Question Number 33175 by prof Abdo imad last updated on 11/Apr/18
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$
Commented by prof Abdo imad last updated on 13/Apr/18
![let put I = ∫_0 ^1 (dt/((1+t^2 )^2 )) .changement t=tanθ give I = ∫_0 ^(π/4) ((1+tan^2 θ)/((1+tan^2 θ)^2 )) dθ = ∫_0 ^(π/4) (dθ/(1+tan^2 θ)) = ∫_0 ^(π/4) cos^2 θ dθ = (1/2) ∫_0 ^(π/4) (1+cos(2θ))dθ = (π/8) + (1/4)[ sin(2θ)]_0 ^(π/4) = (π/8) + (1/4) .](https://www.tinkutara.com/question/Q33233.png)
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:.{changement}\:{t}={tan}\theta\:{give} \\ $$$${I}\:\:=\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:{d}\theta\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{{d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:{cos}^{\mathrm{2}} \theta\:{d}\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta \\ $$$$=\:\frac{\pi}{\mathrm{8}}\:\:+\:\frac{\mathrm{1}}{\mathrm{4}}\left[\:{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:=\:\frac{\pi}{\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$