`int_(0)^(pi) x log sinx\ dx`

Explanation

To solve the integral \( I = \int_{0}^{\pi} x \log(\sin x) \, dx \), we can use a property of definite integrals. Here’s the step-by-step solution:

### Step 1: Apply the Property of Definite Integrals
We can use the property that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we have \( a = \pi \) and \( f(x) = x \log(\sin x) \). Therefore:
\[ I = \int_{0}^{\pi} x \log(\sin x) \, dx = \int_{0}^{\pi} (\pi - x) \log(\sin(\pi - x)) \, dx \]

### Step 2: Simplify the Integral
Using the identity \( \sin(\pi - x) = \sin x \), we can rewrite the integral:
\[ I = \int_{0}^{\pi} (\pi - x) \log(\sin x) \, dx \]
Now, we can expand this integral:
\[ I = \int_{0}^{\pi} \pi \log(\sin x) \, dx - \int_{0}^{\pi} x \log(\sin x) \, dx \]
This gives us:
\[ I = \pi \int_{0}^{\pi} \log(\sin x) \, dx - I \]

### Step 3: Combine Like Terms
Now, we can add \( I \) to both sides:
\[ 2I = \pi \int_{0}^{\pi} \log(\sin x) \, dx \]
Thus, we can express \( I \) as:
\[ I = \frac{\pi}{2} \int_{0}^{\pi} \log(\sin x) \, dx \]

### Step 4: Use a Known Result
There is a known result for the integral:
\[ \int_{0}^{\pi} \log(\sin x) \, dx = -\pi \log(2) \]
Substituting this result into our expression for \( I \):
\[ I = \frac{\pi}{2} (-\pi \log(2)) = -\frac{\pi^2}{2} \log(2) \]

### Step 5: Final Result
Thus, the value of the integral is:
\[ I = -\frac{\pi^2}{2} \log(2) \]