---
title: "Beta-Gamma: ∫ sin⁴x cos²x dx = (3π−4)/192"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:beta-gamma-sin4-cos2-pi4"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/beta-gamma-sin4-cos2-pi4
site: BCT Engineering
---

# Beta-Gamma: ∫ sin⁴x cos²x dx = (3π−4)/192

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2073 Chaitra; 2071 Shrawan, 5 marks (asked 2 times)
- Video: [INTEGRAL CALCULUS BETA GAMMA FUNCTION LECTURE 9](https://www.youtube.com/watch?v=Oh0f_EkYA5g) — TIKLE'S ACADEMY OF MATHS
- 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-sin4-cos2-pi4
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Using Beta-Gamma function, show that ∫₀^{π/4} sin⁴x cos²x dx = (3π−4)/192.

## Short answer

Substitute u=2x to change limits to 0 to π/2, rewrite integrand as (1/16)(1−cos u) sin²u. Use Beta-Gamma: ∫₀^{π/2} sin²u du = π/4, ∫₀^{π/2} sin²u cos u du = 1/3. Then I = (1/16)(π/4 − 1/3) = (3π−4)/192.

## Answer

The Beta-Gamma formula applies directly to limits 0 to π/2, so first change the variable so that the limits become 0 and π/2.

**Rewrite the integrand.** sin⁴x cos²x = sin²x (sin x cos x)² = sin²x · (sin²2x)/4, and sin²x = (1 − cos 2x)/2:

sin⁴x cos²x = (1/8)(1 − cos 2x) sin²2x.

Put u = 2x, dx = du/2. Limits: x = 0 → u = 0; x = π/4 → u = π/2.

```text
I = ∫₀^(π/4) sin⁴x cos²x dx = (1/16) ∫₀^(π/2) (1 − cos u) sin²u du
  = (1/16) [ ∫₀^(π/2) sin²u du − ∫₀^(π/2) sin²u cos u du ]
```
**Use ∫₀^(π/2) sin^p u cos^q u du = Γ((p+1)/2) Γ((q+1)/2) / (2 Γ((p+q+2)/2)).**

- ∫ sin²u du (p = 2, q = 0): Γ(3/2)Γ(1/2)/(2Γ(2)) = (√π/2)(√π)/2 = π/4.
- ∫ sin²u cos u du (p = 2, q = 1): Γ(3/2)Γ(1)/(2Γ(5/2)) = (√π/2)/(2 · 3√π/4) = 1/3.

I = (1/16)(π/4 − 1/3) = (1/16)(3π − 4)/12

**I = (3π − 4)/192.** Proved.

## Check yourself

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

1. The Beta-Gamma formula cannot be applied directly here because
   A. the powers are even
   B. the integrand is not positive
   C. the limits are 0 to π/4, not 0 to π/2
   D. sin and cos have different powers
   Answer: C. The formula requires limits 0 to π/2. (Key point: Limits issue)

2. sin^4 x cos^2 x is rewritten as
   A. (1/4) sin^2 2x cos 2x
   B. (1/8)(1 + cos 2x) sin^2 2x
   C. (1/2) sin^2 2x
   D. (1/8)(1 - cos 2x) sin^2 2x
   Answer: D. sin^2 x · sin^2(2x)/4 with sin^2 x = (1-cos 2x)/2. (Key point: Rewriting)

3. The two Beta-Gamma values used are
   A. integral sin^2 u = π/4 and integral sin^2 u cos u = 1/3
   B. π/4 and 1/2
   C. π/2 and 1/3
   D. π/4 and 1/4
   Answer: A. The second is Γ(3/2)Γ(1)/(2Γ(5/2)) = 1/3. (Key point: Parts)

4. The integral equals
   A. (3π-4)/96
   B. (3π-4)/192
   C. (3π+4)/192
   D. (π-4)/192
   Answer: B. (1/16)(π/4 - 1/3) = (3π-4)/192. (Key point: Result)

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