Evaluate: `int_(0)^(pi) xlog (sin x) dx`

Answer

To evaluate the integral \( I = \int_{0}^{\pi} x \log(\sin x) \, dx \), we can use a property of definite integrals. The property states that:

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]

### Step 1: Apply the property
We can express \( I \) in terms of \( \pi - x \):

\[ I = \int_{0}^{\pi} x \log(\sin x) \, dx = \int_{0}^{\pi} (\pi - x) \log(\sin(\pi - x)) \, dx \]

### Step 2: Simplify the integral
Using the identity \( \sin(\pi - x) = \sin x \), we can rewrite the integral:

\[ I = \int_{0}^{\pi} (\pi - x) \log(\sin x) \, dx \]

### Step 3: Expand the integral
Now we can expand this integral:

\[ I = \int_{0}^{\pi} \pi \log(\sin x) \, dx - \int_{0}^{\pi} x \log(\sin x) \, dx \]

### Step 4: Combine the integrals
Let’s denote the second integral as \( I \):

\[ I = \pi \int_{0}^{\pi} \log(\sin x) \, dx - I \]

### Step 5: Solve for \( I \)
Now, we can add \( I \) to both sides:

\[ 2I = \pi \int_{0}^{\pi} \log(\sin x) \, dx \]

Thus, we can express \( I \):

\[ I = \frac{\pi}{2} \int_{0}^{\pi} \log(\sin x) \, dx \]

### Step 6: Evaluate the integral \( \int_{0}^{\pi} \log(\sin x) \, dx \)
We know from integral calculus that:

\[ \int_{0}^{\pi} \log(\sin x) \, dx = -\pi \log(2) \]

### Step 7: Substitute back
Now we substitute this result back into our expression for \( I \):

\[ I = \frac{\pi}{2} \cdot (-\pi \log(2)) = -\frac{\pi^2}{2} \log(2) \]

### Final Result
Thus, the final answer is:

\[ I = -\frac{\pi^2}{2} \log(2) \]