Evaluate `int sin x cos x cos2x cos 4x cos 8x dx`

Explanation

To evaluate the integral \( \int \sin x \cos x \cos 2x \cos 4x \cos 8x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \sin x \cos x \cos 2x \cos 4x \cos 8x \, dx \]
We can multiply and divide by 2:
\[ I = \frac{1}{2} \int 2 \sin x \cos x \cos 2x \cos 4x \cos 8x \, dx \]

### Step 2: Use the Double Angle Identity
Using the identity \( 2 \sin x \cos x = \sin 2x \), we can rewrite the integral:
\[ I = \frac{1}{2} \int \sin 2x \cos 2x \cos 4x \cos 8x \, dx \]

### Step 3: Apply the Double Angle Identity Again
Again, we can multiply and divide by 2:
\[ I = \frac{1}{4} \int 2 \sin 2x \cos 2x \cos 4x \cos 8x \, dx \]
Using the identity \( 2 \sin 2x \cos 2x = \sin 4x \), we have:
\[ I = \frac{1}{4} \int \sin 4x \cos 4x \cos 8x \, dx \]

### Step 4: Repeat the Process
We repeat the process once more:
\[ I = \frac{1}{8} \int 2 \sin 4x \cos 4x \cos 8x \, dx \]
Using the identity \( 2 \sin 4x \cos 4x = \sin 8x \), we get:
\[ I = \frac{1}{8} \int \sin 8x \cos 8x \, dx \]

### Step 5: Final Application of the Identity
We apply the identity again:
\[ I = \frac{1}{16} \int 2 \sin 8x \cos 8x \, dx \]
Using \( 2 \sin 8x \cos 8x = \sin 16x \), we have:
\[ I = \frac{1}{16} \int \sin 16x \, dx \]

### Step 6: Integrate
Now we can integrate:
\[ I = \frac{1}{16} \left( -\frac{1}{16} \cos 16x \right) + C = -\frac{1}{256} \cos 16x + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = -\frac{1}{256} \cos 16x + C \]