If `int_(0)^(a)sqrt((a-x)/(x))dx=(K)/(2),` then K=. . .

Options

A. `(pia)/(2)`
B. `(5pia)/(2)`
C. `(3pia)/(2)`
D. `pia`

Explanation

To solve the integral equation given in the problem, we start with the expression:

\[ \int_{0}^{a} \sqrt{\frac{a-x}{x}} \, dx = \frac{K}{2} \]

### Step 1: Substitution
We will use the substitution \( x = a \sin^2 \theta \). This implies that:

\[ dx = 2a \sin \theta \cos \theta \, d\theta \]

### Step 2: Change of Limits
When \( x = 0 \), \( \sin^2 \theta = 0 \) which gives \( \theta = 0 \).

When \( x = a \), \( \sin^2 \theta = 1 \) which gives \( \theta = \frac{\pi}{2} \).

### Step 3: Substitute in the Integral
Now substituting \( x \) and \( dx \) into the integral:

\[ \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{a - a \sin^2 \theta}{a \sin^2 \theta}} \cdot 2a \sin \theta \cos \theta \, d\theta \]

This simplifies to:

\[ \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{a(1 - \sin^2 \theta)}{a \sin^2 \theta}} \cdot 2a \sin \theta \cos \theta \, d\theta \]

### Step 4: Simplify the Square Root
Since \( 1 - \sin^2 \theta = \cos^2 \theta \), we can rewrite the integral as:

\[ \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{a \cos^2 \theta}{a \sin^2 \theta}} \cdot 2a \sin \theta \cos \theta \, d\theta \]

This simplifies to:

\[ \int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta} \cdot 2a \sin \theta \cos \theta \, d\theta = 2a \int_{0}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta \]

### Step 5: Evaluate the Integral
Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \):

\[ 2a \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} \, d\theta = a \int_{0}^{\frac{\pi}{2}} (1 + \cos 2\theta) \, d\theta \]

This evaluates to:

\[ a \left[ \theta + \frac{\sin 2\theta}{2} \right]_{0}^{\frac{\pi}{2}} = a \left[ \frac{\pi}{2} + 0 \right] = \frac{a\pi}{2} \]

### Step 6: Set Equal to Given Expression
Now we have:

\[ \frac{a\pi}{2} = \frac{K}{2} \]

### Step 7: Solve for K
Multiplying both sides by 2 gives:

\[ K = a\pi \]

### Conclusion
Thus, the value of \( K \) is:

\[ \boxed{a\pi} \]