0-2pi-1-a-2-cos-2-t-b-2-sin-2-t-dt-2pi-ab-

Question Number 61465 by arcana last updated on 02/Jun/19

\int_{0}^{2 π} \frac{1}{a^{2} {c o s}^{2} (t) + b^{2} {s i n}^{2} (t)} d t = \frac{2 π}{a b} ?

Commented by maxmathsup by imad last updated on 03/Jun/19

l e t A = \int_{0}^{2 π} \frac{d t}{a^{2} {c o s}^{2} t + b^{2} {s i n}^{2} t} \Rightarrow A = \int_{0}^{2 π} \frac{2 d t}{a^{2} (1 + c o s (2 t)) + b^{2} (1 - c o s (2 t)}

= \int_{0}^{2 π} \frac{2 d t}{a^{2} + b^{2} + (a^{2} - b^{2}) c o s (2 t)} =_{2 t = x} \int_{0}^{4 π} \frac{d x}{a^{2} + b^{2} + (a^{2} - b^{2}) c o s (x)}

= \int_{0}^{2 π} \frac{d x}{a^{2} + b^{2} + (a^{2} - b^{2}) c o s (x)} + \int_{2 π}^{4 π} \frac{d x}{a^{2} + b^{2} + (a^{2} - b^{2}) c o s (x)} = H + K

H =_{e^{i x} = z} \int_{∣ z ∣= 1} \frac{d z}{i z {a^{2} + b^{2} + (a^{2} - b^{2}) \frac{z + z^{- 1}}{2}}} = \int_{∣ z ∣= 1} \frac{2 d z}{i z {2 (a^{2} + b^{2}) + (a^{2} - b^{2}) (z + z^{- 1})}}

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

= \int \frac{- 2 i d z}{(a^{2} - b^{2}) z^{2} + 2 (a^{2} + b^{2}) z + a^{2} - b^{2}} l e t φ (z) = \frac{- 2 i}{(a^{2} - b^{2}) z^{2} + 2 (a^{2} + b^{2}) z + a^{2} - b^{2}}

Δ_{d}^{^{'}} = = {(a^{2} + b^{2})}^{2} - {(a^{2} - b^{2})}^{2} = a^{4} + 2 a^{2} b^{2} + b^{4} - a^{4} + 2 a^{2} b^{2} - b^{4} = 4 a^{2} b^{2} \Rightarrow

z_{1} = \frac{- a^{2} - b^{2} + 2 ∣ a b ∣}{a^{2} - b^{2}} = - \frac{{(∣ a ∣ - ∣ b ∣)}^{2}}{∣ a ∣^{2} - ∣ b ∣^{2}} = - \frac{∣ a ∣ - ∣ b ∣}{∣ a ∣ + ∣ b ∣} (a \neq b)

z_{2} = \frac{- a^{2} - b^{2} - 2 ∣ a b ∣}{a^{2} - b^{2}} = - \frac{{(∣ a ∣ + ∣ b ∣)}^{2}}{a^{2} - b^{2}} = - \frac{∣ a ∣ + ∣ b ∣}{∣ a ∣ - ∣ b ∣} (= \frac{1}{z_{1}}) ∣ z

∣ z_{1} ∣ - 1 = \frac{∣∣ a ∣ - ∣ b ∣∣}{∣ a ∣ + ∣ b ∣} - 1 = \frac{∣∣ a ∣ - ∣ b ∣∣ - (∣ a ∣ + ∣ b ∣)}{(\dots .)} < 0

∣ z_{2} ∣ - 1 > 0 (t o e l i m i n a t e f r o m r e s i d u s \Rightarrow

\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1}) w e h a v e φ (z) = \frac{- 2 i}{(a^{2} - b^{2}) (z - z_{1}) (z - z_{2})} ⇎

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

= \frac{- 2 i}{4 ∣ a b ∣} = - \frac{i}{2 ∣ a b ∣} \Rightarrow \int_{∣ z ∣= 1} φ (z) d z = 2 i π \frac{- i}{2 ∣ a b ∣} = \frac{π}{∣ a b ∣} = H

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

\int_{0}^{2 π} \frac{d t}{a^{2} {c o s}^{2} t + b^{2} {s i n}^{2} t} = 2 H = \frac{2 π}{∣ a b ∣}

i f a = b w e g e t \int_{0}^{2 π} \frac{d t}{a^{2} {c o s}^{2} t + b^{2} {s i n}^{2} t} = \frac{1}{a^{2}} (2 π) = \frac{2 π}{a^{2}} .

Answered by tanmay last updated on 03/Jun/19

\int_{0}^{2 π} \frac{{s e c}^{2} t}{a^{2} + b^{2} {t a n}^{2} t} d t

\frac{1}{b^{2}} \int_{0}^{2 π} \frac{d (t a n t)}{{(\frac{a}{b^{}})}^{2} + {(t a n t)}^{2}} d t

\int_{0}^{2 a} f (t) d t = 2 \int_{0}^{a} f (t) d t w h e n f (2 a - t) = f (t)

\frac{2}{b^{2}} \int_{0}^{π} \frac{d (t a n t)}{{(\frac{a}{b})}^{2} + {(t a n t)}^{2}}

= \frac{2}{b^{2}} \times 2 \int_{0}^{\frac{π}{2}} \frac{d (t a n t)}{{(\frac{a}{b})}^{2} + {(t a n t)}^{2}}

= \frac{4}{b^{2}} \times \frac{1}{(\frac{a}{b})} \times ∣ {t a n}^{- 1} (\frac{t a n t}{\frac{a}{b}}) ∣_{0}^{\frac{π}{2}}

= \frac{4}{a b} \times [{t a n}^{- 1} (\infty) - {t a n}^{- 1} (0)]

= \frac{4}{a b} \times \frac{π}{2} = \frac{2 π}{a b}

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