1-find-f-a-0-2pi-dt-a-cos-2-t-sin-2-t-with-a-0-2-find-g-a-0-2pi-cos-2-t-a-cos-2-t-sin-2-t-2-dt-

Question Number 35225 by abdo mathsup 649 cc last updated on 16/May/18

1) f i n d f (a) = \int_{0}^{2 π} \frac{d t}{a {c o s}^{2} t + {s i n}^{2} t} w i t h a \neq 0

2) f i n d g (a) = \int_{0}^{2 π} \frac{{c o s}^{2} t}{{(a {c o s}^{2} t + {s i n}^{2} t)}^{2}} d t

Commented by abdo mathsup 649 cc last updated on 16/May/18

a > 0

Commented by prof Abdo imad last updated on 20/May/18

w e h a v e f (a) = \int_{0}^{2 π} \frac{d t}{a \frac{1 + c o s (2 t)}{2} + \frac{1 - c o s (t)}{2}}

= 2 \int_{0}^{2 π} \frac{d t}{a + a c o s (2 t) + 1 - c o s (2 t)}

= 2 \int_{0}^{2 π} \frac{d t}{a + 1 + (a - 1) c o s (2 t)}

=_{2 t = x} 2 \int_{0}^{4 π} \frac{1}{a + 1 + (a - 1) c o s x} \frac{d x}{2}

= \int_{0}^{2 π} \frac{d x}{a + 1 + (a - 1) c o s x} + \int_{2 π}^{4 π} \frac{d x}{a + 1 + (a - 1) c o s x}

\int_{2 π}^{4 π} \frac{d x}{a + 1 + (a - 1) c o s x} =_{x = t + 2 π} \int_{0}^{2 π} \frac{d t}{a + 1 + (a - 1) c o s t}

\Rightarrow f (a) = 2 \int_{0}^{2 π} \frac{d x}{a + 1 + (a - 1) c o s x}

c h a n g e m e n t e^{i x} = z g i v e

f (a) = 2 \int_{∣ z ∣= 1} \frac{1}{a + 1 + (a - 1) \frac{z + z^{- 1}}{2}} \frac{d z}{i z}

f (a) = 4 \int_{∣ z ∣= 1} \frac{d z}{i z (2 + 2 a + (a - 1) (z + z^{- 1}))}

= \int_{∣ z ∣= 1} \frac{- 4 i d z}{2 (1 + a) z + (a - 1) z^{2} + a - 1}

l e t i n y r o d u c e t b e c o m p l e x f u n c t i o n

φ (z) = \frac{- 4 i}{(a - 1) z^{2} + 2 (a + 1) z + a - 1} . p o l e s o f φ ?

Commented by abdo mathsup 649 cc last updated on 20/May/18

Δ^{^{'}} = {(a + 1)}^{2} - {(a - 1)}^{2} = a^{2} + 2 a + 1 - a^{2} + 2 a - 1 = 4 a

z_{1} = \frac{- a - 1 + 2 \sqrt{a}}{a - 1} = \frac{a - 2 \sqrt{a} + 1}{1 - a} = \frac{{(1 - \sqrt{a})}^{2}}{(1 - \sqrt{a}) (1 + \sqrt{a})}

= \frac{1 - \sqrt{a}}{1 + \sqrt{a}} a n d z_{2} = \frac{- a - 1 - 2 \sqrt{a}}{a - 1} = \frac{a + 2 \sqrt{a} + 1}{1 - a}

= \frac{{(1 + \sqrt{a})}^{2}}{(1 - \sqrt{a}) (1 + \sqrt{a})} = \frac{1 + \sqrt{a}}{1 - \sqrt{a}}

c a s e 1 a > 1 ∣ z_{1} ∣ - 1 = \frac{\sqrt{a} - 1}{\sqrt{a} + 1} - 1 = \frac{\sqrt{a} - 1 - \sqrt{a} - 1}{\sqrt{a} + 1}

= \frac{- 2}{\sqrt{a} + 1} < 0 \Rightarrow ∣ z_{1} ∣< 1

∣ z_{2} ∣ - 1 = \frac{1}{∣ z_{1} ∣} - 1 = \frac{1 - ∣ z_{1} ∣}{∣ z_{1} ∣} > 0 \Rightarrow ∣ z_{2} ∣> 1 (t o

e l i m i n a t e f r o m r e s i d u s)

\int_{∣ z ∣= 1} φ (2) d z = 2 i π R e s (φ, z_{1}) w e h a v e

φ (z) = \frac{- 4 i}{(a - 1) (z - z_{1}) (z - z_{2})} \Rightarrow

R e s (φ, z_{1}) = \frac{- 4 i}{(a - 1) (z_{1} - z_{2})}

= \frac{- 4 i}{(a - 1) (z_{1} - \frac{1}{z_{1}})} = \frac{- 4 i z_{1}}{(a - 1) (z_{1}^{2} - 1)}

= \frac{- 4 i \frac{1 - \sqrt{a}}{1 + \sqrt{a}}}{(a - 1) (\frac{{(1 - \sqrt{a})}^{2}}{{(1 + \sqrt{a})}^{2}} - 1)}

= \frac{- 4 i (1 - \sqrt{a})}{(a - 1) (1 + \sqrt{a}) {{(1 - \sqrt{a})}^{2} - {(1 + \sqrt{a})}^{2}}} {(1 + \sqrt{a})}^{2}

= \frac{4 i}{a - 2 \sqrt{a} + 1 - a - 2 \sqrt{a} - 1} = \frac{4 i}{- 4 \sqrt{a}} = \frac{- i}{\sqrt{a}}

\int_{∣ z ∣= 1} φ (z) d z = 2 i π {\frac{- i}{\sqrt{a}}} = \frac{2 π}{\sqrt{a}} \Rightarrow

f (a) = \frac{2 π}{\sqrt{a}} w i t h a > 1

c a s e 2 0 < a < 1 ∣ z_{1} ∣ - 1 = \frac{1 - \sqrt{a}}{1 + \sqrt{a}} - 1

= \frac{1 - \sqrt{a} - 1 - \sqrt{a}}{1 + \sqrt{a}} = \frac{- 2 \sqrt{a}}{1 + a} < 0 \Rightarrow ∣ z_{1} ∣< 1

a n d ∣ z_{2} ∣> 1 s o t h e v a l u e o f f (a) d o n t c h a n g e a n d

f (a) = \frac{2 π}{\sqrt{a}} .

Commented by abdo mathsup 649 cc last updated on 20/May/18

w e h a v e f^{^{'}} (a) = \int_{0}^{2 π} \frac{\partial}{\partial a} {\frac{1}{a {c o s}^{2} t + {s i n}^{2} t}} d t

= - \int_{0}^{2 π} \frac{{c o s}^{2} t}{{({a c o s}^{2} t + {s i n}^{2} t)}^{2}} d t = - g (a) \Rightarrow

g (a) = - f^{^{'}} (a) o r f (a) = \frac{2 π}{\sqrt{a}} \Rightarrow f^{^{'}} (a) = 2 π {- \frac{{(\sqrt{a})}^{^{'}}}{a}}

= - 2 π \frac{1}{2 a \sqrt{a}} = \frac{- π}{a \sqrt{a}} \Rightarrow g (a) = - \frac{π}{a \sqrt{a}} .