---
title: "Fourier series: x sin x"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:fourier-series-x-sin-x"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-x-sin-x
site: BCT Engineering
---

# Fourier series: x sin x

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2074 Asoj, 5 marks; 2071 Chaitra, 5 marks (asked 2 times)
- Video: [Expand f(x)=xsinx as a Fourier Series in (0,2π) Example 2 exercise 4.1 explained by Yogeshwar Sir](https://www.youtube.com/watch?v=LlsXAC5greQ) — Anant Academy of Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-x-sin-x
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Expand f(x) = x sin x as a Fourier series in −π ≤ x ≤ π.

## Short answer

Since x sin x is even, b_n=0. a_0 = 2, a_1 = -1/2, and for n ≥ 2, a_n = 2(-1)^(n+1)/(n^2-1). The Fourier series is x sin x = 1 - (1/2)cos x + 2 sum_{n=2}^infinity (-1)^(n+1)/(n^2-1) cos(nx).

## Answer

f(x) = x sin x is even on (−π, π) (x odd × sin x odd), so bₙ = 0.

#### Coefficients
```text
a₀ = (2/π) ∫₀^π x sin x dx = (2/π)[−x cos x + sin x]₀^π = (2/π)(π) = 2
```

For n = 1:

```text
a₁ = (2/π) ∫₀^π x sin x cos x dx = (1/π) ∫₀^π x sin 2x dx
   = (1/π)[−x cos 2x/2 + sin 2x/4]₀^π = (1/π)(−π/2) = −1/2
```

For n ≥ 2, use 2 sin x cos nx = sin(n+1)x − sin(n−1)x and ∫₀^π x sin mx dx = π(−1)^(m+1)/m:

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

#### Series
```text
x sin x = 1 − (1/2) cos x + 2 Σ_{n=2}^∞ [ (−1)ⁿ⁺¹ / (n² − 1) ] cos nx
        = 1 − (1/2) cos x − (2/3) cos 2x + (2/8) cos 3x − (2/15) cos 4x + ...
```

Check at x = π: π sin π = 0. The series gives 1 + 1/2 − 2 Σ_{n≥2} 1/(n² − 1) = 3/2 − 2(3/4) = 0 ✓.

## Check yourself

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

1. x sin x is even because
   A. x is even
   B. x and sin x are both odd
   C. sin x is even
   D. it is periodic
   Answer: B. Odd*odd = even. (Key point: Symmetry)

2. a0 =
   A. 1
   B. pi
   C. 2
   D. 0
   Answer: C. (2/pi)(pi). (Key point: a0)

3. a1 =
   A. 1/2
   B. -1
   C. 0
   D. -1/2
   Answer: D. (1/pi) integral x sin 2x = -1/2. (Key point: a1)

4. For n >= 2, an =
   A. 2(-1)^(n+1)/(n^2-1)
   B. 2(-1)^n/(n^2-1)
   C. 2(-1)^(n+1)/(n^2+1)
   D. (-1)^(n+1)/n
   Answer: A. Product-to-sum then integrate. (Key point: an)

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