`int x cos 2x sin 4x dx`

Answer

To solve the integral \( \int x \cos(2x) \sin(4x) \, dx \), we can follow these steps:

### Step 1: Use the Product-to-Sum Identity
We can use the identity \( 2 \sin A \cos B = \sin(A + B) - \sin(A - B) \). Here, let \( A = 4x \) and \( B = 2x \).

\[ 2 \sin(4x) \cos(2x) = \sin(6x) - \sin(2x) \]

Thus, we can rewrite the integral:

\[ \int x \cos(2x) \sin(4x) \, dx = \frac{1}{2} \int x (\sin(6x) - \sin(2x)) \, dx \]

### Step 2: Split the Integral
Now we can split the integral into two parts:

\[ \frac{1}{2} \int x \sin(6x) \, dx - \frac{1}{2} \int x \sin(2x) \, dx \]

### Step 3: Apply Integration by Parts
For both integrals, we will use integration by parts, which states:

\[ \int u \, dv = uv - \int v \, du \]

#### For \( \int x \sin(6x) \, dx \):
Let \( u = x \) and \( dv = \sin(6x) \, dx \).

Then, \( du = dx \) and \( v = -\frac{1}{6} \cos(6x) \).

Now applying integration by parts:

\[ \int x \sin(6x) \, dx = -\frac{1}{6} x \cos(6x) - \int -\frac{1}{6} \cos(6x) \, dx \]

The integral of \( \cos(6x) \) is:

\[ \int \cos(6x) \, dx = \frac{1}{6} \sin(6x) \]

Thus,

\[ \int x \sin(6x) \, dx = -\frac{1}{6} x \cos(6x) + \frac{1}{36} \sin(6x) \]

#### For \( \int x \sin(2x) \, dx \):
Let \( u = x \) and \( dv = \sin(2x) \, dx \).

Then, \( du = dx \) and \( v = -\frac{1}{2} \cos(2x) \).

Applying integration by parts:

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

The integral of \( \cos(2x) \) is:

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

Thus,

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

### Step 4: Combine Results
Now we can substitute back into our original expression:

\[ \int x \cos(2x) \sin(4x) \, dx = \frac{1}{2} \left( -\frac{1}{6} x \cos(6x) + \frac{1}{36} \sin(6x) \right) - \frac{1}{2} \left( -\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) \right) \]

This simplifies to:

\[ = -\frac{1}{12} x \cos(6x) + \frac{1}{72} \sin(6x) + \frac{1}{4} x \cos(2x) - \frac{1}{8} \sin(2x) + C \]

### Final Answer
Thus, the final answer is:

\[ \int x \cos(2x) \sin(4x) \, dx = -\frac{1}{12} x \cos(6x) + \frac{1}{72} \sin(6x) + \frac{1}{4} x \cos(2x) - \frac{1}{8} \sin(2x) + C \]