solve-by-contour-integrstion-0-2pi-dx-1-acosx-

Question Number 200130 by universe last updated on 14/Nov/23

s o l v e b y c o n t o u r i n t e g r s t i o n

\int_{0}^{2 π} \frac{d x}{1 + a \cos x}

Answered by Mathspace last updated on 14/Nov/23

I = \int_{0}^{2 π} \frac{d x}{1 + a \frac{e^{i x} + e^{- i x}}{2}} (e^{i x} = z)

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

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

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

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

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

Δ^{^{'}} = 1 - a^{2}

c a s e 1 ∣ a ∣< 1 \Rightarrow z_{1} = \frac{- 1 + \sqrt{1 - a^{2}}}{a}

z_{2} = \frac{- 1 - \sqrt{1 - a^{2}}}{a} (a \neq 0)

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

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

{(1 - ∣ a ∣)}^{2} - (1 - a^{2})

= 1 - 2 ∣ a ∣ + a^{2} - 1 + a^{2}

= 2 a^{2} - 2 ∣ a ∣= 2 ∣ a ∣ (∣ a ∣ - 1) < 0

\Rightarrow∣ z_{1} ∣< 1

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

= \frac{1 - ∣ a ∣ + \sqrt{1 - a^{2}}}{∣ a ∣} > 0 \Rightarrow∣ z_{2} ∣> 1

f (z) = \frac{- 2 i}{a (z - z_{1}) (z - z_{2})}

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

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

= \frac{2 π}{\sqrt{1 - a^{2}}} \Rightarrow

I = \frac{2 π}{\sqrt{1 - a^{2}}}

r e s t t h e c a s e \Rightarrow∣ a ∣> 1

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

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

\dots . f o l l o w t h e s a m e p a t h \dots

Answered by MM42 last updated on 15/Nov/23

I = \int \frac{d x}{1 + a c o s x}

i f a = 0 \Rightarrow I = x + c

i f a = 1 \Rightarrow I = \frac{1}{2} \int (1 + {t a n}^{2} \frac{x}{2}) d x \Rightarrow I = t a n \frac{x}{2} + c

i f a = - 1 \Rightarrow I = \frac{1}{2} \int (1 + {c o t a n}^{2} \frac{x}{2}) d x \Rightarrow I = - c o t a n \frac{x}{2} + c

o t h e r w i s e

\Rightarrow l e t t a n \frac{x}{2} = u \Rightarrow \int \frac{2 d u}{(1 + a) + (1 - a) u^{2}} = \frac{2}{1 - a} \int \frac{d u}{\frac{1 + a}{1 - a} + u^{2}}

l e t \frac{1 + a}{1 - a} = k \Rightarrow I = \frac{2}{1 - a} \int \frac{d u}{u^{2} + k}

i f ∣ a ∣< 1 \Rightarrow k > 0 \Rightarrow I = \frac{2}{1 - a} \times \frac{1}{\sqrt{k}} \times {t a n}^{- 1} (\frac{u}{\sqrt{k}}) + c

= \frac{2}{\sqrt{1 - a^{2}}} \times {t a n}^{- 1} (\sqrt{\frac{1 - a}{1 + a}} t a n (\frac{x}{2} \times \sqrt{\frac{1 - a}{1 + a}})) + c

i f ∣ a ∣> 1 \Rightarrow k < 0 \Rightarrow I = \frac{1}{1 - a} \times \frac{1}{k} \times l n (\frac{u - k}{u + k}) + c

= \frac{1}{\sqrt{1 - a^{2}}} \times l n (\frac{(1 - a) t a n \frac{x}{2} - \sqrt{1 + a}}{(1 - a) t a n \frac{x}{2} + \sqrt{1 + a}}) + c

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