calculate-0-2pi-d-2-cos-2-

Question Number 36200 by prof Abdo imad last updated on 30/May/18

c a l c u l a t e \int_{0}^{2 π} \frac{d θ}{{(2 + c o s θ)}^{2}}

Commented by prof Abdo imad last updated on 31/May/18

l e t u t I = \int_{0}^{2 π} \frac{d θ}{{(2 + c o s θ)}^{2}}

I = \int_{0}^{π} \frac{d θ}{{(2 + c o s θ)}^{2}} + \int_{π}^{2 π} \frac{d θ}{{(2 + c o s θ)}^{2}} = I_{1} + I_{2}

c h a n g e m e n t t a n (\frac{θ}{2}) = t g i v e

I_{1} = \int_{0}^{\infty} \frac{1}{{(2 + \frac{1 - t^{2}}{1 + t^{2}})}^{2}} \frac{2 d t}{1 + t^{2}}

= 2 \int_{0}^{\infty} \frac{d t}{(1 + t^{2}) {(\frac{2 + 2 t^{2} + 1 - t^{2}}{1 + t^{2}})}^{2}}

= 2 \int_{0}^{\infty} \frac{{(1 + t^{2})}^{2}}{(1 + t^{2}) {(3 + t^{2})}^{2}} d t

= 2 \int_{0}^{\infty} \frac{t^{2} + 1}{{(t^{2} + 3)}^{2}} d t = \int_{- \infty}^{+ \infty} \frac{t^{2} + 1}{{(t^{2} + 3)}^{2}} d t l e t

c o n s i d e r t h e c o m p l e x f u n c t i o n

φ (z) = \frac{z^{2} + 1}{{(z^{2} + 3)}^{2}}

φ (z) = \frac{z^{2} + 1}{{(z - i \sqrt{3})}^{2} {(z + i \sqrt{3})}^{2}} t h e p o l e s o f φ a r e

i \sqrt{3} a n d - i \sqrt{3} (d o u b l e s) R e s i d u s t h e o r e m g i v e

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i \sqrt{3})

R e s (φ, i \sqrt{3}) = {l i m}_{z \to i \sqrt{3}} \frac{1}{(2 - 1)!} {{(z - i \sqrt{3})}^{2} φ (z)}^{(1)}

= {l i m}_{z \to i \sqrt{3}} {\frac{z^{2} + 1}{{(z + i \sqrt{3})}^{2}}}^{(1)}

{l i m}_{z \to i \sqrt{3}} {\frac{2 z {(z + i \sqrt{3})}^{2} - (z^{2} + 1) 2 (z + i \sqrt{3})}{{(z + i \sqrt{3})}^{4}}}

= {l i m}_{z \to i \sqrt{3}} {\frac{2 z (z + i \sqrt{3}) - 2 (z^{2} + 1)}{{(z + i \sqrt{3})}^{3}}}

= {l i m}_{z \to i \sqrt{3}} {\frac{2 z^{2} + 2 i z \sqrt{3} - 2 z^{2} - 2}{{(z + i \sqrt{3})}^{3}}}

= \frac{2 i (i \sqrt{3}) \sqrt{3} - 2}{{(2 i \sqrt{3})}^{3}} = \frac{- 8}{- 8 i 3 \sqrt{3}} = \frac{1}{i 3 \sqrt{3}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{i 3 \sqrt{3}} = \frac{2 π}{3 \sqrt{3}} = I_{1}

Commented by abdo.msup.com last updated on 31/May/18

I_{2} = \int_{π}^{2 π} \frac{d θ}{{(2 + c o s θ)}^{2}}

=_{θ = π + t} \int_{0}^{π} \frac{d t}{{(2 - c o s t)}^{2}}

=_{t a n (\frac{t}{2}) = x} \int_{0}^{+ \infty} \frac{1}{{(2 - \frac{1 - x^{2}}{1 + x^{2}})}^{2}} \frac{2 d x}{1 + x^{2}}

= 2 \int_{0}^{+ \infty} \frac{d x}{(1 + x^{2}) {\frac{2 + 2 x^{2} - 1 + x^{2}}{1 + x^{2}}}^{2}}

= \int_{- \infty}^{+ \infty} \frac{1 + x^{2}}{{(3 x^{2} + 1)}^{2}} d x l e t c o n s i d e r

t h e c o m p l e x f u n c t i o n

φ (z) = \frac{z^{2} + 1}{{(3 z^{2} + 1)}^{2}}

φ (z) = \frac{z^{2} + 1}{9 (z - \frac{i}{\sqrt{3}}) (z + \frac{i}{\sqrt{3}})} t h e p o l e s o f

φ a r e \frac{i}{\sqrt{3}} a n d \frac{- i}{\sqrt{3}} (d o u b l e)

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, \frac{i}{\sqrt{3}})

R e s (φ, \frac{i}{\sqrt{3}}) = {l i m}_{z \to \frac{i}{\sqrt{3}}} \frac{1}{(2 - 1)!} {(z - \frac{i}{\sqrt{3}}) φ (z)}^{(1)}

\dots b e c o n t i n u e d \dots

Commented by prof Abdo imad last updated on 01/Jun/18

R e s (φ, \frac{i}{\sqrt{3}}) = {l i m}_{z \to \frac{i}{\sqrt{3}}} {\frac{z^{2} + 1}{9 {(z + \frac{i}{\sqrt{3}})}^{2}}}^{(1)}

= \frac{1}{9} {\frac{2 z {(z + \frac{i}{\sqrt{3}})}^{2} - (z^{2} + 1) 2 (z + \frac{i}{\sqrt{3}})}{{(z + \frac{i}{\sqrt{3}})}^{4}}}

= {l i m}_{z \to \frac{i}{\sqrt{3}}} \frac{1}{9} {\frac{2 z (z + \frac{i}{\sqrt{3}}) - 2 (z^{2} + 1)}{{(z + \frac{i}{\sqrt{3}})}^{3}}}

= {l i m}_{z \to \frac{i}{\sqrt{3}}} \frac{1}{9} {\frac{\frac{2}{\sqrt{3}} i z - 2}{{(z + \frac{i}{\sqrt{3}})}^{3}}}

= \frac{2}{9} (\frac{\frac{i}{\sqrt{3}} \frac{i}{\sqrt{3}} - 1}{{(\frac{2 i}{\sqrt{3}})}^{3}}) = \frac{2}{9} \frac{4}{3 \frac{8}{3 \sqrt{3}} i} = \frac{\sqrt{3}}{9 i}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{\sqrt{3}}{9 i} = \frac{2 π \sqrt{3}}{9} = I_{2}

I = I_{1} + I_{2}

Commented by prof Abdo imad last updated on 01/Jun/18

I = \frac{2 π}{3 \sqrt{3}} + \frac{2 π \sqrt{3}}{9} = \frac{2 π \sqrt{3}}{9} + \frac{2 π \sqrt{3}}{9} \Rightarrow

I = \frac{4 π \sqrt{3}}{9} .