The value of `int(sinx.cosx.cos2x.cos4x.cos8x.cos16x)`dx is equal to

Options

A. `(sin16x)/(1024) + C`
B. `-(cos32x)/(1024)+C`
C. `(cos32x)/(1096)+C`
D. `-(cos32x)/(1096)+C`

Answer

To solve the integral \( I = \int \sin x \cos x \cos 2x \cos 4x \cos 8x \cos 16x \, dx \), we will follow a systematic approach using trigonometric identities.

### Step-by-Step Solution:

1. **Multiply and Divide by 2**:
We start by multiplying and dividing the integral by 2:
\[ I = \int \sin x \cos x \cos 2x \cos 4x \cos 8x \cos 16x \, dx = \frac{1}{2} \int 2 \sin x \cos x \cos 2x \cos 4x \cos 8x \cos 16x \, dx \]

2. **Use the Identity**:
We use the identity \( 2 \sin x \cos x = \sin 2x \):
\[ I = \frac{1}{2} \int \sin 2x \cos 2x \cos 4x \cos 8x \cos 16x \, dx \]

3. **Repeat the Process**:
Again, we multiply and divide by 2:
\[ I = \frac{1}{2} \cdot \frac{1}{2} \int 2 \sin 2x \cos 2x \cos 4x \cos 8x \cos 16x \, dx = \frac{1}{4} \int \sin 2x \cos 2x \cos 4x \cos 8x \cos 16x \, dx \]
Using the identity again:
\[ I = \frac{1}{4} \cdot \frac{1}{2} \int \sin 4x \cos 4x \cos 8x \cos 16x \, dx = \frac{1}{8} \int \sin 4x \cos 4x \cos 8x \cos 16x \, dx \]

4. **Continue the Process**:
We repeat this process:
\[ I = \frac{1}{8} \cdot \frac{1}{2} \int \sin 8x \cos 8x \cos 16x \, dx = \frac{1}{16} \int \sin 8x \cos 8x \cos 16x \, dx \]
\[ I = \frac{1}{16} \cdot \frac{1}{2} \int \sin 16x \cos 16x \, dx = \frac{1}{32} \int \sin 16x \cos 16x \, dx \]

5. **Final Step**:
Finally, we apply the identity one last time:
\[ I = \frac{1}{32} \cdot \frac{1}{2} \int \sin 32x \, dx = \frac{1}{64} \int \sin 32x \, dx \]
The integral of \( \sin 32x \) is:
\[ \int \sin 32x \, dx = -\frac{1}{32} \cos 32x + C \]
Therefore:
\[ I = \frac{1}{64} \left(-\frac{1}{32} \cos 32x\right) + C = -\frac{1}{2048} \cos 32x + C \]

### Final Answer:
\[ I = -\frac{1}{2048} \cos 32x + C \]