find-the-value-of-0-2pi-dx-cos-2-x-3-sin-2-x-

Question Number 36197 by prof Abdo imad last updated on 30/May/18

f i n d t h e v a l u e o f \int_{0}^{2 π} \frac{d x}{{c o s}^{2} x + 3 {s i n}^{2} x}

Commented by maxmathsup by imad last updated on 14/Aug/18

l e t A = \int_{0}^{2 π} \frac{d x}{{c o s}^{2} x + 3 {s i n}^{2} x}

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

=_{2 x = t} \int_{0}^{4 π} \frac{d t}{2 + c o s t - s i n t} = \int_{0}^{2 π} \frac{d t}{2 + c o s t - s i n t} + \int_{2 π}^{4 π} \frac{d t}{2 + c o s t - s i n t}

b u t \int_{2 π}^{4 π} \frac{d t}{2 + c o s t - s i n t} =_{t = 2 π + α} \int_{0}^{2 π} \frac{d α}{2 + c o s α - s i n α} \Rightarrow

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

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

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

= \int_{∣ z ∣= 1} \frac{- 4 i z}{(1 - i) z^{2} + 4 z + 1 + i} l e t c o n s i d e r t h e c o m p o l e x f u n c t i o n

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

Δ^{^{'}} = 4 - (1 - i) (1 + i) = 4 - 2 = 2 \Rightarrow z_{1} = \frac{- 2 + \sqrt{2}}{1 - i} = \frac{\sqrt{2} (1 - \sqrt{2})}{\sqrt{2} e^{- \frac{i π}{4}}} = (1 - \sqrt{2}) e^{\frac{i π}{4}}

z_{2} = \frac{- 2 - \sqrt{2}}{1 - i} = \frac{\sqrt{2} (- 1 - \sqrt{2})}{\sqrt{2} e^{- \frac{i π}{4}}} = (- 1 - \sqrt{2}) e^{\frac{i π}{4}}

∣ z_{1} ∣ = \sqrt{2} - 1 < 1 a n d ∣ z_{2} ∣ = \sqrt{2} + 1 > 1 (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}) b u t φ (z) = \frac{- 4 i}{(1 - i) (z - z_{1}) (z - z_{2})}

R e s (φ, z_{1}) = \frac{- 4 i}{(1 - i) (z_{1} - z_{2})} = \frac{- 4 i}{(1 - i) 2 e^{i \frac{π}{4}}} = \frac{- 2 i}{\sqrt{2}} = \frac{- i}{\sqrt{2}} \Rightarrow

\int_{∣ z ∣= 1} φ (z) d z = 2 i π (\frac{- i}{\sqrt{2}}) = \frac{2 π}{\sqrt{2}} \Rightarrow A = \frac{2 π}{\sqrt{2}} \Rightarrow A = π \sqrt{2} .