---
title: "Half range sine series: e^(2x)"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:half-range-sine-series-e-2x"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-e-2x
site: BCT Engineering
---

# Half range sine series: e^(2x)

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2073 Chaitra, 5 marks
- Video: [Half Range sine series f(x)=e^x](https://www.youtube.com/watch?v=BwjvnPvIdWg) — Btech Maths Hub
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-e-2x
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Find the half range sine series for f(x) = e^(2x) in 0 < x < π.

## Short answer

Compute bₙ = (2/π)∫₀^π e^(2x) sin nx dx = (2/π) n[1 − (−1)ⁿ e^(2π)]/(n²+4). The half-range sine series is e^(2x) = (2/π) Σ_{n=1}^∞ { n[1−(−1)ⁿ e^(2π)]/(n²+4) } sin nx.

## Answer

The half range sine series on (0, π) is f(x) = Σ bₙ sin nx, with bₙ = (2/π)∫₀^π f(x) sin nx dx.

**Coefficient.** Use ∫ eᵃˣ sin nx dx = eᵃˣ(a sin nx − n cos nx)/(a² + n²) with a = 2:

```text
bₙ = (2/π)∫₀^π e^(2x) sin nx dx
   = (2/π)[ e^(2x)(2 sin nx − n cos nx)/(4 + n²) ]₀^π
   = (2/π)·[ −n(−1)ⁿ e^(2π) + n ]/(n² + 4)
   = (2/π)·n[1 − (−1)ⁿ e^(2π)]/(n² + 4)
```

**Series**

e^(2x) = (2/π) Σ (n=1 to ∞) { n[1 − (−1)ⁿ e^(2π)] / (n² + 4) } sin nx,  0 < x < π

= (2/π)[ (1 + e^(2π))/5 · sin x + 2(1 − e^(2π))/8 · sin 2x + 3(1 + e^(2π))/13 · sin 3x + … ]

For n = 1, 2, 3 the denominators n² + 4 are 5, 8 and 13.

## Check yourself

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

1. For a half-range sine series on (0, pi), what is bn?
   A. (1/2pi) integral of f(x) sin nx dx
   B. (2/pi) integral of f(x) sin nx dx from 0 to pi
   C. (1/pi) integral of f(x) cos nx dx
   D. (2/pi) integral of f(x) cos nx dx
   Answer: B. Half-range sine series uses sin nx with factor 2/pi. (Key point: Half-range sine formula)

2. Which integral formula is used with a = 2?
   A. e^(ax)(a cos nx + n sin nx)/(a^2+n^2)
   B. e^(ax)(sin nx - n cos nx)/(a^2-n^2)
   C. e^(ax)(a sin nx - n cos nx)/(a^2+n^2)
   D. e^(ax)(n sin nx - a cos nx)/(a^2+n^2)
   Answer: C. This is the standard antiderivative of e^(ax) sin nx. (Key point: Integral used)

3. What is bn?
   A. (2/pi) n[1 - (-1)^n e^(2 pi)]/(n^2+4)
   B. (2/pi) n[1 + (-1)^n e^(2 pi)]/(n^2-4)
   C. (1/pi) n[(-1)^n e^(2 pi) - 1]/(n^2+4)
   D. (2/pi)[1 - (-1)^n e^(2 pi)]/(n^2+4)
   Answer: A. At pi sin n pi = 0 and cos n pi = (-1)^n, leaving n[1 - (-1)^n e^(2pi)]/(n^2+4). (Key point: Coefficient)

## Close variants set in other exams

- [Half range sine series: x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x-x2-three) — Mathematics III, chapter 4, asked 11×
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- [Half range sine series: e^(ax)](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-e-ax) — Mathematics III, chapter 4, asked 5×
- [Half range sine and cosine series: eˣ](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-and-cosine-series) — Mathematics III, chapter 4, asked 2×

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