`int (1-x^2) sin 2x dx`

Answer

To solve the integral \( \int (1 - x^2) \sin(2x) \, dx \), we can break it down into two parts:

1. \( \int \sin(2x) \, dx \)
2. \( -\int x^2 \sin(2x) \, dx \)

### Step 1: Solve \( \int \sin(2x) \, dx \)

The integral of \( \sin(2x) \) can be computed using a simple substitution.

\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C_1 \]

### Step 2: Solve \( -\int x^2 \sin(2x) \, dx \) using Integration by Parts

For the integral \( -\int x^2 \sin(2x) \, dx \), we will use integration by parts. We choose:

- \( u = x^2 \) → \( du = 2x \, dx \)
- \( dv = \sin(2x) \, dx \) → \( v = -\frac{1}{2} \cos(2x) \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):

\[ -\int x^2 \sin(2x) \, dx = -\left( x^2 \cdot -\frac{1}{2} \cos(2x) - \int -\frac{1}{2} \cos(2x) \cdot 2x \, dx \right) \]

This simplifies to:

\[ \frac{1}{2} x^2 \cos(2x) + \int x \cos(2x) \, dx \]

### Step 3: Solve \( \int x \cos(2x) \, dx \) using Integration by Parts again

For the integral \( \int x \cos(2x) \, dx \), we apply integration by parts again:

- \( u = x \) → \( du = dx \)
- \( dv = \cos(2x) \, dx \) → \( v = \frac{1}{2} \sin(2x) \)

Using the integration by parts formula again:

\[ \int x \cos(2x) \, dx = x \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \, dx \]

Calculating the integral \( \int \frac{1}{2} \sin(2x) \, dx \):

\[ \int \frac{1}{2} \sin(2x) \, dx = -\frac{1}{4} \cos(2x) \]

Putting it all together:

\[ \int x \cos(2x) \, dx = \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x) \]

### Step 4: Combine all parts

Now we can substitute back into our expression:

\[ -\int x^2 \sin(2x) \, dx = \frac{1}{2} x^2 \cos(2x) + \left( \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x) \right) \]

Combining everything gives:

\[ \int (1 - x^2) \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + \frac{1}{2} x^2 \cos(2x) + \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x) + C \]

### Final Answer

\[ \int (1 - x^2) \sin(2x) \, dx = \left( \frac{1}{2} x^2 - \frac{1}{4} \right) \cos(2x) + \frac{1}{2} x \sin(2x) + C \]