`intxsin^2xdx =`

Options

A. `1/8[2x^2 - 2x sin 2x - cos 2x]`
B. `1/8[2x^2 + 2x sin 2x - cos 2x]`
C. `1/8[2x^2 - 2x sin 2x + cos 2x]`
D. None of these

Answer

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

### Step 1: Use the identity for \( \sin^2 x \)
We know that:
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
Thus, we can rewrite the integral as:
\[ \int x \sin^2 x \, dx = \int x \left( \frac{1 - \cos 2x}{2} \right) \, dx = \frac{1}{2} \int x (1 - \cos 2x) \, dx \]

### Step 2: Split the integral
Now, we can split the integral into two parts:
\[ \frac{1}{2} \int x (1 - \cos 2x) \, dx = \frac{1}{2} \left( \int x \, dx - \int x \cos 2x \, dx \right) \]

### Step 3: Integrate the first part
The first integral \( \int x \, dx \) can be calculated as:
\[ \int x \, dx = \frac{x^2}{2} \]

### Step 4: Integrate the second part using integration by parts
For the integral \( \int x \cos 2x \, dx \), we will use integration by parts. Let:
- \( u = x \) (thus \( du = dx \))
- \( dv = \cos 2x \, dx \) (thus \( v = \frac{\sin 2x}{2} \))

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int x \cos 2x \, dx = x \cdot \frac{\sin 2x}{2} - \int \frac{\sin 2x}{2} \, dx \]
Now, we need to integrate \( \int \frac{\sin 2x}{2} \, dx \):
\[ \int \frac{\sin 2x}{2} \, dx = -\frac{\cos 2x}{4} \]
Thus,
\[ \int x \cos 2x \, dx = \frac{x \sin 2x}{2} + \frac{\cos 2x}{4} \]

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

### Step 6: Final result
Thus, the final result of the integral is:
\[ \int x \sin^2 x \, dx = \frac{1}{8} \left( 2x^2 - 2x \sin 2x - \cos 2x \right) + C \]