Question Number 32714 by caravan msup abdo. last updated on 31/Mar/18
$${calculate}\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:. \\ $$
Commented by abdo imad last updated on 03/Apr/18
![ch.t=tanθ give I = ∫_(π/4) ^(π/2) (1/(tan^2 θ .))cosθ .(1+tan^2 θ)dθ = ∫_(π/4) ^(π/2) (dθ/(cosθ tan^2 θ)) = ∫_(π/4) ^(π/2) ((cosθ)/(sin^2 θ)) dθ =[ −(1/(sinθ))]_(π/4) ^(π/2) = (1/((√2)/2)) −1 = (2/( (√2))) −1 =(√2) −1 .](https://www.tinkutara.com/question/Q32852.png)
$${ch}.{t}={tan}\theta\:{give}\:\:{I}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\:\frac{\mathrm{1}}{{tan}^{\mathrm{2}} \theta\:.}{cos}\theta\:.\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{d}\theta}{{cos}\theta\:{tan}^{\mathrm{2}} \theta}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{cos}\theta}{{sin}^{\mathrm{2}} \theta}\:{d}\theta\:\:=\left[\:−\frac{\mathrm{1}}{{sin}\theta}\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:−\mathrm{1}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\:−\mathrm{1}\:=\sqrt{\mathrm{2}}\:−\mathrm{1}\:. \\ $$