1-calculate-f-a-0-pi-dx-a-sin-2-x-cos-2-x-with-a-gt-0-2-find-the-value-of-g-a-0-pi-sin-2-x-a-sin-2-x-cos-2-x-2-dx-

Question Number 35226 by abdo mathsup 649 cc last updated on 16/May/18

1) c a l c u l a t e f (a) = \int_{0}^{π} \frac{d x}{a {s i n}^{2} x + {c o s}^{2} x}

w i t h a > 0

2) f i n d t h e v a l u e o f g (a) = \int_{0}^{π} \frac{{s i n}^{2} x}{{(a {s i n}^{2} x + {c o s}^{2} x)}^{2}} d x

Commented by prof Abdo imad last updated on 19/May/18

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

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

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

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

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

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

e^{i t} = z g i v e

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

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

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

l e t c o n s i d e r φ (z) = \frac{- 2 i}{{(1 - a)}^{} z^{2} + 2 (1 + a) z + 1 - a}

p o l e s o f φ ?

Commented by prof Abdo imad last updated on 19/May/18

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

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

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

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

= \frac{\sqrt{a} + 1}{\sqrt{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 o m i n a t e f r o m r r s i d u s)

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

R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} (z - z_{1}) φ (z)

b u t φ (z) = \frac{- 2 i}{(1 - a) (z - z_{1}) (z - z_{2})}

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

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

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{- 2 {i z}_{1}}{(1 - a) (z_{1}^{2} - 1)}

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

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

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{4 π}{1 - a} \frac{\frac{\sqrt{a} - 1}{\sqrt{a} + 1}}{\frac{{(\sqrt{a} - 1)}^{2}}{{(\sqrt{a} + 1)}^{2}} - 1}

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

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

= \frac{4 π}{4 \sqrt{a}} = \frac{π}{\sqrt{a}} s o

f (a) = \frac{π}{\sqrt{a}} .

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

2) w e h a v e f^{^{'}} (a) = \int_{0}^{π} \frac{\partial}{\partial a} {\frac{1}{{a s i n}^{2} x + {c o s}^{2} x}} d x

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

g (a) = - f^{^{'}} (a) b u t f (a) = \frac{π}{\sqrt{a}} \Rightarrow f^{^{'}} (a) = - π \frac{{(\sqrt{a})}^{^{'}}}{a}

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