find-the-value-of-dt-t-2-2t-2-3-2-

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

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 .