Prove that ∫
2
0
x³(8−x³)^1/3
1

in form of beta functions

To prove that the integral \( \int_{0}^{2} x^3 (8 - x^3)^{\frac{1}{3}} \, dx \) can be expressed in terms of Beta functions, we'll first manipulate the integral into a standard form that can be recognized as a Beta function.

### Step 1: Change of Variables
Let's start with the substitution:
\[
u = x^3
\]
Then, the differential \( du \) is:
\[
du = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{du}{3x^2}
\]
Since \( u = x^3 \), we have \( x = u^{1/3} \). Therefore, \( x^2 = u^{2/3} \), and the differential becomes:
\[
dx = \frac{du}{3u^{2/3}}
\]

### Step 2: Rewriting the Integral
Substitute into the integral:
\[
\int_{0}^{2} x^3 (8 - x^3)^{\frac{1}{3}} \, dx = \int_{0}^{8} u (8 - u)^{\frac{1}{3}} \cdot \frac{du}{3u^{2/3}}
\]
Simplifying, we get:
\[
\int_{0}^{8} \frac{u^{1 - \frac{2}{3}} (8 - u)^{\frac{1}{3}}}{3} \, du = \frac{1}{3} \int_{0}^{8} u^{\frac{1}{3}} (8 - u)^{\frac{1}{3}} \, du
\]

### Step 3: Further Substitution
Next, we introduce another substitution to bring the integral into the standard Beta function form. Let:
\[
u = 8v
\]
Then, \( du = 8 dv \), and the integral limits change from \( u = 0 \) to \( u = 8 \) to \( v = 0 \) to \( v = 1 \). The integral becomes:
\[
\frac{1}{3} \int_{0}^{1} (8v)^{\frac{1}{3}} (8(1 - v))^{\frac{1}{3}} \cdot 8 \, dv
\]
Simplify this:
\[
\frac{8}{3} \int_{0}^{1} 8^{\frac{2}{3}} v^{\frac{1}{3}} (1 - v)^{\frac{1}{3}} \, dv
\]
Since \( 8^{\frac{2}{3}} = 4 \), this simplifies to:
\[
\frac{32}{3} \int_{0}^{1} v^{\frac{1}{3}} (1 - v)^{\frac{1}{3}} \, dv
\]

### Step 4: Expressing the Integral as a Beta Function
The integral \( \int_{0}^{1} v^{\frac{1}{3}} (1 - v)^{\frac{1}{3}} \, dv \) is the standard form of the Beta function \( B(x, y) \) where:
\[
B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, dt
\]
Here, \( x = \frac{4}{3} \) and \( y = \frac{4}{3} \). Therefore:
\[
\int_{0}^{1} v^{\frac{1}{3}} (1 - v)^{\frac{1}{3}} \, dv = B\left(\frac{4}{3}, \frac{4}{3}\right)
\]
Thus, the original integral can be expressed as:
\[
\int_{0}^{2} x^3 (8 - x^3)^{\frac{1}{3}} \, dx = \frac{32}{3} B\left(\frac{4}{3}, \frac{4}{3}\right)
\]

### Final Answer:
The integral \( \int_{0}^{2} x^3 (8 - x^3)^{\frac{1}{3}} \, dx \) can be written in terms of the Beta function as:
\[
\int_{0}^{2} x^3 (8 - x^3)^{\frac{1}{3}} \, dx = \frac{32}{3} B\left(\frac{4}{3}, \frac{4}{3}\right)
\]

Would you like a deeper explanation of any step? Here are 5 related questions to consider:

1. What are the properties of the Beta function?
2. How can the Beta function be related to the Gamma function?
3. What are other examples where integrals can be solved using Beta functions?
4. How does the substitution method help in solving integrals?
5. Can you explain the relationship between Beta and Gamma functions in more detail?

**Tip:** Understanding the properties of special functions like Beta and Gamma functions can greatly simplify complex integrals.

Learn how to prove the integral ∫ 2 0 x³(8−x³)^1/3 using Beta functions. This problem involves integral calculus and the application of Beta function theory. Suitable for undergraduate level mathematics.

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