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Question Number 60960 by maxmathsup by imad last updated on 27/May/19
find ∫_(−∞) ^(+∞) tan((1/(1+x^2 )))dx
$${find}\:\int_{−\infty} ^{+\infty} \:\:{tan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx}\: \\ $$
Commented by mathsolverby Abdo last updated on 28/May/19
let I =∫_(−∞) ^(+∞) tan((1/(1+x^2 )))dx ⇒ I =2 ∫_0 ^(+∞) tan((1/(1+x^2 )))dx =_(x=tanθ) 2∫_0 ^(π/2) tan(cos^2 θ)(1+tan^2 θ)dθ =2 ∫_0 ^(π/2) ((tan(cos^2 θ))/(cos^2 θ)) dθ if we take tan(u)∼ u+(u^3 /3) we get tan(cos^2 θ)∼cos^2 θ +((cos^6 θ)/3) ⇒ ((tan(cos^2 θ))/(cos^2 θ)) ∼ 1+((cos^4 θ)/3) ⇒ I ∼ 2 ∫_0 ^(π/2) (1+((cos^4 θ)/3))dθ =π +(2/3) ∫_0 ^(π/2) cos^4 θ dθ ∫_0 ^(π/2) cos^4 θdθ =(1/4) ∫_0 ^(π/2) (1+cos(2θ)^2 dθ =(1/4) ∫_0 ^(π/2) (1+2cosθ +cos^2 (2θ))dθ =(π/8) +(1/2) ∫_0 ^(π/2) cosθdθ +(1/8) ∫_0 ^(π/2) (1+cos(4θ))dθ =(π/8) +(1/2) +(π/(16)) +0 =((3π)/(16)) +(1/2) ⇒I ∼ π +(2/3)(((3π)/(16)) +(1/2)) ⇒ I ∼ π +(π/8) +(1/3) ⇒I ∼ ((9π)/8) +(1/3)
$${let}\:\:{I}\:=\int_{−\infty} ^{+\infty} \:{tan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx}\:\Rightarrow \\ $$$${I}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{+\infty} \:{tan}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx}\:=_{{x}={tan}\theta} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{tan}\left({cos}^{\mathrm{2}} \theta\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{tan}\left({cos}^{\mathrm{2}} \theta\right)}{{cos}^{\mathrm{2}} \theta}\:{d}\theta\:\:\:{if}\:{we}\:{take} \\ $$$${tan}\left({u}\right)\sim\:{u}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\:\:{we}\:{get}\: \\ $$$${tan}\left({cos}^{\mathrm{2}} \theta\right)\sim{cos}^{\mathrm{2}} \theta\:+\frac{{cos}^{\mathrm{6}} \theta}{\mathrm{3}}\:\Rightarrow \\ $$$$\frac{{tan}\left({cos}^{\mathrm{2}} \theta\right)}{{cos}^{\mathrm{2}} \theta}\:\sim\:\mathrm{1}+\frac{{cos}^{\mathrm{4}} \theta}{\mathrm{3}}\:\Rightarrow \\ $$$${I}\:\sim\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\frac{{cos}^{\mathrm{4}} \theta}{\mathrm{3}}\right){d}\theta\:=\pi\:+\frac{\mathrm{2}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{4}} \theta\:{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{4}} \theta{d}\theta\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)^{\mathrm{2}} {d}\theta\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}+\mathrm{2}{cos}\theta\:+{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\theta{d}\theta\:\:+\frac{\mathrm{1}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)\right){d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{16}}\:+\mathrm{0} \\ $$$$=\frac{\mathrm{3}\pi}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{I}\:\sim\:\pi\:+\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{3}\pi}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${I}\:\sim\:\pi\:+\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow{I}\:\sim\:\frac{\mathrm{9}\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by mathsolverby Abdo last updated on 28/May/19
I ∼ 3.86
$${I}\:\sim\:\mathrm{3}.\mathrm{86} \\ $$

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