Evaluate: `int_0^pi cos 2x log sin x dx`

Answer

To evaluate the integral \( I = \int_0^\pi \cos(2x) \log(\sin x) \, dx \), we will use integration by parts.

### Step 1: Set up integration by parts
Let:
- \( u = \log(\sin x) \) (which we will differentiate)
- \( dv = \cos(2x) \, dx \) (which we will integrate)

Then, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{\sin x} \cos x \, dx = \cot x \, dx \]
- Integrate \( dv \):
\[ v = \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) \]

### Step 2: Apply integration by parts formula
The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ I = \left[ \log(\sin x) \cdot \frac{1}{2} \sin(2x) \right]_0^\pi - \int_0^\pi \frac{1}{2} \sin(2x) \cot x \, dx \]

### Step 3: Evaluate the boundary term
Evaluate \( \left[ \log(\sin x) \cdot \frac{1}{2} \sin(2x) \right]_0^\pi \):
- At \( x = 0 \):
\[ \sin(2 \cdot 0) = 0 \quad \text{and} \quad \log(\sin(0)) \text{ is undefined, but approaches } -\infty \]
- At \( x = \pi \):
\[ \sin(2 \cdot \pi) = 0 \quad \text{and} \quad \log(\sin(\pi)) \text{ is also undefined, but approaches } -\infty \]
Thus, both terms vanish, leading to:
\[ \left[ \log(\sin x) \cdot \frac{1}{2} \sin(2x) \right]_0^\pi = 0 \]

### Step 4: Simplify the integral
Now we have:
\[ I = 0 - \frac{1}{2} \int_0^\pi \sin(2x) \cot x \, dx \]
This can be rewritten as:
\[ I = -\frac{1}{2} \int_0^\pi \frac{\sin(2x)}{\sin x} \cos x \, dx \]

### Step 5: Evaluate the integral
Using the identity \( \sin(2x) = 2 \sin x \cos x \):
\[ \int_0^\pi \frac{\sin(2x)}{\sin x} \cos x \, dx = \int_0^\pi 2 \cos^2 x \, dx \]
Now, we can evaluate:
\[ \int_0^\pi \cos^2 x \, dx = \frac{1}{2} \int_0^\pi (1 + \cos(2x)) \, dx = \frac{1}{2} \left[ x + \frac{1}{2} \sin(2x) \right]_0^\pi = \frac{1}{2} \left[ \pi + 0 \right] = \frac{\pi}{2} \]

### Step 6: Substitute back
Thus, we have:
\[ I = -\frac{1}{2} \cdot 2 \cdot \frac{\pi}{2} = -\frac{\pi}{2} \]

### Conclusion
The final result is:
\[ \int_0^\pi \cos(2x) \log(\sin x) \, dx = -\frac{\pi}{2} \]