The value of `int_(0)^(a) sqrt((a-x)/x )` dx is

Options

A. `(a)/(2)`
B. `(a)/(4)`
C. `(pia)/(2)`
D. `(pia)/(4)`

Explanation

To solve the integral \( I = \int_{0}^{a} \sqrt{\frac{a-x}{x}} \, dx \), we will use a substitution method. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( x = a \sin^2 \theta \). Then, we need to find \( dx \):
\[ dx = 2a \sin \theta \cos \theta \, d\theta = a \sin(2\theta) \, d\theta \]

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ 0 = a \sin^2 \theta \implies \sin \theta = 0 \implies \theta = 0 \]
When \( x = a \):
\[ a = a \sin^2 \theta \implies \sin^2 \theta = 1 \implies \theta = \frac{\pi}{2} \]

### Step 3: Rewrite the integral
Now we can rewrite the integral in terms of \( \theta \):
\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{a - a \sin^2 \theta}{a \sin^2 \theta}} \cdot a \sin(2\theta) \, d\theta \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{a(1 - \sin^2 \theta)}{a \sin^2 \theta}} \cdot a \sin(2\theta) \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{a \cos^2 \theta}{a \sin^2 \theta}} \cdot a \sin(2\theta) \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta} \cdot a \sin(2\theta) \, d\theta \]

### Step 4: Simplifying the integral
Using \( \sin(2\theta) = 2 \sin \theta \cos \theta \):
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta} \cdot a \cdot 2 \sin \theta \cos \theta \, d\theta \]
\[ = 2a \int_{0}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta \]

### Step 5: Evaluate the integral
Using the identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \):
\[ I = 2a \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta \]
\[ = a \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{0}^{\frac{\pi}{2}} \]
Evaluating the limits:
\[ = a \left[ \frac{\pi}{2} + 0 - (0 + 0) \right] = a \cdot \frac{\pi}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{a \pi}{2} \]

---