The value of the integral `int_(0) ^(pi) cos 2 x log _(e) sin x dx` is

Options

A. 0
B. `-pi/2`
C. `pi/2`
D. `-pi`

Answer

To solve 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
We will let:
- \( u = \log(\sin x) \) (which we will differentiate)
- \( dv = \cos(2x) \, dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now we differentiate \( u \) and integrate \( dv \):
- \( du = \frac{1}{\sin x} \cos x \, dx \) (using the derivative of \( \log(\sin x) \))
- \( v = \frac{1}{2} \sin(2x) \) (integrating \( \cos(2x) \))

### Step 3: Apply integration by parts formula
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I = \left[ \log(\sin x) \cdot \frac{1}{2} \sin(2x) \right]_{0}^{\pi} - \int_{0}^{\pi} \frac{1}{2} \sin(2x) \cdot \frac{\cos x}{\sin x} \, dx \]

### Step 4: Evaluate the boundary term
Now we evaluate the boundary term:
\[ \left[ \log(\sin x) \cdot \frac{1}{2} \sin(2x) \right]_{0}^{\pi} \]
At \( x = 0 \) and \( x = \pi \), \( \sin(2x) = 0 \), hence the boundary term evaluates to:
\[ 0 - 0 = 0 \]

### Step 5: Simplify the integral
Now we simplify the remaining integral:
\[ I = - \frac{1}{2} \int_{0}^{\pi} \sin(2x) \cot(x) \, dx \]
Since \( \cot(x) = \frac{\cos x}{\sin x} \), we can rewrite the integral:
\[ I = - \frac{1}{2} \int_{0}^{\pi} \frac{\sin(2x) \cos x}{\sin x} \, dx \]

### Step 6: Use the identity for \( \sin(2x) \)
Using the identity \( \sin(2x) = 2 \sin x \cos x \):
\[ I = - \frac{1}{2} \int_{0}^{\pi} 2 \cos^2 x \, dx = - \int_{0}^{\pi} \cos^2 x \, dx \]

### Step 7: Evaluate the integral of \( \cos^2 x \)
Using the identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \):
\[ I = - \int_{0}^{\pi} \frac{1 + \cos(2x)}{2} \, dx = -\frac{1}{2} \left[ x + \frac{1}{2} \sin(2x) \right]_{0}^{\pi} \]
Evaluating this gives:
\[ = -\frac{1}{2} \left[ \pi + 0 - (0 + 0) \right] = -\frac{\pi}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{-\frac{\pi}{2}} \]