`inte^(x)(cotx-"cosec"^(2)x)dx=?`

Options

A. `-e^(x)"cosec"^(2)x+C`
B. `e^(x)"cotx+C`
C. `-e^(x)"cotx+C`
D. none of these

Explanation

To solve the integral \( \int e^x \left( \cot x - \csc^2 x \right) dx \), we can break it down into two separate integrals:

\[ \int e^x \cot x \, dx - \int e^x \csc^2 x \, dx \]

Let's denote these two integrals as \( I_1 \) and \( I_2 \):

1. **First Integral**: \( I_1 = \int e^x \cot x \, dx \)
2. **Second Integral**: \( I_2 = \int e^x \csc^2 x \, dx \)

### Step 1: Solve \( I_1 = \int e^x \cot x \, dx \)

To solve \( I_1 \), we will use integration by parts. We can let:

- \( u = \cot x \) (first function)
- \( dv = e^x \, dx \) (second function)

Now we need to find \( du \) and \( v \):

- \( du = -\csc^2 x \, dx \)
- \( v = e^x \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):

\[ I_1 = e^x \cot x - \int e^x (-\csc^2 x) \, dx \]

This simplifies to:

\[ I_1 = e^x \cot x + \int e^x \csc^2 x \, dx \]

### Step 2: Substitute \( I_2 \) into \( I_1 \)

Now we can substitute \( I_2 \) back into the equation for \( I_1 \):

\[ I_1 = e^x \cot x + I_2 \]

### Step 3: Solve for \( I_2 \)

Now we have:

\[ I = I_1 - I_2 \]

Substituting \( I_1 \):

\[ I = (e^x \cot x + I_2) - I_2 \]

This simplifies to:

\[ I = e^x \cot x \]

### Step 4: Final Result

Thus, we have:

\[ \int e^x \left( \cot x - \csc^2 x \right) dx = e^x \cot x + C \]

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