---
title: "Half range sine series: πx − x²"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:half-range-sine-series-x-x2-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-x-x2-two
site: BCT Engineering
---

# Half range sine series: πx − x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2081 Bhadra, 5 marks
- Video: [Half range Fourier sine series of x^2 in (0, π) || 18mat31 || Dr Prashant Patil](https://www.youtube.com/watch?v=pXffA97BkiE) — Dr Prashant Patil
- 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-x-x2-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Expand f(x) = πx − x² in a half range sine series in the interval (0, π) up to the first three terms.

## Short answer

Half-range sine series: f(x) = Σ b_n sin nx with b_n = 4(1−(−1)^n)/(π n^3), non-zero for odd n. First three terms: (8/π)(sin x + sin 3x/27 + sin 5x/125).

## Answer

Half-range sine series on (0, π): f(x) = Σ bₙ sin nx, bₙ = (2/π) ∫₀^π (πx − x²) sin nx dx.

Integrate by parts twice:

```text
∫₀^π (πx − x²) sin nx dx = [ −(πx − x²) cos nx / n ]₀^π + (1/n) ∫₀^π (π − 2x) cos nx dx
                         = 0 + (1/n) [ (π − 2x) sin nx/n ]₀^π + (2/n²) ∫₀^π sin nx dx
                         = (2/n²) (1 − (−1)ⁿ)/n
                         = 2 (1 − (−1)ⁿ)/n³
```

```text
bₙ = (2/π) · 2 (1 − (−1)ⁿ)/n³ = 4 (1 − (−1)ⁿ)/(π n³)
```

So bₙ = 0 for even n, and bₙ = 8/(π n³) for odd n.

```text
πx − x² = (8/π) [ sin x/1³ + sin 3x/3³ + sin 5x/5³ + ... ]
```

**First three terms:**

```text
πx − x² ≈ (8/π) [ sin x + (1/27) sin 3x + (1/125) sin 5x ]
        ≈ 2.546 sin x + 0.0943 sin 3x + 0.0204 sin 5x
```

(Check at x = π/2: exact value π²/4 = 2.467; three terms give 2.546 − 0.0943 + 0.0204 = 2.472 ✓.)

## Check yourself

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

1. bn =
   A. (2/pi) integral ... cos nx
   B. (1/pi) integral from -pi to pi
   C. 2 integral from 0 to 1
   D. (2/pi) integral from 0 to pi (pi x - x^2) sin nx dx
   Answer: D. Half-range sine. (Key point: Formula)

2. bn =
   A. 8/(pi n^3) for odd n, 0 for even n
   B. 8/(pi n^3) for even n
   C. 4/(pi n) for odd n
   D. 8/(n^3) for odd n
   Answer: A. 4(1-(-1)^n)/(pi n^3). (Key point: bn)

3. The first three non-zero terms are
   A. (8/pi)[sin x + sin 2x/8 + sin 3x/27]
   B. (8/pi)[sin x + sin 3x/27 + sin 5x/125]
   C. (8/pi)[sin x + sin 3x/9 + sin 5x/25]
   D. (4/pi)[sin x + sin 3x/27 + sin 5x/125]
   Answer: B. Odd n only, with 1/n^3. (Key point: First three terms)

## Related in Mathematics III

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