find-pi-pi-2dt-2-sint-cost-

Question Number 28882 by abdo imad last updated on 31/Jan/18

f i n d \int_{- π}^{π} \frac{2 d t}{2 + s i n t + c o s t} .

Commented by abdo imad last updated on 02/Feb/18

l e t u s e t h e c h . e^{i t} = z

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

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

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

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

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

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

f u n c t i o n w (z) = \frac{4}{(1 + i) z^{2} + 4 i z - 1 + i} p o l e s o f w ?

Δ^{^{'}} = {(2 i)}^{2} - (1 + i) (- 1 + i) = - 4 - (- 2) = - 2 = {(i \sqrt{2})}^{2}

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

∣ z_{1} ∣= \frac{∣ - 2 i + i \sqrt{2} ∣}{∣ 1 + i ∣} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1 a n d ∣ z_{1} ∣ - 1 = \sqrt{2} - 2 < 0

s o ∣ z_{1} ∣< 1

Commented by abdo imad last updated on 02/Feb/18

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

\int_{∣ z ∣= 1} w (z) d z = 2 i π R e s (w, z_{1}) b u t

w (z) = \frac{4}{(1 + i) (z - z_{1}) (z - z_{2})} \Rightarrow

R e s (w, z_{1}) = \frac{4}{(1 + i) (z_{1} - z_{2})} = \frac{4}{(1 + i) \frac{2 i \sqrt{2}}{1 + i}} = \frac{2}{i \sqrt{2}} = \frac{\sqrt{2}}{i}

\int_{∣ z ∣= 1} w (z) d z = 2 i π . \frac{\sqrt{2}}{i} = 2 π \sqrt{2} . \Rightarrow I = 2 π \sqrt{2} .