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

Explanation

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

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

### Step 1: Apply the property of definite integrals

Let \( I = \int_{0}^{\pi} x \log(\sin x) \, dx \).

Now, we can change the variable by letting \( x = \pi - t \). Then, \( dx = -dt \) and the limits change as follows:
- When \( x = 0 \), \( t = \pi \)
- When \( x = \pi \), \( t = 0 \)

Thus, we have:

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

### Step 2: Simplify the integral

Using the property \( \sin(\pi - t) = \sin t \), we can rewrite the integral as:

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

Now, we can expand this integral:

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

### Step 3: Combine the integrals

Notice that the second integral is just \( I \):

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

### Step 4: Solve for \( I \)

Now, we can add \( I \) to both sides:

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

Thus,

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

### Step 5: Evaluate the integral \( \int_{0}^{\pi} \log(\sin t) \, dt \)

We can use the known result:

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

### Step 6: Substitute back into the equation for \( I \)

Substituting this result back into our equation for \( I \):

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

### Final Result

Thus, the value of the integral is:

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

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