calculate-0-2pi-dx-2cos-2-x-3-sin-2-x-

Question Number 37587 by prof Abdo imad last updated on 15/Jun/18

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

Commented by math khazana by abdo last updated on 16/Jun/18

I = \int_{0}^{2 π} \frac{d x}{2 \frac{1 + c o s (2 x)}{2} + \sqrt{3} \frac{1 - c o s (2 x)}{2}}

= \int_{0}^{2 π} \frac{d x}{1 + c o s (2 x) + \frac{\sqrt{3}}{2} (1 - c o s (2 x))}

= \int_{0}^{2 π} \frac{2 d x}{2 + 2 c o s (2 x) + \sqrt{3} - \sqrt{3} c o s (2 x)}

= \int_{0}^{2 π} \frac{2 d x}{2 + \sqrt{3} + (2 - \sqrt{3}) c o s (2 x)}

=_{2 x = t} \int_{0}^{4 π} \frac{2 d x}{2 + \sqrt{3} + (2 - \sqrt{3}) c o s t} \frac{d t}{2}

= \int_{0}^{4 π} \frac{d t}{2 + \sqrt{3} + (2 - \sqrt{3}) c o s t}

= \int_{0}^{2 π} \frac{d t}{2 + \sqrt{3} + (2 - \sqrt{3}) c o s t} + \int_{2 π}^{4 π} \frac{d t}{2 + \sqrt{3} + (2 - \sqrt{3}) c o s t}

= K + H

c h a n g . e^{i t} = z g i v e

K = \int_{∣ z ∣ = 1} \frac{1}{2 + \sqrt{3} + (2 - \sqrt{3}) \frac{z + z^{- 1}}{2}} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{2 d z}{i z (4 + 2 \sqrt{3} + (2 - \sqrt{3}) (z + z^{- 1})}

= \int_{∣ z ∣= 1} \frac{- 2 i d z}{(2 - \sqrt{3}) z^{2} + (4 + 2 \sqrt{3}) z + 2 - \sqrt{3}}

l e t φ (z) = \frac{- 2 i}{(2 - \sqrt{3}) z^{2} + (4 + 2 \sqrt{3}) z + 2 - \sqrt{3}}

p o l e s o f φ ?

Δ^{^{'}} = {(2 + \sqrt{3})}^{2} - {(2 - \sqrt{3})}^{2} = 4 + 4 \sqrt{3} + 3 - 4 + 4 \sqrt{3} - 3

= 8 \sqrt{3}

z_{1} = \frac{- 2 - \sqrt{3} + 2 \sqrt{2} \sqrt{\sqrt{3}}}{2 - \sqrt{3}}

z_{2} = \frac{- 2 - \sqrt{3} - 2 \sqrt{2} \sqrt{\sqrt{3}}}{2 - \sqrt{3}}

∣ z_{1} ∣ - 1 = \frac{∣ 2 + \sqrt{3} - 2 \sqrt{2} \sqrt{\sqrt{3}} ∣}{2 - \sqrt{3}}

{(2 + \sqrt{3})}^{2} - 8 \sqrt{3} = 7 - 4 \sqrt{3} - 8 \sqrt{3} = 7 - 12 \sqrt{3} < 0

\Rightarrow∣ z_{1} ∣< 1 a n d w e v e r i f y t h a t ∣ z_{2} ∣> 1

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

R e s (φ, z_{1}) = \frac{- 2 i}{(2 - \sqrt{3}) (z_{1} - z_{2})}

= \frac{- 2 i}{(2 - \sqrt{3}) \frac{4 \sqrt{2 \sqrt{3}}}{2 - \sqrt{3}}} = \frac{- i}{2 \sqrt{2 \sqrt{3}}} . \Rightarrow

\int_{∣ z ∣= 1} φ (z) d z = 2 i π \frac{- i}{2 \sqrt{2 \sqrt{3}}} = \frac{π}{\sqrt{2 \sqrt{3}}} = K .

Commented by math khazana by abdo last updated on 16/Jun/18

H = \int_{2 π}^{4 π} \frac{d t}{2 + \sqrt{3} + (2 - \sqrt{3}) c o s t}

=_{t = 2 π + x} \int_{0}^{2 π} \frac{d x}{2 + \sqrt{3} + (2 - \sqrt{3}) c o s x} = K

= \frac{π}{\sqrt{2 \sqrt{3}}} \Rightarrow

I = H + K = \frac{2 π}{\sqrt{2 \sqrt{3}}} = \frac{π \sqrt{2}}{\sqrt{\sqrt{3}}}

I = \frac{π \sqrt{2}}{\sqrt{\sqrt{3}}} .

Answered by ajfour last updated on 15/Jun/18

\int_{0}^{2 π} \frac{\sec^{2} x d x}{2 + \sqrt{3} \tan^{2} x}

= \frac{4}{\sqrt{3}} \int_{0}^{\infty} \frac{d t}{t^{2} + \frac{2}{\sqrt{3}}}

= \frac{4}{\sqrt{3}} \times \frac{3^{1 / 4}}{\sqrt{2}} (\frac{π}{2}) = π \sqrt{\frac{2}{\sqrt{3}}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18

i n p l a c e o f \frac{Π}{2} i p u t \frac{Π}{4}