Evaluate the integral `int_0^4 dx / sqrt(4x – x^2)` using beta and gamma functions.

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Evaluation of the Integral Using Beta and Gamma Functions Introduction We are given the integral \(I=\int_0^4 \frac{d x}{\sqrt{4 x-x^2}}\), and we aim to evaluate it using the beta and gamma functions. Solution First, we rewrite the integral in a more convenient form: \[ I = \int_0^4 \frac{d x}{\sqrRead more

## Evaluation of the Integral Using Beta and Gamma Functions

## Introduction

We are given the integral

\(I=\int_0^4 \frac{d x}{\sqrt{4 x-x^2}}\),

and we aim to evaluate it using the beta and gamma functions.

## Solution

First, we rewrite the integral in a more convenient form:

\[

I = \int_0^4 \frac{d x}{\sqrt{x(4-x)}} = \int_0^4 x^{-1/2}(4-x)^{-1/2} d x = \int_0^4 x^{\frac{1}{2}-1}(4-x)^{\frac{1}{2}-1} d x

\]

Next, we apply the substitution \(x=4t\), which implies \(dx = 4dt\). The limits of integration change accordingly: when \(x=0\), \(t=0\), and when \(x=4\), \(t=1\). Therefore, the integral becomes:

\[

\begin{aligned}

I &= \int_0^1 (4t)^{\frac{1}{2}-1}(4-4t)^{\frac{1}{2}-1} \cdot 4 dt \\

&= \int_0^1 4^{\frac{1}{2}-1} t^{\frac{1}{2}-1} \cdot 4^{\frac{1}{2}-1}(1-t)^{\frac{1}{2}-1} \cdot 4 dt \\

&= 4^{\frac{1}{2}+\frac{1}{2}-1} \int_0^1 t^{\frac{1}{2}-1}(1-t)^{\frac{1}{2}-1} dt \\

&= 4^0 B\left(\frac{1}{2}, \frac{1}{2}\right)

\end{aligned}

\]

Using the definition of the beta function \(B(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx\), we have:

\[

I = B\left(\frac{1}{2}, \frac{1}{2}\right)

\]

Now, using the relationship between the beta and gamma functions, \(B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\), we can express the integral in terms of gamma functions:

\[

I = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2} + \frac{1}{2}\right)} = \frac{\sqrt{\pi} \cdot \sqrt{\pi}}{1} = \pi

\]

## Conclusion

Therefore, the value of the integral \(I=\int_0^4 \frac{d x}{\sqrt{4 x-x^2}}\) is \(\pi\).

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