`inte^(x)(cotx+logsinx)dx=?`

Options

A. `e^(x)cotx+C`
B. `e^(x)logsinx+C`
C. `e^(x)sinx+C`
D. none of these

Explanation

To solve the integral \( \int e^x (\cot x + \log \sin x) \, dx \), we will follow a systematic approach using substitution and integration techniques.

### Step-by-Step Solution:

1. **Set Up the Integral**:
Let \( I = \int e^x (\cot x + \log \sin x) \, dx \).

2. **Use Substitution**:
We will use the substitution \( t = e^x \log \sin x \). To differentiate this, we will apply the product rule.

3. **Differentiate \( t \)**:
\[ \frac{dt}{dx} = e^x \frac{d}{dx}(\log \sin x) + \log \sin x \frac{d}{dx}(e^x) \]
Using the chain rule, we have:
\[ \frac{d}{dx}(\log \sin x) = \frac{1}{\sin x} \cos x = \cot x \]
Thus,
\[ \frac{dt}{dx} = e^x \cot x + e^x \log \sin x \]

4. **Rearranging the Equation**:
From the differentiation, we can write:
\[ dt = (e^x \cot x + e^x \log \sin x) \, dx \]
This implies:
\[ dt = e^x (\cot x + \log \sin x) \, dx \]
Therefore, we can express \( dx \) in terms of \( dt \):
\[ dx = \frac{dt}{e^x (\cot x + \log \sin x)} \]

5. **Integrating Both Sides**:
Now substituting back into the integral:
\[ I = \int dt \]
This simplifies to:
\[ I = t + C \]

6. **Substituting Back for \( t \)**:
Recall that \( t = e^x \log \sin x \), so we have:
\[ I = e^x \log \sin x + C \]

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