---
title: "Gamma function: ∫ x⁶√(1−x²) dx"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:gamma-x6-sqrt-1-minus-x2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/gamma-x6-sqrt-1-minus-x2
site: BCT Engineering
---

# Gamma function: ∫ x⁶√(1−x²) dx

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2068 Chaitra
- Video: [Beta and Gamma Functions Relationship Explained | Full Derivation with Formula](https://www.youtube.com/watch?v=a8VxN-EL0u8) — Mathematics Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/gamma-x6-sqrt-1-minus-x2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Use Gamma function to evaluate ∫₀¹ x⁶ √(1−x²) dx.

## Short answer

Substitute x² = t to obtain a Beta integral (1/2)B(7/2, 3/2) = (1/2)Γ(7/2)Γ(3/2)/Γ(5), which evaluates to 5π/256.

## Answer

Let I = ∫₀¹ x⁶ √(1 − x²) dx. Put x² = t, so x = t^(1/2), dx = (1/2) t^(−1/2) dt. Limits stay 0 to 1.

```text
I = ∫₀¹ t³ (1 − t)^(1/2) · (1/2) t^(−1/2) dt
  = (1/2) ∫₀¹ t^(5/2) (1 − t)^(1/2) dt
  = (1/2) B(7/2, 3/2)            [m − 1 = 5/2, n − 1 = 1/2]
  = (1/2) Γ(7/2) Γ(3/2) / Γ(5)
```
Values: Γ(7/2) = 15√π/8, Γ(3/2) = √π/2, Γ(5) = 24.

I = (1/2) · (15√π/8)(√π/2) / 24 = (1/2)(15π/16)/24 = 15π/768

**I = 5π/256.**

## Check yourself

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

1. Put
   A. x = sin t
   B. x = t^2
   C. x^2 = t
   D. x^6 = t
   Answer: C. Then dx = (1/2) t^(-1/2) dt. (Key point: Substitution)

2. The integral becomes (1/2) B(m,n) with
   A. m = 3, n = 1/2
   B. m = 7/2, n = 1/2
   C. m = 5/2, n = 1/2
   D. m = 7/2, n = 3/2
   Answer: D. t^(5/2): m-1 = 5/2; (1-t)^(1/2): n = 3/2. (Key point: Beta form)

3. The value is
   A. 5π/256
   B. 15π/256
   C. 5π/128
   D. 5π/768
   Answer: A. (1/2)(15π/16)/24 = 15π/768 = 5π/256. (Key point: Result)

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