---
title: "Beta-Gamma: ∫ x⁵√(2ax−x²) dx"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:beta-gamma-x5-sqrt-2ax-x2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/beta-gamma-x5-sqrt-2ax-x2
site: BCT Engineering
---

# Beta-Gamma: ∫ x⁵√(2ax−x²) dx

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2074 Asoj; 2074 Chaitra, 5 marks; 2073 Shrawan; 2072 Kartik; 2069 Chaitra (asked 5 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/beta-gamma-x5-sqrt-2ax-x2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Use Beta and Gamma functions to evaluate ∫₀^{2a} x⁵ √(2ax − x²) dx.

## Short answer

Substitute x = 2a sin²θ to obtain I = 256a⁷ ∫₀^{π/2} sin¹²θ cos²θ dθ. Using Beta function, this equals (1/2)B(13/2,3/2). Evaluating Gamma functions yields 33πa⁷/16.

## Answer

#### Formulas
Beta function B(m, n) = 2∫₀^(π/2) sin^(2m−1)θ cos^(2n−1)θ dθ = Γ(m)Γ(n)/Γ(m+n). Γ(n+1) = nΓ(n), Γ(1/2) = √π.

#### Substitution
I = ∫₀^(2a) x⁵ √(2ax − x²) dx. Put x = 2a sin²θ, so dx = 4a sin θ cos θ dθ. Limits: x = 0 → θ = 0; x = 2a → θ = π/2.

2ax − x² = x(2a − x) = 2a sin²θ · 2a cos²θ = 4a² sin²θ cos²θ, so the square root is 2a sin θ cos θ.

```text
I = ∫ (2a)⁵ sin¹⁰θ · 2a sinθ cosθ · 4a sinθ cosθ dθ
  = 256 a⁷ ∫₀^(π/2) sin¹²θ cos²θ dθ
```
#### Using Beta function
2m − 1 = 12 gives m = 13/2; 2n − 1 = 2 gives n = 3/2.

∫₀^(π/2) sin¹²θ cos²θ dθ = (1/2) B(13/2, 3/2) = Γ(13/2)Γ(3/2) / (2Γ(8))

Values:
- Γ(13/2) = (11/2)(9/2)(7/2)(5/2)(3/2)(1/2)√π = (10395/64)√π
- Γ(3/2) = √π/2
- Γ(8) = 7! = 5040

```text
I = 256 a⁷ · Γ(13/2)Γ(3/2) / (2Γ(8))
  = 128 a⁷ · (10395/64)√π · (√π/2) / 5040
  = 128 a⁷ · 10395 π / (128 · 5040)
  = (10395/5040) π a⁷
  = 33π a⁷/16
```
**Answer: ∫₀^(2a) x⁵√(2ax − x²) dx = 33πa⁷/16.**

## Check yourself

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

1. Which substitution makes the integrand a trigonometric Beta form?
   A. x = 2a sin θ
   B. x = a tan θ
   C. x = 2a cos θ
   D. x = 2a sin^2 θ
   Answer: D. Then 2ax - x^2 = 4a^2 sin^2 θ cos^2 θ, a perfect square. (Key point: Substitution)

2. The integral reduces to
   A. 256 a^7 integral of sin^12 θ cos^2 θ dθ over 0 to π/2
   B. 64 a^7 integral of sin^10 θ cos^2 θ dθ
   C. 256 a^7 integral of sin^11 θ cos θ dθ
   D. 32 a^6 integral of sin^12 θ cos^2 θ dθ
   Answer: A. (2a)^5 sin^10 · 2a sin θ cos θ · 4a sin θ cos θ gives 256 a^7 sin^12 cos^2. (Key point: Reduced integral)

3. Matching sin^12 θ cos^2 θ with 2m-1=12, 2n-1=2 gives
   A. m = 6, n = 1
   B. m = 13/2, n = 3/2
   C. m = 12, n = 2
   D. m = 7, n = 2
   Answer: B. 2m-1 = 12 gives m = 13/2; 2n-1 = 2 gives n = 3/2. (Key point: Beta parameters)

4. The value of the integral is
   A. 33 π a^7/8
   B. 11 π a^7/16
   C. 33 π a^7/16
   D. 33 π a^6/16
   Answer: C. 256 a^7 · Γ(13/2)Γ(3/2)/(2·5040) = 33πa^7/16. (Key point: Result)

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