---
title: "Beta-Gamma: ∫ cos²6θ sin⁴3θ dθ = 7π/192"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:beta-gamma-cos2-6theta-sin4-3theta"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/beta-gamma-cos2-6theta-sin4-3theta
site: BCT Engineering
---

# Beta-Gamma: ∫ cos²6θ sin⁴3θ dθ = 7π/192

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2072 Chaitra
- 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-cos2-6theta-sin4-3theta
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Apply Beta and Gamma function to evaluate ∫₀^{π/6} cos²6θ sin⁴3θ dθ = 7π/192.

## Short answer

Substitute 3θ = φ to express the integral in terms of sin powers. Apply Beta-Gamma formula to evaluate ∫₀^{π/2} sinⁿφ dφ, yielding 7π/192.

## Answer

Let I = ∫₀^(π/6) cos²6θ sin⁴3θ dθ. Put φ = 3θ, dθ = dφ/3; limits 0 to π/2. Then cos 6θ = cos 2φ = 1 − 2 sin²φ.

```text
I = (1/3) ∫₀^(π/2) (1 − 2 sin²φ)² sin⁴φ dφ
  = (1/3) ∫₀^(π/2) (sin⁴φ − 4 sin⁶φ + 4 sin⁸φ) dφ
```
**Beta-Gamma formula:** ∫₀^(π/2) sinⁿφ dφ = Γ((n+1)/2) Γ(1/2) / (2Γ((n+2)/2)).

- n = 4: Γ(5/2)Γ(1/2)/(2Γ(3)) = (3√π/4)(√π)/(4) = 3π/16
- n = 6: Γ(7/2)Γ(1/2)/(2Γ(4)) = (15√π/8)(√π)/12 = 5π/32
- n = 8: Γ(9/2)Γ(1/2)/(2Γ(5)) = (105√π/16)(√π)/48 = 35π/256

```text
I = (1/3) [ 3π/16 − 4(5π/32) + 4(35π/256) ]
  = (1/3) [ 12π/64 − 40π/64 + 35π/64 ]
  = (1/3)(7π/64)
```
**I = 7π/192.** Shown.

## Check yourself

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

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

2. cos 6θ equals
   A. 2 sin φ cos φ
   B. 2 cos^2 φ + 1
   C. sin^2 φ - 1
   D. 1 - 2 sin^2 φ
   Answer: D. cos 2φ = 1 - 2 sin^2 φ. (Key point: Identity)

3. (1 - 2 sin^2 φ)^2 sin^4 φ expands to
   A. sin^4 - 4 sin^6 + 4 sin^8
   B. sin^4 - 2 sin^6 + sin^8
   C. sin^4 + 4 sin^6 + 4 sin^8
   D. sin^4 - 4 sin^6 - 4 sin^8
   Answer: A. (1-2s)^2 = 1 - 4s + 4s^2. (Key point: Expansion)

4. The integral equals
   A. 7π/64
   B. 7π/192
   C. 5π/192
   D. 7π/96
   Answer: B. (1/3)(7π/64). (Key point: Result)

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