Question Number 66345 by mathmax by abdo last updated on 12/Aug/19
$${find}\:{the}\:{value}\:{of}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$
Commented by mathmax by abdo last updated on 13/Aug/19
![let I =∫_(−∞) ^(+∞) (dt/((t^2 −2t+2)^(3/2) )) ⇒I =∫_(−∞) ^(+∞) (dt/({(t−1)^2 +1}^(3/2) )) =_(t−1 =u) ∫_(−∞) ^(+∞) (du/((1+u^2 )^(3/2) )) changement u =tanθ give I =∫_(−(π/2)) ^(π/2) (((1+tan^2 θ)dθ)/((1+tan^2 θ)^(3/2) )) =∫_(−(π/2)) ^(π/2) (dθ/( (√(1+tan^2 θ)))) =∫_(−(π/2)) ^(π/2) cosθ dθ =[sinθ]_(−(π/2)) ^(π/2) =2 .](https://www.tinkutara.com/question/Q66376.png)
$${let}\:{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\Rightarrow{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left\{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$=_{{t}−\mathrm{1}\:={u}} \:\:\:\int_{−\infty} ^{+\infty} \:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{changement}\:{u}\:={tan}\theta\:{give} \\ $$$${I}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}}\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\theta\:{d}\theta \\ $$$$=\left[{sin}\theta\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{2}\:. \\ $$