---
title: "Real integral by contour integration"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:real-integral-by-contour-integration"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/real-integral-by-contour-integration
site: BCT Engineering
---

# Real integral by contour integration

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2082 Chaitra (new course), 4 marks
- Video: [Contour Integration.  Type :1 ( Trigonometric Functions)  in Complex Analysis.](https://www.youtube.com/watch?v=4Ze5-9jUCvU) — Maths preparation corner
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/real-integral-by-contour-integration
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∫_0^{2π} cos 2θ/(5 + 4 sin θ) dθ using the method of contour integration in the complex plane.

## Short answer

Substitute z=e^{iθ} to get contour integral over unit circle. Poles inside: z=0 (order 2) and z=-i/2. Sum of residues = i/12, so I=2πi·(i/12)=−π/6.

## Answer

Let I = ∫₀^2π cos 2θ/(5 + 4 sin θ) dθ. Put z = e^(iθ) on the unit circle |z| = 1:

```text
cos 2θ = (z² + z⁻²)/2,   sin θ = (z − 1/z)/(2i),   dθ = dz/(iz)
5 + 4 sin θ = 5 − 2iz + 2i/z = (−2i z² + 5z + 2i)/z = −i (2z² + 5iz − 2)/z
```

So

```text
I = ∮_C g(z) dz,   g(z) = (z⁴ + 1) / [2 z² (2z² + 5iz − 2)]
```

Factor: 2z² + 5iz − 2 = 2(z + 2i)(z + i/2). Poles: z = 0 (double), z = −i/2 (simple), z = −2i (simple). Inside |z| = 1: z = 0 and z = −i/2.

```text
Res at z = −i/2:  (z⁴ + 1) / [2z² · 2(z + 2i)] at z = −i/2
                = (17/16) / [(−1/2)(3i)] = 17i/24

Res at z = 0 (double pole): d/dz [ (z⁴ + 1) / (2(2z² + 5iz − 2)) ] at 0
                = [0 − 1·2·(5i)] / [4·(−2)²] = −5i/8
```

Sum of residues = 17i/24 − 15i/24 = 2i/24 = i/12.

```text
I = 2πi × (i/12) = −π/6
```

**Answer: I = −π/6** (numerical integration gives −0.5236, which agrees).

## Check yourself

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

1. The substitution used is:
   A. z = e^theta, dtheta = dz/z
   B. z = cos theta
   C. z = e^(i theta), dtheta = dz/(iz)
   D. z = tan(theta/2)
   Answer: C. Maps the integral to the unit circle. (Key point: Substitution)

2. Poles inside |z|=1 are:
   A. z = -2i and z = 0
   B. z = -2i only
   C. z = -i/2 only
   D. z = 0 (double) and z = -i/2
   Answer: D. -2i is outside the unit circle. (Key point: Poles inside)

3. Sum of residues is:
   A. i/12
   B. 17i/24
   C. -5i/8
   D. 2i
   Answer: A. 17i/24 - 15i/24 = i/12. (Key point: Residue sum)

4. The integral equals:
   A. pi/6
   B. -pi/6
   C. pi/12
   D. -pi/12
   Answer: B. 2 pi i * i/12 = -pi/6. (Key point: Result)

## Close variants set in other exams

- [Real integral by contour integration](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Areal-integral-by-contour-integration-two) — Mathematics III, chapter 2, asked 2×

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