Evaluate `int sin x cosx * cos 2x * cos 4x dx`

Answer

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

### Step 1: Rewrite the integral
We start by rewriting the integral and factoring out constants:
\[ \int \sin x \cos x \cos 2x \cos 4x \, dx = \frac{1}{2} \int 2 \sin x \cos x \cos 2x \cos 4x \, dx \]
**Hint:** Remember that \( \sin x \cos x = \frac{1}{2} \sin 2x \).

### Step 2: Use the identity for \( \sin 2x \)
Using the identity \( \sin 2x = 2 \sin x \cos x \), we can rewrite the integral:
\[ = \frac{1}{2} \int \sin 2x \cos 2x \cos 4x \, dx \]

### Step 3: Rewrite \( \sin 2x \cos 2x \)
Now, we can use the identity \( \sin 2x \cos 2x = \frac{1}{2} \sin 4x \):
\[ = \frac{1}{2} \int \frac{1}{2} \sin 4x \cos 4x \, dx = \frac{1}{4} \int \sin 4x \cos 4x \, dx \]

### Step 4: Use the identity for \( \sin 4x \cos 4x \)
Again, we apply the identity \( \sin 4x \cos 4x = \frac{1}{2} \sin 8x \):
\[ = \frac{1}{4} \int \frac{1}{2} \sin 8x \, dx = \frac{1}{8} \int \sin 8x \, dx \]

### Step 5: Integrate \( \sin 8x \)
The integral of \( \sin 8x \) is:
\[ \int \sin 8x \, dx = -\frac{1}{8} \cos 8x + C \]
Thus,
\[ = \frac{1}{8} \left(-\frac{1}{8} \cos 8x + C\right) = -\frac{1}{64} \cos 8x + \frac{C}{8} \]

### Final Result
Therefore, the final answer is:
\[ \int \sin x \cos x \cos 2x \cos 4x \, dx = -\frac{1}{64} \cos 8x + C \]