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Question Number 38210 by prof Abdo imad last updated on 22/Jun/18

l e t f (a) = \int_{0}^{π} \frac{d θ}{a + {s i n}^{2} θ} (a f r o m R)

1) f i n d f (a)

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

3) c a l c u l a t e \int_{0}^{π} \frac{d θ}{1 + {s i n}^{2} θ} a n d \int_{0}^{π} \frac{d θ}{2 + {s i n}^{2} θ}

4) c a l c u l a t e \int_{0}^{π} \frac{d θ}{{(3 + {s i n}^{2} θ)}^{2}} .

Commented by math khazana by abdo last updated on 23/Jun/18

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

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

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

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

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

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

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

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

c a s e 1 a (a + 1) > 0 \Rightarrow z_{1} = 2 a + 1 + 2 \sqrt{a^{2} + a}

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

∣ z_{2} ∣ - 1 = 2 a - 2 \sqrt{a^{2} + a} < 0 a n d i t s c l e a r t h a t

∣ z_{1} ∣ - 1 > 0 (t o e l o m i n a t e f r o m r e s i d u s)

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

= \frac{- 4 π}{- 4 \sqrt{a^{2} + a}} = \frac{π}{\sqrt{a^{2} + a}} \Rightarrow

f (a) = \frac{π}{\sqrt{a^{2} + a}}

c a s e 2 a (a + 1) < 0 \Leftrightarrow - 1 < a < 0 \Rightarrow Δ^{,} = - 4 (- a (a + 1))

= {(2 i \sqrt{- a^{2} - a})}^{2}

z_{1} = 2 a + 1 + 2 i \sqrt{- a^{2} - a}

z_{2} = 2 a + 1 - 2 i \sqrt{- a^{2} - a}

∣ z_{1} ∣= \sqrt{{(2 a + 1)}^{2} + 4 (- a^{2} - a)}

= \sqrt{4 a^{2} + 4 a + 1 - 4 a^{2} - 4 a} = 1

∣ z_{2} ∣= 1 \Rightarrow

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

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

R e s (φ, z_{2}) = \frac{2 i}{z_{2} - z_{1}} \Rightarrow \int_{∣ z ∣= 1} φ (z) d z = 0 \Rightarrow f (a) = 0

Commented by math khazana by abdo last updated on 23/Jun/18

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

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

g (a) = - f^{^{'}} (a)

i f a (a + 1) < 0 \Rightarrow g (a) = 0

i f a (a + 1) > 0 f (a) = \frac{π}{\sqrt{a^{2} + a}} \Rightarrow

f^{^{'}} (a) = π {{(a^{2} + a)}^{- \frac{1}{2}}}^{^{'}} = - \frac{π}{2} (2 a + 1) {(a^{2} + a)}^{- \frac{3}{2}}

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

g (a) = \frac{(2 a + 1) π}{2 (a^{2} + a) \sqrt{a^{2} + a}} .

Commented by math khazana by abdo last updated on 23/Jun/18

3) w e h a v e p r o v e d t h a t f (a) = \frac{π}{\sqrt{a^{2} + a}} i f a (a + 1) > 0

s o \int_{0}^{π} \frac{d θ}{1 + {s i n}^{2} θ} = f (1) = \frac{π}{\sqrt{2}}

\int_{0}^{π} \frac{d θ}{2 + {s i n}^{2} θ} = f (2) = \frac{π}{\sqrt{6}}

\frac{}{}

Commented by math khazana by abdo last updated on 23/Jun/18

4) \int_{0}^{π} \frac{d θ}{{(3 + {s i n}^{2} θ)}^{2}} = g (3) = \frac{7 π}{24 \sqrt{12}} = \frac{7 π}{48 \sqrt{3}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18

\int_{0}^{Π} \frac{d θ}{a + {s i n}^{2} θ}

\int \frac{{s e c}^{2} θ}{{a s e c}^{2} θ + {t a n}^{2} θ} d θ

t = t a n θ d t = {s e c}^{2} θ d θ

\int \frac{d t}{a + {a t}^{2} + t^{2}}

\int \frac{d t}{a + t^{2} (1 + a)}

\frac{1}{1 + a} \int \frac{d t}{\frac{a}{1 + a} + t^{2}}

\frac{1}{1 + a} \times \frac{1}{\frac{\sqrt{a}}{\sqrt{1 + a}}} {t a n}^{- 1} (\frac{t}{\frac{\sqrt{a}}{\sqrt{1 + a}}})

= 2 ∣ \frac{1}{\sqrt{a} \times \sqrt{1 + a}} \times {t a n}^{- 1} (\frac{t a n θ}{\frac{\sqrt{a}}{\sqrt{1 + a}}}) ∣_{0}^{\frac{Π}{2}}

= 2 \times \frac{1}{\sqrt{a} \times \sqrt{1 + a}} \times \frac{Π}{2}

= \frac{Π}{\sqrt{a + a^{2}}} A n s

\int_{0}^{Π} f (θ) d θ = 2 \int_{0}^{\frac{Π}{2}} f (θ) d θ w h e n f (Π - θ) = f (θ)

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18

g (a) = \int_{0}^{Π} \frac{d θ}{{(a + {s i n}^{2} θ)}^{2}}

f (a) = \int_{0}^{Π} \frac{d θ}{a + {s i n}^{2} θ}

\frac{d f (a)}{d a} = \int_{0}^{Π} \frac{\partial}{\partial a} \frac{1}{(a + {s i n}^{2} θ)} d θ

= \int_{0}^{Π} \frac{- 1}{{(a + {s i n}^{2} θ)}^{2}} d θ

s o \frac{d f (a)}{d a} = - g (a)

g (a) = - \frac{d}{d a} (\frac{Π}{\sqrt{a + a^{2}}})

g (a) = - Π \times \frac{- 1}{2 {(a + a^{2})}^{\frac{3}{2}}} \times (1 + 2 a)

g (a) = \frac{Π (1 + 2 a)}{2 {(a + a^{2})}^{\frac{3}{2}}} A N S

i h a v e p r o v e d \int_{0}^{Π} \frac{d θ}{a + {s i n}^{2} θ} = \frac{Π}{\sqrt{a + a^{2}}}