The value of the integral `int_(0)^(a) (1)/(x+sqrt(a^(2)-x^(2)))dx`, is

Options

A. `pi`
B. `(pi)/(2)`
C. `(pi)/(4)`
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{a} \frac{1}{x + \sqrt{a^2 - x^2}} \, dx \), we will use the substitution \( x = a \sin \theta \). Let's go through the steps:

### Step 1: Substitute \( x = a \sin \theta \)
When \( x = 0 \), \( \theta = 0 \) and when \( x = a \), \( \theta = \frac{\pi}{2} \). Thus, the limits of integration change from \( 0 \) to \( \frac{\pi}{2} \).

### Step 2: Calculate \( dx \)
Differentiating \( x = a \sin \theta \) gives:
\[ dx = a \cos \theta \, d\theta \]

### Step 3: Substitute in the integral
Now substituting \( x \) and \( dx \) into the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{a \cos \theta}{a \sin \theta + \sqrt{a^2 - (a \sin \theta)^2}} \, d\theta \]
The term under the square root simplifies as follows:
\[ \sqrt{a^2 - a^2 \sin^2 \theta} = \sqrt{a^2 (1 - \sin^2 \theta)} = \sqrt{a^2 \cos^2 \theta} = a \cos \theta \]
Thus, the integral becomes:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{a \cos \theta}{a \sin \theta + a \cos \theta} \, d\theta \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta + \cos \theta} \, d\theta \]

### Step 4: Set up the second integral
Let’s denote this integral as \( I \):
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta + \cos \theta} \, d\theta \]
Now, we will consider the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta + \cos \theta} \, d\theta \]

### Step 5: Add the two integrals
Adding the two expressions for \( I \):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\cos \theta}{\sin \theta + \cos \theta} + \frac{\sin \theta}{\sin \theta + \cos \theta} \right) d\theta \]
This simplifies to:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin \theta + \cos \theta}{\sin \theta + \cos \theta} \, d\theta = \int_{0}^{\frac{\pi}{2}} 1 \, d\theta \]

### Step 6: Evaluate the integral
The integral evaluates to:
\[ \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2} \]
Thus, we have:
\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4} \]

### Final Answer
The value of the integral is:
\[ \boxed{\frac{\pi}{4}} \]