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

# Half range sine series: πx − x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2070 Chaitra, 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-four
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Obtain the half range sine series for f(x) = πx − x² in (0, π).

## Short answer

Use bₙ = (2/π)∫₀^π (πx-x²) sin nx dx. Integration gives bₙ=0 for even n, bₙ=8/(πn³) for odd n. Hence πx - x² = (8/π)(sin x + sin 3x/3³ + sin 5x/5³ + ...), 0<x<π.

## 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** for f(x) = πx − x². Integrate by parts with u = πx − x²:

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

The first bracket is 0 (the factor πx − x² vanishes at 0 and π). For the remaining integral:

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

So

```text
bₙ = (2/π)(1/n)·2(1 − (−1)ⁿ)/n² = 4(1 − (−1)ⁿ)/(π n³)
```

- n even: bₙ = 0
- n odd: bₙ = 8/(π n³)

**Series**

πx − x² = (8/π) Σ (n odd) sin nx / n³

= (8/π)[ sin x + sin 3x/3³ + sin 5x/5³ + … ],  0 < x < π

Check: at x = π/2 the left side is π²/4; the right side is (8/π)(1 − 1/27 + 1/125 − …) = (8/π)(π³/32) = π²/4.

## Check yourself

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

1. On (0, π) the sine series coefficient is:
   A. bn = (2/π)∫0^π f(x) sin nx dx
   B. bn = (1/π)∫0^π f cos nx dx
   C. bn = (2/π)∫ f cos nx dx
   D. bn = ∫ f sin nx dx
   Answer: A. Half-range sine formula. (Key point: Formula)

2. The boundary term [−(πx − x^2)cos nx/n] is zero because:
   A. cos nπ = 0
   B. πx − x^2 vanishes at 0 and π
   C. sin nπ = 1
   D. n is odd
   Answer: B. The factor is zero at both ends. (Key point: Boundary term)

3. bn equals:
   A. 8/(πn) for odd n
   B. 0 for odd n
   C. 8/(πn^3) for odd n and 0 for even n
   D. 4/(πn^3) for all n
   Answer: C. 4(1 − (−1)^n)/(πn^3). (Key point: Odd/even n)

4. At x = π/2, the series gives π^2/4 using:
   A. Σ1/n^2 = π^2/6
   B. 1 − 1/3 + 1/5 = π/4
   C. Σ1/n^4 = π^4/90
   D. 1 − 1/27 + 1/125 − ... = π^3/32
   Answer: D. (8/π)(π^3/32) = π^2/4. (Key point: Check)

## Close variants set in other exams

- [Half range sine series: eˣ](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-e) — Mathematics III, chapter 4, asked 12×
- [Half range sine series: π − x](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x) — Mathematics III, chapter 4, asked 13×
- [Half range sine series: x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x-x2) — Mathematics III, chapter 4, asked 14×
- [Half range sine series: x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x2-three) — Mathematics III, chapter 4, asked 9×

## Related in Mathematics III

- [Half range sine series: x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x2-two) — Mathematics III, chapter 4, asked 9×
- [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×
- [Half range sine series: eˣ](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-e-two) — Mathematics III, chapter 4, asked 6×
- [Half range sine series: x](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x-three) — Mathematics III, chapter 4, asked 6×
- [Half range sine series: e^(2x)](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-e-2x) — Mathematics III, chapter 4, asked 6×
- [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×

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