---
title: "Contour ∫dθ/(2+sinθ)"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:contour-1-over-2-plus-sin"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/contour-1-over-2-plus-sin
site: BCT Engineering
---

# Contour ∫dθ/(2+sinθ)

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2069 Poush
- Video: [By Using Contour Integration Type 01 Problems 02 -Complex Integration](https://www.youtube.com/watch?v=dvwFlCOqrmA) — SK Maths Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/contour-1-over-2-plus-sin
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Using contour integration to evaluate ∫₀^{2π} dθ/(2 + sinθ).

## Short answer

Set z=e^{iθ}. Integrand becomes 2/(z²+4iz-1). Pole inside unit circle at z₁=i(-2+√3), residue 1/(i√3). Value = 2πi·1/(i√3) = 2π/√3.

## Answer

**Method.** Let z = e^{iθ}, so the integral becomes one around the unit circle |z| = 1:

dθ = dz/(iz),  sin θ = (z² − 1)/(2iz)

**Transform:**

2 + sin θ = 2 + (z² − 1)/(2iz) = (z² + 4iz − 1)/(2iz)

I = ∮_C [ 2iz/(z² + 4iz − 1) ] · dz/(iz) = ∮_C 2 dz/(z² + 4iz − 1)

**Poles:** solve z² + 4iz − 1 = 0:

z = [ −4i ± √(−16 + 4) ] / 2 = −2i ± i√3 = i(−2 ± √3)

- z₁ = i(−2 + √3), |z₁| = 0.268 < 1 : **inside** C
- z₂ = −i(2 + √3), |z₂| = 3.73 > 1 : outside C

**Residue at z₁** (simple pole):

Res = 2/(z₁ − z₂) = 2/(2i√3) = 1/(i√3)

**Result:** I = 2πi · 1/(i√3) = 2π/√3.

**Answer: ∫₀^{2π} dθ/(2 + sin θ) = 2π/√3 ≈ 3.628**

(Check: 2π/√(a² − b²) with a = 2, b = 1.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. After substitution, what is the integrand?
   A. 2 dz/(z²+4iz-1)
   B. dz/(z²+4iz-1)
   C. 2 dz/(z²-4iz+1)
   D. dz/(2iz)
   Answer: A. I = ∮ 2iz/(z²+4iz-1) × dz/(iz) = ∮ 2 dz/(z²+4iz-1). (Key point: Transformation)

2. Which pole is inside the unit circle?
   A. z₂ = -i(2+√3)
   B. z₁ = i(-2+√3), modulus 0.268
   C. Both
   D. None
   Answer: B. z₂ has modulus 3.73. (Key point: Poles)

3. What is the value?
   A. 2π/3
   B. π/√3
   C. 2π/√3
   D. 2π
   Answer: C. Residue 1/(i√3) times 2πi. (Key point: Result)

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