The value of the integral `int_alpha^beta (1)/(sqrt((x-alpha)(beta-x)))dx " for "beta gt alpha`, is

Options

A. `sin^(-1)(alpha//beta)`
B. `pi//2`
C. `sin^(-1)(beta//2alpha)`
D. `pi`

Answer

To solve the integral

\[ I = \int_{\alpha}^{\beta} \frac{1}{\sqrt{(x - \alpha)(\beta - x)}} \, dx \]

for \( \beta > \alpha \), we will follow these steps:

### Step 1: Substitution
Let us denote

\[ f(x) = (x - \alpha)(\beta - x) \]

This gives us

\[ f(x) = -x^2 + (\alpha + \beta)x - \alpha\beta \]

### Step 2: Completing the Square
We can rewrite \( f(x) \) by completing the square:

\[ f(x) = -\left(x^2 - (\alpha + \beta)x + \alpha\beta\right) \]

To complete the square, we add and subtract \(\left(\frac{\alpha + \beta}{2}\right)^2\):

\[ f(x) = -\left(x - \frac{\alpha + \beta}{2}\right)^2 + \left(\frac{\alpha + \beta}{2}\right)^2 - \alpha\beta \]

### Step 3: Simplifying the Expression
Now, we simplify the expression:

\[ f(x) = -\left(x - \frac{\alpha + \beta}{2}\right)^2 + \frac{(\alpha + \beta)^2}{4} - \alpha\beta \]

We can express \(\frac{(\alpha + \beta)^2}{4} - \alpha\beta\) as:

\[ \frac{\alpha^2 + 2\alpha\beta + \beta^2 - 4\alpha\beta}{4} = \frac{(\alpha - \beta)^2}{4} \]

Thus, we have:

\[ f(x) = \frac{(\alpha - \beta)^2}{4} - \left(x - \frac{\alpha + \beta}{2}\right)^2 \]

### Step 4: Substituting Back into the Integral
Now, substituting back into the integral:

\[ I = \int_{\alpha}^{\beta} \frac{1}{\sqrt{\frac{(\alpha - \beta)^2}{4} - \left(x - \frac{\alpha + \beta}{2}\right)^2}} \, dx \]

### Step 5: Change of Variable
Let

\[ u = x - \frac{\alpha + \beta}{2} \implies dx = du \]

When \( x = \alpha \), \( u = \alpha - \frac{\alpha + \beta}{2} = \frac{\alpha - \beta}{2} \) and when \( x = \beta \), \( u = \beta - \frac{\alpha + \beta}{2} = \frac{\beta - \alpha}{2} \).

Thus, the limits change from \( \frac{\alpha - \beta}{2} \) to \( \frac{\beta - \alpha}{2} \):

\[ I = \int_{\frac{\alpha - \beta}{2}}^{\frac{\beta - \alpha}{2}} \frac{1}{\sqrt{\frac{(\alpha - \beta)^2}{4} - u^2}} \, du \]

### Step 6: Recognizing the Integral Form
This integral is of the form:

\[ \int \frac{1}{\sqrt{a^2 - u^2}} \, du \]

where \( a = \frac{\alpha - \beta}{2} \).

### Step 7: Evaluating the Integral
The integral evaluates to:

\[ \sin^{-1}\left(\frac{u}{a}\right) \bigg|_{\frac{\alpha - \beta}{2}}^{\frac{\beta - \alpha}{2}} = \sin^{-1}(1) - \sin^{-1}(-1) \]

### Step 8: Final Calculation
Calculating the values:

\[ \sin^{-1}(1) = \frac{\pi}{2}, \quad \sin^{-1}(-1) = -\frac{\pi}{2} \]

Thus,

\[ I = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

### Conclusion
The value of the integral is

\[ \boxed{\pi} \]