Evaluate: `inte^x\ ((sinxcosx-1)/(sin^2x))\ dx`

Answer

To evaluate the integral

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

we can follow these steps:

### Step 1: Separate the terms in the integral
We can rewrite the integral by separating the numerator:

\[ I = \int e^x \left( \frac{\sin x \cos x}{\sin^2 x} - \frac{1}{\sin^2 x} \right) \, dx. \]

### Step 2: Simplify the fractions
Now, we can simplify the fractions:

\[ I = \int e^x \left( \frac{\cos x}{\sin x} - \frac{1}{\sin^2 x} \right) \, dx. \]

This can be rewritten as:

\[ I = \int e^x \left( \cot x - \csc^2 x \right) \, dx. \]

### Step 3: Use the identity for integration
We can use the integration identity:

\[ \int e^x f(x) + e^x f'(x) \, dx = e^x f(x) + C, \]

where \( f(x) = \cot x \) and \( f'(x) = -\csc^2 x \).

### Step 4: Apply the identity
From the identity, we can apply it to our integral:

\[ I = e^x \cot x + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ I = e^x \cot x + C. \]