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

Options

A. `(pi)/(2) log 2`
B. `(pi^(2))/(2) log 2`
C. `-(pi^(2))/(2)` log 2`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{\pi} x \log(\sin x) \, dx \), we will use the property of definite integrals and some symmetry.

### Step 1: Set up the integral
Let
\[ I = \int_{0}^{\pi} x \log(\sin x) \, dx \]

### Step 2: Use the property of integrals
Using the property of integrals, we know that:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \pi \), so:
\[ I = \int_{0}^{\pi} (\pi - x) \log(\sin(\pi - x)) \, dx \]

### Step 3: Simplify the integral
Since \( \sin(\pi - x) = \sin x \), we can rewrite the integral as:
\[ I = \int_{0}^{\pi} (\pi - x) \log(\sin x) \, dx \]
This gives us:
\[ I = \int_{0}^{\pi} \pi \log(\sin x) \, dx - \int_{0}^{\pi} x \log(\sin x) \, dx \]
Thus, we have:
\[ I = \pi \int_{0}^{\pi} \log(\sin x) \, dx - I \]

### Step 4: Solve for \( I \)
Adding \( I \) to both sides, we get:
\[ 2I = \pi \int_{0}^{\pi} \log(\sin x) \, dx \]
So,
\[ I = \frac{\pi}{2} \int_{0}^{\pi} \log(\sin x) \, dx \]

### Step 5: Evaluate \( \int_{0}^{\pi} \log(\sin x) \, dx \)
It is known that:
\[ \int_{0}^{\pi} \log(\sin x) \, dx = -\pi \log(2) \]
Thus, substituting this result back into our expression for \( I \):
\[ I = \frac{\pi}{2} (-\pi \log(2)) = -\frac{\pi^2}{2} \log(2) \]

### Final Result
The value of the integral \( \int_{0}^{\pi} x \log(\sin x) \, dx \) is:
\[ \boxed{-\frac{\pi^2}{2} \log(2)} \]