0-2-dt-4-2-sint-6-

Question Number 149917 by mathdanisur last updated on 08/Aug/21

\underset{0}{\int^{2 π}} \frac{dt}{4 \sqrt{2} \sin t + 6} = ?

Answered by mathmax by abdo last updated on 08/Aug/21

Ψ = \int_{0}^{2 π} \frac{dt}{4 \sqrt{2} sint + 6} \Rightarrow Ψ = \int_{0}^{2 π} \frac{dt}{4 \sqrt{2} (\frac{e^{it} - e^{- it}}{2 i}) + 6}

= \int_{0}^{2 π} \frac{2 idt}{4 \sqrt{2} (e^{it} - e^{- it}) + 12 i} =_{e^{it} = z} \int_{∣ z ∣= 1} \frac{2 i}{4 \sqrt{2} (z - z^{- 1}) + 12 i} \frac{dz}{iz}

= \int_{∣ z ∣= 1} \frac{1}{z (2 \sqrt{2} z - 2 \sqrt{2} z^{- 1} + 6 i)} dz

= \int_{∣ z ∣= 1} \frac{dz}{2 \sqrt{2} z^{2} + 6 iz - 2 \sqrt{2}} let Λ (z) = \frac{1}{2 \sqrt{2} z^{2} + 6 iz - 2 \sqrt{2}} poles of Λ ?

Δ^{^{'}} = {(3 i)}^{2} + 8 = - 9 + 8 = - 1

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

z_{2} = \frac{- 3 i - i}{2 \sqrt{2}} = \frac{- 4 i}{2 \sqrt{2}} = \frac{- 2 i}{\sqrt{2}} = - \sqrt{2} i ona

∣ z_{1} ∣= \frac{1}{2} < 1 and ∣ z_{2} ∣= \sqrt{2} > 1 (out of circle) \Rightarrow

\Rightarrow \int_{∣ z ∣= 1} Λ (z) dz = 2 i π Res (Λ, z_{1}) we have

Λ (z) = \frac{1}{2 \sqrt{2} (z - z_{1}) (z - z_{2})} \Rightarrow Res (Λ, z_{1}) = \frac{1}{2 \sqrt{2} (z_{1} - z_{2})}

= \frac{1}{2 \sqrt{2} (- \frac{i}{\sqrt{2}} + \sqrt{2} i)} = \frac{1}{2 i \sqrt{2} (\sqrt{2} - \frac{1}{\sqrt{2}})} = \frac{\sqrt{2}}{2 i \sqrt{2}} = \frac{1}{2 i} \Rightarrow

\int_{∣ z ∣= 1} Λ (z) dz = 2 i π \times \frac{1}{2 i} = π \Rightarrow Ψ = π

Commented by mathdanisur last updated on 08/Aug/21

S er, Thank You