find-the-value-of-dt-1-t-t-2-2-

Question Number 33202 by prof Abdo imad last updated on 12/Apr/18

f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{d t}{{(1 + t + t^{2})}^{2}} .

Commented by prof Abdo imad last updated on 13/Apr/18

l e t p u t I = \int_{- \infty}^{+ \infty} \frac{d t}{{(1 + t + t^{2})}^{2}} w e h a v e

I = \int_{- \infty}^{+ \infty} \frac{d t}{{({(t + \frac{1}{2})}^{2} + \frac{3}{4})}^{2}} . c h a n g e m e n t

t + \frac{1}{2} = \frac{\sqrt{3}}{2} t a n θ g i v e

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1}{{(\frac{3}{4} {t a n}^{2} θ + \frac{3}{4})}^{2}} \frac{\sqrt{3}}{2} (1 + {t a n}^{2} θ) d θ

= \frac{\sqrt{3}}{2} . \frac{16}{9} \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d θ}{1 + {t a n}^{2} θ} = \frac{16 \sqrt{3}}{9} \int_{0}^{\frac{π}{2}} {c o s}^{2} θ d θ

= \frac{16 \sqrt{3}}{9} \int_{0}^{\frac{π}{2}} \frac{1 + c o s (2 θ)}{2} d θ

= \frac{8 \sqrt{3}}{9} \int_{0}^{\frac{π}{2}} (1 + c o s (2 θ)) d θ

= \frac{4 \sqrt{3} π}{9} + \frac{4 \sqrt{3}}{9} {[s i n (2 θ)]}_{0}^{\frac{π}{2}} = \frac{4 π \sqrt{3}}{9} + 0 \Rightarrow

I = \frac{4 π \sqrt{3}}{9} .