The value of `int_(-pi//4)^(pi//4)(e^(x)sec^(2)x)/(e^(2x)-1)dx`, is

Options

A. 0
B. 2
C. 4
D. none of these

Explanation

To solve the integral

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{e^x \sec^2 x}{e^{2x} - 1} \, dx, \]

we can use the property of definite integrals that states:

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

### Step 1: Define the function
Let

\[ f(x) = \frac{e^x \sec^2 x}{e^{2x} - 1}. \]

### Step 2: Find \(f(-x)\)
Next, we need to compute \(f(-x)\):

\[ f(-x) = \frac{e^{-x} \sec^2(-x)}{e^{-2x} - 1}. \]

Since \(\sec(-x) = \sec(x)\) and \(\sec^2(-x) = \sec^2(x)\), we can rewrite \(f(-x)\):

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

### Step 3: Combine \(f(x)\) and \(f(-x)\)
Now we can add \(f(x)\) and \(f(-x)\):

\[ f(x) + f(-x) = \frac{e^x \sec^2 x}{e^{2x} - 1} + \frac{e^x \sec^2 x}{1 - e^{2x}}. \]

### Step 4: Simplify the expression
To combine the fractions, we find a common denominator:

\[ f(x) + f(-x) = e^x \sec^2 x \left( \frac{1}{e^{2x} - 1} + \frac{1}{1 - e^{2x}} \right). \]

Notice that \(1 - e^{2x} = -(e^{2x} - 1)\), so:

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

Thus,

\[ f(x) + f(-x) = e^x \sec^2 x \left( \frac{1 - 1}{e^{2x} - 1} \right) = 0. \]

### Step 5: Evaluate the integral
Now we can substitute back into our integral:

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) \, dx = \int_{0}^{\frac{\pi}{4}} (f(x) + f(-x)) \, dx = \int_{0}^{\frac{\pi}{4}} 0 \, dx = 0. \]

Thus, the value of the integral is

\[ \boxed{0}. \]