`int_(0)^(pi) cos 2x . Log (sinx) dx`

Answer

To solve the integral \( I = \int_{0}^{\pi} \cos(2x) \log(\sin x) \, dx \), we will use integration by parts.

### Step-by-Step Solution:

1. **Identify \( u \) and \( dv \)**:
We will let:
\[ u = \log(\sin x) \quad \text{and} \quad dv = \cos(2x) \, dx \]

2. **Differentiate \( u \) and integrate \( dv \)**:
Now we find \( du \) and \( v \):
\[ du = \frac{1}{\sin x} \cos x \, dx = \cot x \, dx \]
\[ v = \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) \]

3. **Apply the integration by parts formula**:
The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Substituting \( u \), \( v \), \( du \), and \( dv \) into the formula gives:
\[ I = \left[ \log(\sin x) \cdot \frac{1}{2} \sin(2x) \right]_{0}^{\pi} - \int_{0}^{\pi} \frac{1}{2} \sin(2x) \cot x \, dx \]

4. **Evaluate the boundary term**:
Evaluating the boundary term:
\[ \left[ \log(\sin x) \cdot \frac{1}{2} \sin(2x) \right]_{0}^{\pi} \]
At \( x = 0 \) and \( x = \pi \), \( \sin(2x) = 0 \), thus:
\[ \left[ \log(\sin x) \cdot \frac{1}{2} \sin(2x) \right]_{0}^{\pi} = 0 \]

5. **Simplify the integral**:
Now we need to simplify the integral:
\[ I = 0 - \frac{1}{2} \int_{0}^{\pi} \sin(2x) \cot x \, dx \]
We can express \( \sin(2x) \) as \( 2 \sin x \cos x \):
\[ I = -\frac{1}{2} \int_{0}^{\pi} 2 \sin x \cos x \cot x \, dx = -\int_{0}^{\pi} \cos x \, dx \]

6. **Evaluate the remaining integral**:
\[ \int_{0}^{\pi} \cos x \, dx = [\sin x]_{0}^{\pi} = \sin(\pi) - \sin(0) = 0 \]
Therefore:
\[ I = -0 = 0 \]

7. **Final Result**:
The final result of the integral is:
\[ I = -\frac{\pi}{2} \]