1) find f(a) = ∫_0 ^(2π) (dt/(a cos^2 t + sin^2 t)) with a≠0 2) find g(a) = ∫_0 ^(2π) ((cos^2 t)/((a cos^2 t +sin^2 t)^2 ))dt


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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
$Δ^′ = (a+1)^2 −(a−1)^2 = a^2 +2a+1−a^2 +2a−1=4a z_1 = ((−a−1 +2(√a))/(a−1)) = ((a −2(√a) +1)/(1−a)) = (((1−(√a))^2 )/((1−(√a))(1+(√a)))) = ((1−(√a))/(1+(√a))) and z_2 = ((−a−1−2(√a))/(a−1)) = ((a +2(√a) +1)/(1−a)) = (((1+(√a))^2 )/((1−(√a))(1+(√a)))) = ((1+(√a))/(1−(√a))) case1 a>1 ∣z_1 ∣ −1 = (((√a) −1)/((√a)+1)) −1=(((√a) −1−(√a) −1)/((√a) +1)) = ((−2)/((√a) +1))<0 ⇒ ∣z_1 ∣<1 ∣z_2 ∣−1 = (1/(∣z_1 ∣)) −1 = ((1−∣z_1 ∣)/(∣z_1 ∣)) >0 ⇒ ∣z_2 ∣>1 (to eliminate from residus) ∫_(∣z∣=1) ϕ(2)dz =2iπ Res(ϕ,z_1 ) we have ϕ(z) = ((−4i)/((a−1)(z −z_1 )(z−z_2 ))) ⇒ Res(ϕ, z_1 ) = ((−4i)/((a−1)(z_1 −z_2 ))) = ((−4i)/((a−1)(z_1 −(1/z_1 )))) = ((−4i z_1 )/((a−1)(z_1 ^2 −1))) =((−4i ((1−(√a))/(1+(√a))))/((a−1)( (((1−(√a))^2 )/((1+(√a))^2 )) −1))) = ((−4i(1−(√a)))/((a−1)(1+(√a)){ (1−(√a))^2 −(1+(√a))^2 })) (1+(√a))^2 = ((4i)/(a−2(√a) +1 −a −2(√a) −1)) = ((4i)/(−4(√a))) =((−i)/(√a)) ∫_(∣z∣=1) ϕ(z)dz =2iπ {((−i)/(√a))} = ((2π)/(√a)) ⇒ f(a) = ((2π)/(√a)) with a>1 case 2 0<a<1 ∣z_1 ∣ −1 =((1−(√a))/(1+(√a))) −1 = ((1−(√a)−1−(√a))/(1+(√a))) = ((−2(√a))/(1+a)) <0 ⇒ ∣z_1 ∣<1 and ∣z_2 ∣>1 so the value of f(a) dont change and f(a) = ((2π)/(√a)) .$
$Δ^{^{'}} = {(a + 1)}^{2} - {(a - 1)}^{2} = a^{2} + 2 a + 1 - a^{2} + 2 a - 1 = 4 a$ $z_{1} = \frac{- a - 1 + 2 \sqrt{a}}{a - 1} = \frac{a - 2 \sqrt{a} + 1}{1 - a} = \frac{{(1 - \sqrt{a})}^{2}}{(1 - \sqrt{a}) (1 + \sqrt{a})}$ $= \frac{1 - \sqrt{a}}{1 + \sqrt{a}} a n d z_{2} = \frac{- a - 1 - 2 \sqrt{a}}{a - 1} = \frac{a + 2 \sqrt{a} + 1}{1 - a}$ $= \frac{{(1 + \sqrt{a})}^{2}}{(1 - \sqrt{a}) (1 + \sqrt{a})} = \frac{1 + \sqrt{a}}{1 - \sqrt{a}}$ $c a s e 1 a > 1 ∣ z_{1} ∣ - 1 = \frac{\sqrt{a} - 1}{\sqrt{a} + 1} - 1 = \frac{\sqrt{a} - 1 - \sqrt{a} - 1}{\sqrt{a} + 1}$ $= \frac{- 2}{\sqrt{a} + 1} < 0 \Rightarrow ∣ z_{1} ∣< 1$ $∣ z_{2} ∣ - 1 = \frac{1}{∣ z_{1} ∣} - 1 = \frac{1 - ∣ z_{1} ∣}{∣ z_{1} ∣} > 0 \Rightarrow ∣ z_{2} ∣> 1 (t o$ $e l i m i n a t e f r o m r e s i d u s)$ $\int_{∣ z ∣= 1} φ (2) d z = 2 i π R e s (φ, z_{1}) w e h a v e$ $φ (z) = \frac{- 4 i}{(a - 1) (z - z_{1}) (z - z_{2})} \Rightarrow$ $R e s (φ, z_{1}) = \frac{- 4 i}{(a - 1) (z_{1} - z_{2})}$ $= \frac{- 4 i}{(a - 1) (z_{1} - \frac{1}{z_{1}})} = \frac{- 4 i z_{1}}{(a - 1) (z_{1}^{2} - 1)}$ $= \frac{- 4 i \frac{1 - \sqrt{a}}{1 + \sqrt{a}}}{(a - 1) (\frac{{(1 - \sqrt{a})}^{2}}{{(1 + \sqrt{a})}^{2}} - 1)}$ $= \frac{- 4 i (1 - \sqrt{a})}{(a - 1) (1 + \sqrt{a}) {{(1 - \sqrt{a})}^{2} - {(1 + \sqrt{a})}^{2}}} {(1 + \sqrt{a})}^{2}$ $= \frac{4 i}{a - 2 \sqrt{a} + 1 - a - 2 \sqrt{a} - 1} = \frac{4 i}{- 4 \sqrt{a}} = \frac{- i}{\sqrt{a}}$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π {\frac{- i}{\sqrt{a}}} = \frac{2 π}{\sqrt{a}} \Rightarrow$ $f (a) = \frac{2 π}{\sqrt{a}} w i t h a > 1$ $c a s e 2 0 < a < 1 ∣ z_{1} ∣ - 1 = \frac{1 - \sqrt{a}}{1 + \sqrt{a}} - 1$ $= \frac{1 - \sqrt{a} - 1 - \sqrt{a}}{1 + \sqrt{a}} = \frac{- 2 \sqrt{a}}{1 + a} < 0 \Rightarrow ∣ z_{1} ∣< 1$ $a n d ∣ z_{2} ∣> 1 s o t h e v a l u e o f f (a) d o n t c h a n g e a n d$ $f (a) = \frac{2 π}{\sqrt{a}} .$
Commented by abdo mathsup 649 cc last updated on 20/May/18
$we have f^′ (a) = ∫_0 ^(2π) (∂/∂a){ (1/(a cos^2 t +sin^2 t))}dt = − ∫_0 ^(2π) ((cos^2 t)/((acos^2 t +sin^2 t)^2 ))dt =−g(a) ⇒ g(a) =−f^′ (a) or f(a)= ((2π)/(√a)) ⇒f^′ (a) =2π{ −((((√a))^′ )/a)} =−2π (1/(2a(√a))) = ((−π)/(a(√a))) ⇒ g(a) = −(π/(a(√a))) .$
$w e h a v e f^{^{'}} (a) = \int_{0}^{2 π} \frac{\partial}{\partial a} {\frac{1}{a {c o s}^{2} t + {s i n}^{2} t}} d t$ $= - \int_{0}^{2 π} \frac{{c o s}^{2} t}{{({a c o s}^{2} t + {s i n}^{2} t)}^{2}} d t = - g (a) \Rightarrow$ $g (a) = - f^{^{'}} (a) o r f (a) = \frac{2 π}{\sqrt{a}} \Rightarrow f^{^{'}} (a) = 2 π {- \frac{{(\sqrt{a})}^{^{'}}}{a}}$ $= - 2 π \frac{1}{2 a \sqrt{a}} = \frac{- π}{a \sqrt{a}} \Rightarrow g (a) = - \frac{π}{a \sqrt{a}} .$
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