---
title: "Gamma function: ∫ sin⁶(x/2) cos⁶(x/2) dx"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:gamma-sin6-cos6-half-x"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/gamma-sin6-cos6-half-x
site: BCT Engineering
---

# Gamma function: ∫ sin⁶(x/2) cos⁶(x/2) dx

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2069 Asar
- Video: [Definite Integrals by  Gamma Function. Integral of sin^mx)cos^nx) from 0 to pi/2. Reduction Formula.](https://www.youtube.com/watch?v=nOlFqoniFkA) — E for Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/gamma-sin6-cos6-half-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Use Gamma function to prove ∫₀^π sin⁶(x/2) cos⁶(x/2) dx = 5π/2¹¹.

## Short answer

Put x/2 = t, giving 2∫₀^{π/2} sin⁶t cos⁶t dt. Use the Beta function Γ(7/2)²/Γ(7) to obtain 5π/1024.

## Answer

Let I = ∫₀^π sin⁶(x/2) cos⁶(x/2) dx. Put x/2 = t, dx = 2 dt; limits 0 to π/2.

I = 2 ∫₀^(π/2) sin⁶t cos⁶t dt

Use ∫₀^(π/2) sin^p t cos^q t dt = Γ((p+1)/2) Γ((q+1)/2) / (2Γ((p+q+2)/2)), with p = q = 6:

```text
I = 2 · Γ(7/2) Γ(7/2) / (2 Γ(7))
  = Γ(7/2)² / Γ(7)
```
Γ(7/2) = (5/2)(3/2)(1/2)√π = 15√π/8, so Γ(7/2)² = 225π/64. Γ(7) = 720.

I = (225π/64)/720 = 225π/46080 = 5π/1024

**Result: I = 5π/2¹⁰.**

**Note on the printed value:** the question states 5π/2¹¹, but the correct value from the working above is 5π/2¹⁰ = 5π/1024 ≈ 0.01534 (a numerical check of the integral agrees). The printed 2¹¹ appears to be a misprint, or comes from a slightly different integrand. The method (substitute x/2 = t, then Beta-Gamma) is what the exam expects.

## Check yourself

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

1. x/2 = t gives
   A. integral sin^6 t cos^6 t dt over 0 to π
   B. (1/2) integral ... over 0 to π/2
   C. 2 integral sin^6 t cos^6 t dt over 0 to π/2
   D. 2 integral over 0 to π
   Answer: C. dx = 2dt, limits 0 to π/2. (Key point: Substitution)

2. With p = q = 6 the integral simplifies to
   A. Γ(7/2)^2/Γ(6)
   B. Γ(3)^2/Γ(7)
   C. Γ(7/2)/Γ(7)
   D. Γ(7/2)^2/Γ(7)
   Answer: D. 2 · Γ(7/2)Γ(7/2)/(2Γ(7)). (Key point: Gamma form)

3. Γ(7/2) equals
   A. 15 sqrt(π)/8
   B. 15 sqrt(π)/4
   C. 5 sqrt(π)/8
   D. 105 sqrt(π)/16
   Answer: A. (5/2)(3/2)(1/2) sqrt π. (Key point: Value)

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