`int e^x (logsin x + cot x)dx`

Answer

To solve the integral \( \int e^x (\log \sin x + \cot x) \, dx \), we can use the integration technique involving the identity for integrating functions of the form \( e^x f(x) + f'(x) \).

### Step-by-Step Solution:

1. **Identify the Integral**:
\[ I = \int e^x (\log \sin x + \cot x) \, dx \]

2. **Recognize the Structure**:
We can express the integral in a form that allows us to apply the integration identity:
\[ \int e^x f(x) + f'(x) \, dx = e^x f(x) + C \]
Here, we will consider \( f(x) = \log \sin x \).

3. **Differentiate \( f(x) \)**:
We need to find \( f'(x) \):
\[ f(x) = \log \sin x \]
Using the chain rule, we differentiate:
\[ f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x \]

4. **Apply the Identity**:
Since we have \( f(x) = \log \sin x \) and \( f'(x) = \cot x \), we can apply the integration identity:
\[ I = e^x f(x) + C = e^x \log \sin x + C \]

5. **Final Result**:
Thus, the integral evaluates to:
\[ \int e^x (\log \sin x + \cot x) \, dx = e^x \log \sin x + C \]