---
title: "Show ∫ x sin x cos x/(cos⁴x+sin⁴x) dx = π²/16"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:integral-x-sinx-cosx-over-cos4-sin4"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/integral-x-sinx-cosx-over-cos4-sin4
site: BCT Engineering
---

# Show ∫ x sin x cos x/(cos⁴x+sin⁴x) dx = π²/16

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2067 Asoj
- 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-sinx-cosx-over-cos4-sin4
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Show that ∫₀^{π/2} x sin x cos x/(cos⁴x + sin⁴x) dx = π²/16.

## Short answer

Use x→π/2−x to get I = ∫₀^{π/2} (π/2−x) sin x cos x/(sin⁴x+cos⁴x) dx. Adding yields 2I = (π/2) ∫₀^{π/2} sin x cos x/(sin⁴x+cos⁴x) dx. With t=sin²x, the remaining integral equals π/4, so 2I = (π/2)(π/4) = π²/8, giving I = π²/16.

## Answer

Let I = ∫₀^(π/2) x sin x cos x/(cos⁴x + sin⁴x) dx. Replace x by π/2 − x (the denominator and sin x cos x are unchanged):

I = ∫₀^(π/2) (π/2 − x) sin x cos x/(cos⁴x + sin⁴x) dx

Add: 2I = (π/2) ∫₀^(π/2) sin x cos x/(sin⁴x + cos⁴x) dx.

**Evaluate K = ∫₀^(π/2) sin x cos x/(sin⁴x + cos⁴x) dx.** Put t = sin²x, dt = 2 sin x cos x dx. Then cos²x = 1 − t, sin⁴x + cos⁴x = t² + (1 − t)² = 2t² − 2t + 1. Limits 0 to 1.

```text
K = (1/2) ∫₀¹ dt/(2t² − 2t + 1)
  = (1/4) ∫₀¹ dt/((t − 1/2)² + 1/4)
  = (1/4) · 2 [tan⁻¹(2t − 1)]₀¹
  = (1/2)(π/4 + π/4)
  = π/4
```
So 2I = (π/2)(π/4) = π²/8.

**I = π²/16.** Shown.

## Check yourself

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

1. Replacing x by π/2 - x keeps sin x cos x and the denominator, giving 2I =
   A. π K
   B. (π/4) K
   C. K
   D. (π/2) K, K = integral sin x cos x/(sin^4 x + cos^4 x)
   Answer: D. x + (π/2 - x) = π/2. (Key point: Symmetry)

2. Put t = sin^2 x: the denominator becomes
   A. 2t^2 - 2t + 1
   B. t^2 + 1
   C. t^2 - t + 1
   D. 2t^2 + 1
   Answer: A. t^2 + (1-t)^2. (Key point: Substitution)

3. K equals
   A. π/2
   B. π/4
   C. π/8
   D. 1/2
   Answer: B. (1/4)·2·(π/4+π/4)... = π/4. (Key point: K value)

4. I equals
   A. π^2/8
   B. π^2/4
   C. π^2/16
   D. π/16
   Answer: C. 2I = (π/2)(π/4). (Key point: Result)

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