`int_(0)^(pi/2)logsinx=-(pi/2)log2`
`int_(0)^(pi) x log sin x dx`=

Options

A. `(1)/(2) pi^(2) log 2`
B. `-(1)/(2) pi^(2) log2`
C. `(1)/(2) pi^(2) log ((1)/(2))`
D. `-(1)/(2) pi^(2) log ((1)/(2))`

Answer

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

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

### Step 2: Use the Symmetry Property
We can use the property of definite integrals which states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we will let \( a = \pi \) and substitute \( x \) with \( \pi - x \):
\[ I = \int_{0}^{\pi} (\pi - x) \log(\sin(\pi - x)) \, dx \]
Using the identity \( \sin(\pi - x) = \sin x \), we have:
\[ I = \int_{0}^{\pi} (\pi - x) \log(\sin x) \, dx \]

### Step 3: Expand the Integral
Now we can expand this integral:
\[ I = \int_{0}^{\pi} \pi \log(\sin x) \, dx - \int_{0}^{\pi} x \log(\sin x) \, dx \]
This gives us:
\[ I = \pi \int_{0}^{\pi} \log(\sin x) \, dx - I \]

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

### Step 5: Use the Known Result
From the problem statement, we know:
\[ \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2 \]
Using the property of definite integrals:
\[ \int_{0}^{\pi} \log(\sin x) \, dx = 2 \int_{0}^{\frac{\pi}{2}} \log(\sin x) \, dx = 2 \left(-\frac{\pi}{2} \log 2\right) = -\pi \log 2 \]

### Step 6: Substitute Back
Now substituting this result back into our expression for \( I \):
\[ I = \frac{\pi}{2} (-\pi \log 2) = -\frac{\pi^2}{2} \log 2 \]

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