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Question Number 145636 by mathmax by abdo last updated on 06/Jul/21
calculate ∫_0 ^∞  ((arctanx)/((1+x^2 )^2 ))dx
calculate0arctanx(1+x2)2dx
Answered by Dwaipayan Shikari last updated on 06/Jul/21
∫_0 ^∞ ((tan^(−1) (x))/((1+x^2 )^2 ))dx       x=tanθ  =∫_0 ^(π/2) (θ/(sec^4 θ))sec^2 θdθ  =∫_0 ^(π/2) θcos^2 θ dθ=(1/2)∫_0 ^(π/2) θ+θcos2θdθ  =(π^2 /(16))−(1/4)∫_0 ^(π/2) sin2θdθ=(π^2 /(16))−(1/4)
0tan1(x)(1+x2)2dxx=tanθ=0π2θsec4θsec2θdθ=0π2θcos2θdθ=120π2θ+θcos2θdθ=π216140π2sin2θdθ=π21614
Answered by mathmax by abdo last updated on 07/Jul/21
Ψ=∫_0 ^∞  ((arctanx)/((1+x^2 )^2 ))dx  cha7gement x=tant give  Ψ=∫_0 ^(π/2) (t/((1+tan^2 t)^2 ))(1+tan^2 t)dt =∫_0 ^(π/2) (t/(1+tan^2 t))dt  =∫_0 ^(π/2) tcos^2 t dt =∫_0 ^(π/2) t(((1+cos(2t))/2))dt=(1/2)∫_0 ^(π/2)  tdt +(1/2)∫_0 ^(π/2) tcos(2t)dt  ∫_0 ^(π/2)  tdt =[(t^2 /2)]_0 ^(π/2)  =(π^2 /8)  ∫_0 ^(π/2)  tcos(2t)dt =[(t/2)sin(2t)]_0 ^(π/2) −∫_0 ^(π/2) ((sin(2t))/2)dt  =[(1/4)cos(2t)]_0 ^(π/2)  =(1/4)(−2)=−(1/2) ⇒  Ψ=(π^2 /(16))−(1/4)
Ψ=0arctanx(1+x2)2dxcha7gementx=tantgiveΨ=0π2t(1+tan2t)2(1+tan2t)dt=0π2t1+tan2tdt=0π2tcos2tdt=0π2t(1+cos(2t)2)dt=120π2tdt+120π2tcos(2t)dt0π2tdt=[t22]0π2=π280π2tcos(2t)dt=[t2sin(2t)]0π20π2sin(2t)2dt=[14cos(2t)]0π2=14(2)=12Ψ=π21614

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