---
title: "Beta-Gamma: ∫ cos⁴3θ sin²6θ dθ = 5π/192"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:beta-gamma-cos4-3theta-sin2-6theta"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/beta-gamma-cos4-3theta-sin2-6theta
site: BCT Engineering
---

# Beta-Gamma: ∫ cos⁴3θ sin²6θ dθ = 5π/192

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2070 Asar; 2067 Asar (asked 2 times)
- Video: [Beta , Gamma Function Example 12  I Engineering Mathematics](https://www.youtube.com/watch?v=6IPZHLVaXGE) — Rajendra Mahajan
- 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-cos4-3theta-sin2-6theta
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Use Gamma function to prove that ∫₀^{π/6} cos⁴3θ sin²6θ dθ = 5π/192.

## Short answer

Let φ=3θ, then sin²6θ = 4 sin²φ cos²φ, giving I = (4/3) ∫₀^{π/2} sin²φ cos⁶φ dφ. Using Beta-Gamma, this integral equals 5π/256, so I = 5π/192.

## Answer

Let I = ∫₀^(π/6) cos⁴3θ sin²6θ dθ. Put φ = 3θ, dθ = dφ/3; limits: θ = 0 → φ = 0, θ = π/6 → φ = π/2.

sin 6θ = sin 2φ = 2 sin φ cos φ, so sin²6θ = 4 sin²φ cos²φ.

```text
I = (1/3) ∫₀^(π/2) cos⁴φ · 4 sin²φ cos²φ dφ
  = (4/3) ∫₀^(π/2) sin²φ cos⁶φ dφ
```
Use ∫₀^(π/2) sin^p φ cos^q φ dφ = Γ((p+1)/2)Γ((q+1)/2) / (2Γ((p+q+2)/2)) with p = 2, q = 6:

∫₀^(π/2) sin²φ cos⁶φ dφ = Γ(3/2)Γ(7/2) / (2Γ(5)).

Values: Γ(3/2) = √π/2; Γ(7/2) = (5/2)(3/2)(1/2)√π = 15√π/8; Γ(5) = 24.

= (√π/2)(15√π/8) / 48 = (15π/16)/48 = 5π/256.

I = (4/3)(5π/256) = 20π/768

**I = 5π/192.** Proved.

## Check yourself

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

1. Putting φ = 3θ changes the limits to
   A. 0 to π/6 and dθ = 3dφ
   B. 0 to π/3 and dθ = dφ/3
   C. 0 to π/2 and dθ = dφ/3
   D. 0 to π and dθ = dφ/6
   Answer: C. θ = π/6 corresponds to φ = π/2. (Key point: Substitution)

2. sin^2 6θ equals
   A. 2 sin φ cos φ
   B. sin^2 φ cos^2 φ
   C. 2 sin^2 φ
   D. 4 sin^2 φ cos^2 φ
   Answer: D. 6θ = 2φ, so sin 2φ = 2 sin φ cos φ. (Key point: Double angle)

3. The integral becomes
   A. (4/3) integral sin^2 φ cos^6 φ dφ
   B. (1/3) integral sin^2 φ cos^4 φ dφ
   C. (4/3) integral sin^4 φ cos^2 φ dφ
   D. (1/3) integral sin^2 φ cos^6 φ dφ
   Answer: A. cos^4 φ · 4 sin^2 φ cos^2 φ gives sin^2 cos^6. (Key point: Reduced form)

4. The value is
   A. 5π/256
   B. 5π/192
   C. 5π/96
   D. π/192
   Answer: B. (4/3)(5π/256) = 5π/192. (Key point: Result)

## Close variants set in other exams

- [Gamma function: ∫ sin⁶(x/2) cos⁶(x/2) dx](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Agamma-sin6-cos6-half-x) — Mathematics I, chapter 2, asked 3×

## Related in Mathematics I

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- [Beta-Gamma: ∫ cos²6θ sin⁴3θ dθ = 7π/192](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Abeta-gamma-cos2-6theta-sin4-3theta) — Mathematics I, chapter 2
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