The value of `int_(-pi//2)^(pi//2) sin{log(x+sqrt(x^(2)+1)}dx` is

Options

A. 1
B. -1
C. 0
D. none of these

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin\left(\log\left(x + \sqrt{x^2 + 1}\right)\right) \, dx, \]

we will first analyze the function inside the integral.

### Step 1: Define the function
Let

\[ f(x) = \sin\left(\log\left(x + \sqrt{x^2 + 1}\right)\right). \]

### Step 2: Check if the function is odd
To determine if \( f(x) \) is odd, we need to compute \( f(-x) \):

\[ f(-x) = \sin\left(\log\left(-x + \sqrt{(-x)^2 + 1}\right)\right). \]

This simplifies to:

\[ f(-x) = \sin\left(\log\left(-x + \sqrt{x^2 + 1}\right)\right). \]

### Step 3: Simplify \( f(-x) \)
Using the property of logarithms, we can rewrite:

\[ f(-x) = \sin\left(\log\left(\frac{1}{x + \sqrt{x^2 + 1}}\right)\right). \]

Since \( \log\left(\frac{1}{a}\right) = -\log(a) \), we have:

\[ f(-x) = \sin\left(-\log\left(x + \sqrt{x^2 + 1}\right)\right). \]

Using the property of sine, \( \sin(-y) = -\sin(y) \):

\[ f(-x) = -\sin\left(\log\left(x + \sqrt{x^2 + 1}\right)\right) = -f(x). \]

### Step 4: Conclude that \( f(x) \) is odd
Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 5: Evaluate the integral
The integral of an odd function over a symmetric interval around zero is zero:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx = 0. \]

### Final Answer
Thus, the value of the integral is

\[ \boxed{0}. \]