---
title: "Contour ∫cos3θ/(5-4cosθ) dθ"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:contour-cos3theta-over-5-4cos"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/contour-cos3theta-over-5-4cos
site: BCT Engineering
---

# Contour ∫cos3θ/(5-4cosθ) dθ

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2073 Magh, 5 marks; 2069 Bhadra (asked 2 times)
- Video: [Complex Integration| Integral of Cos (3 theta)/[5- 4 Cos (theta)]| Integration of Cos3x/[5-4 Cosx]](https://www.youtube.com/watch?v=fZSJtEEsyyA) — Let's Find the Value of "x"
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/contour-cos3theta-over-5-4cos
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Evaluate real integral ∫₀^{2π} cos3θ/(5 - 4cosθ) dθ by contour integration in the complex plane.

## Short answer

Substitute z=e^{iθ} to convert the real integral to ∮_C i z³/((2z−1)(z−2)) dz over |z|=1. Pole at z=1/2 inside yields residue −i/24. By residue theorem, J = 2πi·(−i/24)=π/12, so the original real integral equals π/12.

## Answer

**Method.** Let z = e^{iθ} so that θ from 0 to 2π traces the unit circle |z| = 1. Then

cos θ = (z + 1/z)/2,  dθ = dz/(iz),  cos 3θ = Re(e^{3iθ}) = Re(z³).

#### Solution
Let I = ∫₀^{2π} cos 3θ/(5 − 4 cos θ) dθ. Write I = Re J, where

J = ∫₀^{2π} e^{3iθ}/(5 − 4 cos θ) dθ  (the imaginary part is sin 3θ/(5 − 4cos θ), which integrates to 0 by symmetry, but we only need the real part).

Denominator: 5 − 4 cos θ = 5 − 2(z + 1/z) = −(2z² − 5z + 2)/z = −(2z − 1)(z − 2)/z.

```text
J = ∮_C z³ · [−z/((2z − 1)(z − 2))] · dz/(iz)
  = ∮_C [ i z³ / ((2z − 1)(z − 2)) ] dz
```

(because −1/i = i.)

Poles: z = 1/2 (inside C) and z = 2 (outside C).

Residue at z = 1/2:

Res = i (1/2)³ / [2 (1/2 − 2)] = i(1/8)/(−3) = −i/24

By the residue theorem:

J = 2πi · (−i/24) = 2π/24 = π/12

This is real, so I = Re J = π/12.

**Answer: ∫₀^{2π} cos 3θ/(5 − 4 cos θ) dθ = π/12 ≈ 0.2618**

(Numerical integration confirms 0.26180.)

## Check yourself

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

1. cos 3θ is handled by:
   A. Setting z³ = 1
   B. Writing I as Re of integral of e^{3iθ}/(5−4cosθ), i.e. z³
   C. Using cos 3θ = 3 cos θ
   D. Ignoring it
   Answer: B. cos 3θ = Re(z³). (Key point: Real part)

2. Poles of the integrand are z = 1/2 and z = 2. Inside |z|=1 is:
   A. neither
   B. both
   C. z = 2
   D. z = 1/2
   Answer: D. |1/2| < 1. (Key point: Pole inside)

3. 5 − 4cosθ in terms of z is:
   A. (z²+1)/z
   B. (2z−1)(z−2)/z
   C. −(2z−1)(z−2)/z
   D. −(2z+1)(z+2)/z
   Answer: C. 5 − 2(z+1/z) = −(2z²−5z+2)/z. (Key point: Denominator)

4. The value is:
   A. π/3
   B. 2π/3
   C. π/6
   D. π/12
   Answer: D. J = 2πi(−i/24) = π/12. (Key point: Result)

## Close variants set in other exams

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