---
title: "Show ∫ x log(sin x) dx"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-x-log-sin-x"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-x-log-sin-x
site: BCT Engineering
---

# Show ∫ x log(sin x) dx

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2073 Shrawan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-x-log-sin-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Show that ∫₀^π x log(sin x) dx = (π²/2) log(1/2).

## Short answer

Use symmetry to get 2I = π∫_0^π log(sin x)dx = 2π∫_0^{π/2}log(sin x)dx. Since ∫_0^{π/2}log(sin x)dx = -(π/2)log 2, we obtain I = (π²/2)log(1/2).

## Answer

Let I = ∫₀^π x log(sin x) dx. Using ∫₀^a f(x) dx = ∫₀^a f(a − x) dx:

I = ∫₀^π (π − x) log(sin x) dx   [since sin(π − x) = sin x]

Add the two forms:

2I = π ∫₀^π log(sin x) dx = 2π ∫₀^(π/2) log(sin x) dx   [sin(π − x) = sin x, symmetric about π/2]

So I = π J, where J = ∫₀^(π/2) log sin x dx.

**Find J.** J = ∫₀^(π/2) log cos x dx as well (replace x by π/2 − x). Add:

```text
2J = ∫₀^(π/2) log(sin x cos x) dx
   = ∫₀^(π/2) [log sin 2x − log 2] dx
   = ∫₀^(π/2) log sin 2x dx − (π/2) log 2
```
Put 2x = u: ∫₀^(π/2) log sin 2x dx = (1/2)∫₀^π log sin u du = ∫₀^(π/2) log sin u du = J. So 2J = J − (π/2) log 2, hence

J = −(π/2) log 2.

I = πJ = −(π²/2) log 2

**I = (π²/2) log(1/2).** Shown.

## Check yourself

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

1. Using f(x) → f(π-x), the integral also equals
   A. integral of x log cos x dx
   B. integral of (π - x) log sin x dx
   C. integral of π log sin x dx
   D. -integral of x log sin x dx
   Answer: B. sin(π - x) = sin x. (Key point: Trick)

2. Adding the two forms gives I =
   A. 2π J
   B. π J/2
   C. π J, J = integral log sin x from 0 to π/2
   D. J
   Answer: C. 2I = π integral over 0..π = 2π J. (Key point: Reduction)

3. J equals
   A. (π/2) log 2
   B. -π log 2
   C. 0
   D. -(π/2) log 2
   Answer: D. From 2J = J - (π/2) log 2. (Key point: J value)

4. I equals
   A. -(π^2/2) log 2 = (π^2/2) log(1/2)
   B. (π^2/2) log 2
   C. -(π/2) log 2
   D. π^2 log 2
   Answer: A. I = πJ. (Key point: Result)

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