---
title: "Contour ∫dθ/(5-4sinθ)"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:contour-1-over-5-4sin"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/contour-1-over-5-4sin
site: BCT Engineering
---

# Contour ∫dθ/(5-4sinθ)

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2072 Asoj, 5 marks
- 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-5-4sin
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Evaluate ∫₀^{2π} 1/(5 - 4 sinθ) dθ by contour integration method in complex plane.

## Short answer

Substitute z=e^{iθ}, unit circle. Poles at z=2i (outside) and z=i/2 (inside). Residue = 1/(3i). Integral = 2πi·1/(3i) = 2π/3.

## Answer

**Method (unit circle substitution).** Put z = e^{iθ}. As θ goes from 0 to 2π, z traverses the unit circle |z| = 1 once. Then

dθ = dz/(iz),  sin θ = (z − 1/z)/(2i)

**Transform the integrand:**

5 − 4 sin θ = 5 − (2/i)(z − 1/z) = 5 + 2i(z − 1/z) = (2i z² + 5z − 2i)/z

I = ∮ [ z/(2i z² + 5z − 2i) ] · dz/(iz) = ∮ dz / ( i (2i z² + 5z − 2i) ) = ∮ dz / ( −2z² + 5iz + 2 )

Factor: −2z² + 5iz + 2 = −2 (z² − (5i/2) z − 1) = −2 (z − 2i)(z − i/2).

So I = ∮_C (−1/2) · 1 / ((z − 2i)(z − i/2)) dz, C: |z| = 1.

**Poles:** z = 2i (|z| = 2, outside C) and z = i/2 (|z| = 1/2, inside C).

**Residue at z = i/2:**

Res = (−1/2) · 1/(i/2 − 2i) = (−1/2) · 1/(−3i/2) = 1/(3i)

**Result:** I = 2πi · 1/(3i) = 2π/3.

**Answer: ∫₀^{2π} dθ/(5 − 4 sin θ) = 2π/3**

(Check with the standard formula ∫₀^{2π} dθ/(a + b sin θ) = 2π/√(a² − b²) = 2π/√(25 − 16) = 2π/3.)

## Check yourself

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

1. What substitution converts the integral to a unit-circle integral?
   A. z = e^{iθ}, dθ = dz/z, sin θ = z-1/z
   B. z = e^{iθ}, dθ = dz/(iz), sin θ = (z-1/z)/(2i)
   C. z = tan θ
   D. z = e^{iθ}, dθ = iz dz
   Answer: B. These hold on |z| = 1. (Key point: Substitution)

2. Which pole is inside |z| = 1?
   A. z = 2i
   B. Both
   C. z = i/2
   D. Neither
   Answer: C. |i/2| = 1/2 < 1; |2i| = 2. (Key point: Poles)

3. What is the value?
   A. 2π/5
   B. π/3
   C. 2π
   D. 2π/3
   Answer: D. Residue 1/(3i), times 2πi gives 2π/3. (Key point: Result)

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