`int_(-pi/4) ^(pi/4) (e^theta(thetasintheta))/(e^(2theta)-1)d theta=`

Options

A. 3
B. 2
C. 1
D. 0

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{e^{\theta} \cdot \theta \sin \theta}{e^{2\theta} - 1} \, d\theta, \]

we will analyze the function inside the integral to determine if it is odd or even.

### Step 1: Define the function

Let

\[ f(\theta) = \frac{e^{\theta} \cdot \theta \sin \theta}{e^{2\theta} - 1}. \]

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

To check if \( f(\theta) \) is odd, we need to compute \( f(-\theta) \):

\[ f(-\theta) = \frac{e^{-\theta} \cdot (-\theta) \sin(-\theta)}{e^{-2\theta} - 1}. \]

Using the properties of sine and exponents, we have:

\[ \sin(-\theta) = -\sin(\theta), \]

and

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

Thus,

\[ f(-\theta) = \frac{e^{-\theta} \cdot (-\theta) \cdot (-\sin \theta)}{\frac{1}{e^{2\theta}} - 1} = \frac{e^{-\theta} \cdot \theta \sin \theta}{\frac{1 - e^{2\theta}}{e^{2\theta}}} = \frac{e^{-\theta} \cdot \theta \sin \theta \cdot e^{2\theta}}{1 - e^{2\theta}}. \]

This simplifies to:

\[ f(-\theta) = \frac{e^{\theta} \cdot \theta \sin \theta}{1 - e^{2\theta}}. \]

### Step 3: Rewrite the denominator

Notice that:

\[ 1 - e^{2\theta} = -(e^{2\theta} - 1). \]

Thus, we can write:

\[ f(-\theta) = -\frac{e^{\theta} \cdot \theta \sin \theta}{e^{2\theta} - 1} = -f(\theta). \]

### Step 4: Conclusion about the function

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

### Step 5: Evaluate the integral

The integral of an odd function over a symmetric interval about the origin is zero:

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(\theta) \, d\theta = 0. \]

Thus, the final answer is:

\[ \boxed{0}. \]