find the value of ∫_(−∞) ^(+∞) (dt/((t^2 −2t +2)^(3/2) ))


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Question Number 66345 by mathmax by abdo last updated on 12/Aug/19

$f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + 2)}^{\frac{3}{2}}}$
Commented by mathmax by abdo last updated on 13/Aug/19
$let I =∫_(−∞) ^(+∞) (dt/((t^2 −2t+2)^(3/2) )) ⇒I =∫_(−∞) ^(+∞) (dt/({(t−1)^2 +1}^(3/2) )) =_(t−1 =u) ∫_(−∞) ^(+∞) (du/((1+u^2 )^(3/2) )) changement u =tanθ give I =∫_(−(π/2)) ^(π/2) (((1+tan^2 θ)dθ)/((1+tan^2 θ)^(3/2) )) =∫_(−(π/2)) ^(π/2) (dθ/(√(1+tan^2 θ))) =∫_(−(π/2)) ^(π/2) cosθ dθ =[sinθ]_(−(π/2)) ^(π/2) =2 .$
$l e t I = \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} - 2 t + 2)}^{\frac{3}{2}}} \Rightarrow I = \int_{- \infty}^{+ \infty} \frac{d t}{{{(t - 1)}^{2} + 1}^{\frac{3}{2}}}$ $=_{t - 1 = u} \int_{- \infty}^{+ \infty} \frac{d u}{{(1 + u^{2})}^{\frac{3}{2}}} c h a n g e m e n t u = t a n θ g i v e$ $I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{(1 + {t a n}^{2} θ) d θ}{{(1 + {t a n}^{2} θ)}^{\frac{3}{2}}} = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d θ}{\sqrt{1 + {t a n}^{2} θ}} = \int_{- \frac{π}{2}}^{\frac{π}{2}} c o s θ d θ$ $= {[s i n θ]}_{- \frac{π}{2}}^{\frac{π}{2}} = 2 .$
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