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Question Number 43909 by abdo.msup.com last updated on 17/Sep/18

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

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

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

Answered by maxmathsup by imad last updated on 24/Sep/18

c h a n g e m e n t e^{i x} = z g i v e f (a, t) = \int_{∣ z ∣= 1} \frac{1}{a + t \frac{z - z^{- 1}}{2 i}} \frac{d z}{i z}

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

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

Δ^{^{'}} = - a^{2} + t^{2} = t^{2} - a^{2}

c a s e 1 Δ^{^{'}} > 0 \Rightarrow z_{1} = \frac{- i a + \sqrt{t^{2} - a^{2}}}{t} a n d z_{2} = \frac{- i a - \sqrt{t^{2} - a^{2}}}{t}

∣ z_{1} ∣ - 1 = \frac{1}{∣ t ∣} \sqrt{a^{2} + t^{2} - a^{2}} = 1 \Rightarrow∣ z_{1} ∣> 1 (o u t o f c i r c l e)

∣ z_{2} ∣ - 1 = \frac{1}{∣ t ∣} \sqrt{a^{2} + t^{2} - a^{2}} = 1 \Rightarrow∣ z_{2} ∣> 1 (o u t o f c i r c l e) \Rightarrow

\int_{∣ z ∣} φ (z) d z = 0

c a s e 2 Δ^{^{'}} < 0 \Rightarrow t^{2} - a^{2} < 0 \Rightarrow Δ^{^{'}} = {(i \sqrt{a^{2} - t^{2}})}^{2} \Rightarrow z_{1} = \frac{- i a + i \sqrt{a^{2} - t^{2}}}{t}

z_{2} = \frac{- i a - i \sqrt{a^{2} - t^{2}}}{t}

∣ z_{1} ∣ - 1 = \frac{∣ a - \sqrt{a^{2} - t^{2}} ∣}{∣ t ∣}

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

i f a > 0 w e g e t ∣ z_{1} ∣> 1 a n d ∣ z_{2} ∣> 1 \Rightarrow \int_{∣ z ∣= 1} φ (z) d z = 0

i f a < 0 ∣ z_{1} ∣> 1 a n d ∣ z_{2} ∣< 1 \Rightarrow \int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{2})

b u t φ (z) = \frac{2}{t (z - z_{1}) (z - z_{2})} \Rightarrow R e s (φ, z_{2}) = \frac{2}{t (z_{2} - z_{1})} = \frac{2}{t (\frac{- 2 i \sqrt{a^{2} - t^{2}}}{t})}

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

Commented by maxmathsup by imad last updated on 24/Sep/18

2) w e h a v e f (a, t) = \int_{0}^{2 π} \frac{d x}{a + t s i n x} \Rightarrow \frac{\partial f}{\partial t} (a, t) = - \int_{0}^{2 π} \frac{s i n x}{{(a + t s i n x)}^{2}} d x \Rightarrow

\int_{0}^{2 π} \frac{s i n x}{{(a + t s i n x)}^{2}} d x = - \frac{\partial f}{\partial t} (a, t)

c a s e 1 ∣ t ∣>∣ a ∣ \Rightarrow f (a, t) = 0 \Rightarrow \int_{0}^{2 π} \frac{s i n x}{{(a + t s i n x)}^{2}} d x = 0

c a s e 2 ∣ t ∣<∣ a ∣ \Rightarrow f (a, t) = \frac{- 2 π}{\sqrt{a^{2} - t^{2}}} \Rightarrow \frac{\partial f}{\partial t} (a, t) = - 2 π {{(a^{2} - t^{2})}^{- \frac{1}{2}}}^{(1) / t}

= - 2 π (- \frac{1}{2}) (- 2 t) {(a^{2} - t^{2})}^{- \frac{3}{2}} = \frac{- 2 π t}{(a^{2} - t^{2}) \sqrt{a^{2} - t^{2}}} \Rightarrow

\int_{0}^{2 π} \frac{s i n x}{{(a + t s i n x)}^{2}} d x = \frac{2 π t}{(a^{2} - t^{2}) \sqrt{a^{2} - t^{2}}} .

Commented by maxmathsup by imad last updated on 24/Sep/18

3) w e h a v e f (a, t) = \int_{0}^{2 π} \frac{d x}{a + t s i n x} \Rightarrow \frac{\partial f}{\partial a} (a, t) = - \int_{0}^{2 π} \frac{d x}{{(a + t s i n x)}^{2}} \Rightarrow

\int_{0}^{2 π} \frac{d x}{{(a + t s i n x)}^{2}} = - \frac{\partial f}{\partial a} (a, t)

c a s e 1 ∣ t ∣>∣ a ∣ \Rightarrow f (a, t) = 0 \Rightarrow \int_{0}^{2 π} \frac{d x}{{(a + t s i n x)}^{2}} = 0

c a s e 2 ∣ t ∣<∣ a ∣ \Rightarrow f (a, t) = \frac{- 2 π}{\sqrt{a^{2} - t^{2}}} \Rightarrow \frac{\partial f}{\partial a} (a, t) = - 2 π {{(a^{2} - t^{2})}^{- \frac{1}{2}}}^{(1) / a}

= - 2 π (2 a) (- \frac{1}{2}) {(a^{2} - t^{2})}^{- \frac{3}{2}} = \frac{2 π}{(a^{2} - t^{2}) \sqrt{a^{2} - t^{2}}} \Rightarrow

\int_{0}^{2 π} \frac{d x}{{(a + t s i n x)}^{2}} = \frac{- 2 π}{(a^{2} - t^{2}) \sqrt{a^{2} - t^{2}}} .