Evaluate `int xsin^(2)x dx`

Explanation

To evaluate the integral \( \int x \sin^2 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
Using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \), we can rewrite the integral:
\[ \int x \sin^2 x \, dx = \int x \left( \frac{1 - \cos(2x)}{2} \right) \, dx = \frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos(2x) \, dx \]

### Step 2: Evaluate the first integral
The first integral can be easily computed:
\[ \frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \]

### Step 3: Evaluate the second integral using integration by parts
For the second integral \( \int x \cos(2x) \, dx \), we will use integration by parts. Let:
- \( u = x \) (which implies \( du = dx \))
- \( dv = \cos(2x) \, dx \) (which implies \( v = \frac{1}{2} \sin(2x) \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int x \cos(2x) \, dx = x \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \, dx \]

### Step 4: Evaluate the integral of \( \sin(2x) \)
The integral of \( \sin(2x) \) is:
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) \]
Thus,
\[ \int x \cos(2x) \, dx = \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) \]

### Step 5: Substitute back into the equation
Now substituting back into our expression for \( \int x \sin^2 x \, dx \):
\[ \int x \sin^2 x \, dx = \frac{x^2}{4} - \frac{1}{2} \left( \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) \right) \]
This simplifies to:
\[ \int x \sin^2 x \, dx = \frac{x^2}{4} - \frac{x}{4} \sin(2x) - \frac{1}{8} \cos(2x) + C \]

### Final Answer
Thus, the final result is:
\[ \int x \sin^2 x \, dx = \frac{x^2}{4} - \frac{x}{4} \sin(2x) + \frac{1}{8} \cos(2x) + C \]