let-a-gt-b-gt-0-calculate-0-2pi-dx-a-bsinx-2-

Question Number 67525 by mathmax by abdo last updated on 28/Aug/19

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

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

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

\int_{0}^{2 π} \frac{d x}{c + s i n x} = 2 i π \sum_{∣ z_{k} ∣< 1} R e s (f (z), z_{k}) w i t h f (z) = \frac{1}{i z} . \frac{1}{c + \frac{z - z^{- 1}}{2 i}} = \frac{2}{z^{2} + 2 i c z - 1} = \frac{2}{(z - z_{0}) (z - z_{1})} w i t h z_{0} = - i (c + \sqrt{c^{2} - 1}) a n d z_{1} = - i (c - \sqrt{c^{2} - 1})

w e h a v e ∣ z_{0} ∣> 1 a n d ∣ z_{1} ∣< 1

\int_{0}^{2 π} \frac{d x}{c + s i n x} = 2 i π \frac{2}{z_{1} - z_{0}} = \frac{2 π}{\sqrt{c^{2} - 1}}

N o w f (a, b) = \frac{1}{b} . \frac{2 π}{\sqrt{{(\frac{a}{b})}^{2} - 1}} = \frac{2 π}{\sqrt{a^{2} - b^{2}}}

\frac{\partial f (a, b)}{\partial a} = \int_{0}^{2 π} \frac{d x}{{(a + b s i n x)}^{2}} = 2 π \frac{- \frac{2 a}{2 \sqrt{a^{2} - b^{2}}}}{(a^{2} - b^{2})} = \frac{- 4 π a}{{(a^{2} - b^{2})}^{\frac{3}{2}}}

Commented by mathmax by abdo last updated on 29/Aug/19

l e t f (a) = \int_{0}^{2 π} \frac{d x}{a + b s i n x} w e h a v e f^{^{'}} (a) = - \int_{0}^{2 π} \frac{d x}{{(a + b s i n x)}^{2}} \Rightarrow

\int_{0}^{2 π} \frac{d x}{{(a + b s i n x)}^{2}} = - f^{^{'}} (a) c h a n g e m e n t e^{i x} = z g i v e

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

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

= \int_{∣ z ∣= 1} \frac{2 d z}{2 i a z + {b z}^{2} - b} l e t W (z) = \frac{2}{{b z}^{2} + 2 i a z - b} p o l e s o f W ?

Δ^{^{'}} = - a^{2} + b^{2} = - (a^{2} - b^{2}) = {(i \sqrt{a^{2} - b^{2}})}^{2} \Rightarrow

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

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

\Rightarrow∣ z_{1} ∣< 1 w e h a v e z_{1} . z_{2} = - 1 \Rightarrow∣ z_{2} ∣= \frac{1}{∣ z_{1} ∣} > 1

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

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

= \frac{1}{i \sqrt{a^{2} - b^{2}}} \Rightarrow \int_{∣ z ∣= 1} W (z) d z = 2 i π \times \frac{1}{i \sqrt{a^{2} - b^{2}}} = \frac{2 π}{\sqrt{a^{2} - b^{2}}} = f (a)

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

= - π {(a^{2} - b^{2})}^{- \frac{3}{2}} \Rightarrow \int_{0}^{2 π} \frac{d x}{{(a + b s i n x)}^{2}} = \frac{π}{{(a^{2} - b^{2})}^{\frac{3}{2}}}

Commented by mathmax by abdo last updated on 29/Aug/19

e r r o r a t f i n a l l i n e s f (a) = 2 π {(a^{2} - b^{2})}^{- \frac{1}{2}} \Rightarrow

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

\int_{0}^{2 π} \frac{d x}{{(a + b s i n x)}^{2}} = \frac{2 π a}{{(a^{2} - b^{2})}^{\frac{3}{2}}}