`int(secx."cosec"x)/(2cotx-secx"cosec x")dx` is equal to

Options

A. `(1)/(2)ln|sec2x+tan2x|+C`
B. `ln|secx+"cosec x"|+C`
C. `ln |secx+Tanx|+C`
D. `(1)/(2)ln|secx+" cosec x"|+C`

Explanation

To solve the integral

\[ I = \int \frac{\sec x \cdot \csc x}{2 \cot x - \sec x \cdot \csc x} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the integrand in terms of sine and cosine

We start by rewriting the trigonometric functions in terms of sine and cosine:

\[ \sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x}, \quad \cot x = \frac{\cos x}{\sin x}. \]

Substituting these into the integral gives:

\[ I = \int \frac{\frac{1}{\cos x} \cdot \frac{1}{\sin x}}{2 \cdot \frac{\cos x}{\sin x} - \frac{1}{\cos x} \cdot \frac{1}{\sin x}} \, dx. \]

### Step 2: Simplify the denominator

The denominator simplifies as follows:

\[ 2 \cot x - \sec x \cdot \csc x = 2 \cdot \frac{\cos x}{\sin x} - \frac{1}{\sin x \cos x}. \]

Finding a common denominator:

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

### Step 3: Substitute back into the integral

Now substituting this back into the integral gives:

\[ I = \int \frac{\frac{1}{\sin x \cos x}}{\frac{2 \cos^2 x - 1}{\sin x \cos x}} \, dx. \]

This simplifies to:

\[ I = \int \frac{1}{2 \cos^2 x - 1} \, dx. \]

### Step 4: Recognize the trigonometric identity

Notice that:

\[ 2 \cos^2 x - 1 = \cos 2x. \]

So we can rewrite the integral as:

\[ I = \int \frac{1}{\cos 2x} \, dx = \int \sec 2x \, dx. \]

### Step 5: Integrate secant function

The integral of \(\sec 2x\) is known:

\[ \int \sec 2x \, dx = \frac{1}{2} \ln | \sec 2x + \tan 2x | + C, \]

where \(C\) is the constant of integration.

### Final Answer

Thus, the final answer for the integral is:

\[ I = \frac{1}{2} \ln | \sec 2x + \tan 2x | + C. \]

---