`int_(a)^(b) sqrt((x-a)/(b-x))dx` is equal to

Answer

To solve the integral \( I = \int_{a}^{b} \sqrt{\frac{x-a}{b-x}} \, dx \), we can use a substitution method and properties of definite integrals. Here’s a step-by-step solution:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{a}^{b} \sqrt{\frac{x-a}{b-x}} \, dx \]

### Step 2: Use substitution
We can use the substitution \( x = a + (b-a)t \), where \( t \) varies from \( 0 \) to \( 1 \) as \( x \) varies from \( a \) to \( b \). The differential \( dx \) becomes:
\[ dx = (b-a) \, dt \]

### Step 3: Change the limits of integration
When \( x = a \), \( t = 0 \) and when \( x = b \), \( t = 1 \). Thus, the integral becomes:
\[ I = \int_{0}^{1} \sqrt{\frac{(a + (b-a)t) - a}{b - (a + (b-a)t)}} (b-a) \, dt \]

### Step 4: Simplify the integrand
Now, simplifying the integrand:
\[ I = (b-a) \int_{0}^{1} \sqrt{\frac{(b-a)t}{b - a - (b-a)t}} \, dt \]
\[ = (b-a) \int_{0}^{1} \sqrt{\frac{(b-a)t}{(b-a)(1-t)}} \, dt \]
\[ = (b-a) \int_{0}^{1} \sqrt{\frac{t}{1-t}} \, dt \]

### Step 5: Evaluate the integral
The integral \( \int_{0}^{1} \sqrt{\frac{t}{1-t}} \, dt \) can be evaluated using the Beta function or recognizing it as a standard integral:
\[ \int_{0}^{1} \sqrt{\frac{t}{1-t}} \, dt = \frac{\pi}{2} \]

### Step 6: Final result
Thus, substituting back, we have:
\[ I = (b-a) \cdot \frac{\pi}{2} \]

### Conclusion
The final answer is:
\[ \int_{a}^{b} \sqrt{\frac{x-a}{b-x}} \, dx = \frac{\pi}{2} (b-a) \]