---
title: "Gamma function: ∫ dx/(1−x⁶)^(1/6) = π/3"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:gamma-dx-over-1-minus-x6"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/gamma-dx-over-1-minus-x6
site: BCT Engineering
---

# Gamma function: ∫ dx/(1−x⁶)^(1/6) = π/3

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2071 Chaitra; 2070 Chaitra; 2068 Shrawan; 2067 Asoj (asked 4 times)
- Video: [Gamma Function - Properties & Its Application | Integral Calculus](https://www.youtube.com/watch?v=SduCO-r_TOA) — Dr.Gajendra Purohit and 2 more
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/gamma-dx-over-1-minus-x6
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Use Gamma function to prove that ∫₀¹ dx/(1−x⁶)^(1/6) = π/3.

## Short answer

Put x⁶ = t, then dx = (1/6) t^{-5/6} dt. The integral becomes (1/6) B(1/6,5/6). Using Γ(1/6)Γ(5/6) = π/sin(π/6) = 2π, we obtain π/3.

## Answer

To show: ∫₀¹ dx/(1 − x⁶)^(1/6) = π/3.

Put x⁶ = t, so x = t^(1/6) and dx = (1/6) t^(−5/6) dt. Limits are unchanged (0 to 1).

```text
I = (1/6) ∫₀¹ t^(−5/6) (1 − t)^(−1/6) dt
  = (1/6) B(1/6, 5/6)          [m − 1 = −5/6 → m = 1/6; n − 1 = −1/6 → n = 5/6]
  = (1/6) Γ(1/6) Γ(5/6) / Γ(1)
```
Since Γ(n)Γ(1 − n) = π/sin(nπ), with n = 1/6:

Γ(1/6)Γ(5/6) = π / sin(π/6) = π/(1/2) = 2π.

I = (1/6)(2π) = π/3. **Proved.**

## Check yourself

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

1. The substitution used is
   A. x = sin t
   B. x^2 = t
   C. x = t^6
   D. x^6 = t
   Answer: D. It converts the integral into a Beta form with limits still 0 to 1. (Key point: Substitution)

2. The integral becomes (1/6) B(m,n) with
   A. m = 1/6, n = 5/6
   B. m = 5/6, n = 1/6 and a different factor
   C. m = 1/6, n = 1/6
   D. m = 6, n = 1/6
   Answer: A. t^(-5/6) gives m-1 = -5/6, so m = 1/6; (1-t)^(-1/6) gives n = 5/6. (Key point: Beta parameters)

3. Γ(1/6)Γ(5/6) is evaluated using
   A. Γ(n+1)=nΓ(n), giving π
   B. Γ(n)Γ(1-n) = π/sin(nπ), giving 2π
   C. Γ(1/2)=sqrt(π), giving sqrt(π)
   D. Γ(n)Γ(1-n) = π cos(nπ), giving π sqrt(3)
   Answer: B. π/sin(π/6) = 2π. (Key point: Reflection formula)

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