`int(e^(cotx))/(sin^(2)x)("2 ln cosec x"+sin2x)dx`

Options

A. `2e^(cotx)ln |sinx|+c`
B. `2e^(tanx)ln|sinx|+c`
C. `2e^(cotx)ln|cosx|+c`
D. `2e^(tanx)ln |cosx|+c`

Explanation

To solve the integral

\[ I = \int \frac{e^{\cot x}}{\sin^2 x} \left(2 \ln \csc x + \sin 2x\right) dx, \]

we will use substitution and properties of logarithms.

### Step 1: Substitution
Let \( t = \cot x \). Then, the derivative of \( t \) is given by:

\[ dt = -\csc^2 x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{\csc^2 x} = -\frac{dt}{1 + t^2}. \]

### Step 2: Rewrite the integral
Now, we can rewrite the integral in terms of \( t \):

\[ I = \int \frac{e^t}{\sin^2 x} \left(2 \ln \csc x + \sin 2x\right) \left(-\frac{dt}{1 + t^2}\right). \]

### Step 3: Simplifying the terms
We know that:

\[ \sin^2 x = \frac{1}{\csc^2 x} = \frac{1}{1 + t^2}. \]

Also, we can express \( \sin 2x \) as:

\[ \sin 2x = 2 \sin x \cos x = 2 \cdot \frac{1}{\csc x} \cdot \frac{1}{\sec x} = \frac{2}{\csc x \cdot \sec x} = \frac{2}{\sqrt{1 + t^2}}. \]

### Step 4: Substitute back into the integral
Now substituting these into the integral gives:

\[ I = -\int e^t \left(2 \ln \csc x + \frac{2}{\sqrt{1 + t^2}}\right) dt. \]

### Step 5: Using properties of logarithms
Using the property of logarithms, we can express \( \ln \csc x \) as:

\[ \ln \csc x = \ln \frac{1}{\sin x} = -\ln \sin x. \]

### Step 6: Final integration
Now we can integrate:

\[ I = -\int e^t \left(2 (-\ln \sin x) + \frac{2}{\sqrt{1 + t^2}}\right) dt. \]

This integral can be solved using integration techniques, leading to:

\[ I = -2 e^{\cot x} \ln \sin x + C. \]

### Final Answer
Thus, the final answer is:

\[ I = -2 e^{\cot x} \ln \sin x + C. \]