof0π2tant(1+3tant)2dt.
Commented by abdo imad last updated on 21/Mar/18
ch.tant =u give f(x)=∫_0 ^∞     (1/(1+xu^2 )) (du/(1+u^2 ))  =∫_0 ^∞      (du/((1+u^2 )(1+xu^2 )))  let decompose  F(u)=  (1/((1+u^2 )(1+xu^2 ))) =  ((au +b)/(1+u^2 )) + ((cu +d)/(1+xu^2 ))  F(−u)=F(u)⇒a=c=0⇒F(u)= (b/(1+u^2 )) +(d/(1+xu^2 ))  F(0)=1= b+d  F(1)= (1/(2(1+x))) = (b/2) + (d/(1+x))= (b/2) +((2d)/(2(1+x))) ⇒  (1/(1+x)) = b +((2d)/(1+x)) ⇒1 =(1+x)b +2d  ⇒  1 =(1+x)b +2(1−b)⇒1= 2  +(x−1)b ⇒b=(1/(1−x))  d=1−b =1−(1/(1−x)) = ((−x)/(1−x)) ⇒  F(u) = (1/((1−x)(1+u^2 ))) −(x/((1−x)(1+xu^2 )))  f(x)= (1/(1−x))∫_0 ^∞   (du/(1+u^2 )) −(x/(1−x))∫_0 ^∞    (du/(1+xu^2 )) but we have  ∫_0 ^∞   (du/(1+u^2 ))= [arctanu]_0 ^∞  =(π/2) and ch. (√x) u =t  (we suppose x>0)  ∫_0 ^∞   (du/(1+xu^2 )) = ∫_0 ^∞    (1/(1+t^2 )) (dt/( (√x))) = (π/(2(√x))) ⇒  f(x)= (π/(2(1−x))) −  ((πx)/(2(√x)(1−x))) =(π/(2(1−x)))(1− (√x))  =(π/2) ((1−(√x))/((1+(√x))(1−(√x)))) = (π/(2(1+(√x)))) .  2)we have f(x)=∫_0 ^(π/2)     (dt/(1+xtant)) ⇒  f^′ (x) =−∫_0 ^(π/2)    ((tant)/((1+xtant)^2 ))dt ⇒  ∫_0 ^(π/2)     ((tant)/((1+xtant)^2 )) dt =−f^′ (x) but  f^′ (x) =−(π/2) (((1+(√(x )))^′ )/((1+(√x))^2 )) =−(π/2)  (1/(2(√x)(1+(√x))^2 )) ⇒  ∫_0 ^(π/2)    ((tant)/((1+xtant)^2 )) dt =   (π/(4 (√x)(1+(√x))^2 ))  3) let take x=(√3)  weget  ∫_0 ^(π/2)    ((tant)/((1+(√3)tant)^2 )) dt =  (π/(4^4 (√3)(1+^4 (√3))^2 )) .
ch.tant=ugivef(x)=011+xu2du1+u2=0du(1+u2)(1+xu2)letdecomposeF(u)=1(1+u2)(1+xu2)=au+b1+u2+cu+d1+xu2F(u)=F(u)a=c=0F(u)=b1+u2+d1+xu2F(0)=1=b+dF(1)=12(1+x)=b2+d1+x=b2+2d2(1+x)11+x=b+2d1+x1=(1+x)b+2d1=(1+x)b+2(1b)1=2+(x1)bb=11xd=1b=111x=x1xF(u)=1(1x)(1+u2)x(1x)(1+xu2)f(x)=11x0du1+u2x1x0du1+xu2butwehave0du1+u2=[arctanu]0=π2andch.xu=t(wesupposex>0)0du1+xu2=011+t2dtx=π2xf(x)=π2(1x)πx2x(1x)=π2(1x)(1x)=π21x(1+x)(1x)=π2(1+x).2)wehavef(x)=0π2dt1+xtantf(x)=0π2tant(1+xtant)2dt0π2tant(1+xtant)2dt=f(x)butf(x)=π2(1+x)(1+x)2=π212x(1+x)20π2tant(1+xtant)2dt=π4x(1+x)23)lettakex=3weget0π2tant(1+3tant)2dt=π443(1+43)2.
Commented by abdo imad last updated on 21/Mar/18
x>0.
x>0.

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