`int_(log 1//2)^(log 2) sin {(e^(x)-1)/(e^(x) +1}dx`=

Options

A. `"cos"(1)/(3)`
B. `"sin"(1)/(2)`
C. 2cos 2
D. 0

Answer

To solve the integral

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

we will use the property of odd functions in definite integrals.

### Step 1: Change the limits of integration

First, we can rewrite the lower limit of integration:

\[ \log \frac{1}{2} = \log 2^{-1} = -\log 2. \]

Thus, we can express the integral as:

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

### Step 2: Check if the function is odd

Next, we define the function:

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

To check if \( f(x) \) is an odd function, we need to evaluate \( f(-x) \):

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

### Step 3: Simplify \( f(-x) \)

We can simplify \( f(-x) \):

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

Now, we can factor out a negative sign:

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

### Step 4: Conclude that \( f(x) \) is odd

Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 5: Use the property of odd functions in definite integrals

The property of definite integrals states that if \( f(x) \) is an odd function over a symmetric interval \([-a, a]\):

\[ \int_{-a}^{a} f(x) \, dx = 0. \]

In our case, we have:

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

### Final Result

Thus, the value of the integral is:

\[ \int_{\log \frac{1}{2}}^{\log 2} \sin\left(\frac{e^x - 1}{e^x + 1}\right) \, dx = 0. \]