`int_(0)^(pi) x sin^(2) x dx`

Answer

To solve the integral \( \int_{0}^{\pi} x \sin^2 x \, dx \), we can follow these steps:

### Step 1: Rewrite \( \sin^2 x \)
We can use the identity for \( \sin^2 x \):
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
Thus, we rewrite the integral:
\[ \int_{0}^{\pi} x \sin^2 x \, dx = \int_{0}^{\pi} x \left(\frac{1 - \cos(2x)}{2}\right) \, dx \]
This simplifies to:
\[ \frac{1}{2} \int_{0}^{\pi} x \, dx - \frac{1}{2} \int_{0}^{\pi} x \cos(2x) \, dx \]

### Step 2: Calculate the first integral
Now we calculate the first integral:
\[ \frac{1}{2} \int_{0}^{\pi} x \, dx \]
The integral of \( x \) is:
\[ \int x \, dx = \frac{x^2}{2} \]
Evaluating from 0 to \( \pi \):
\[ \frac{1}{2} \left[ \frac{x^2}{2} \right]_{0}^{\pi} = \frac{1}{2} \left[ \frac{\pi^2}{2} - 0 \right] = \frac{\pi^2}{4} \]

### Step 3: Calculate the second integral using integration by parts
Next, we need to evaluate:
\[ -\frac{1}{2} \int_{0}^{\pi} x \cos(2x) \, dx \]
We will use integration by parts, where we let:
- \( u = x \) (thus \( du = dx \))
- \( dv = \cos(2x) \, dx \) (thus \( v = \frac{1}{2} \sin(2x) \))

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int x \cos(2x) \, dx = x \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \, dx \]
The integral of \( \sin(2x) \) is:
\[ -\frac{1}{2} \cos(2x) \]
Thus:
\[ \int x \cos(2x) \, dx = \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) \]
Now we evaluate this from 0 to \( \pi \):
\[ \left[ \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) \right]_{0}^{\pi} \]
At \( x = \pi \):
\[ \frac{\pi}{2} \sin(2\pi) + \frac{1}{4} \cos(2\pi) = \frac{\pi}{2} \cdot 0 + \frac{1}{4} \cdot 1 = \frac{1}{4} \]
At \( x = 0 \):
\[ \frac{0}{2} \sin(0) + \frac{1}{4} \cos(0) = 0 + \frac{1}{4} = \frac{1}{4} \]
Thus:
\[ \left[ \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) \right]_{0}^{\pi} = \frac{1}{4} - \frac{1}{4} = 0 \]

### Step 4: Combine results
Putting it all together:
\[ \int_{0}^{\pi} x \sin^2 x \, dx = \frac{\pi^2}{4} - \frac{1}{2} \cdot 0 = \frac{\pi^2}{4} \]

### Final Answer
The final result is:
\[ \int_{0}^{\pi} x \sin^2 x \, dx = \frac{\pi^2}{4} \]