`int sin""(x)/(2) cos""(x)/(2) cos x dx ` is equal to

Options

A. `- ( 1)/( 4) cos 2x + C`
B. `- ( 1)/( 8 ) cos 2x + C`
C. `(1)/(8) cos 2x + C`
D. ` - (1)/( 8 ) sin 2x + C`

Answer

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

### Step-by-Step Solution

1. **Rewrite the Integral**:
We start with the integral:
\[ I = \int \frac{\sin(x/2)}{\cos(x/2) \cos x} \, dx \]

2. **Use the Identity for Sine**:
We know that:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Let \( \theta = \frac{x}{2} \). Then, we can express \( \sin(x/2) \cos(x/2) \) as:
\[ \sin(x) = 2 \sin(x/2) \cos(x/2) \implies \sin(x/2) \cos(x/2) = \frac{1}{2} \sin(x) \]

3. **Substitute in the Integral**:
Substitute \( \sin(x/2) \cos(x/2) \) in the integral:
\[ I = \int \frac{\frac{1}{2} \sin(x)}{\cos(x/2) \cos x} \, dx \]
This simplifies to:
\[ I = \frac{1}{2} \int \frac{\sin(x)}{\cos(x/2) \cos x} \, dx \]

4. **Rewrite the Integral**:
We can further simplify:
\[ I = \frac{1}{2} \int \sin(x) \sec(x/2) \, dx \]

5. **Use Another Identity**:
We can use the identity again to express \( \sin(x) \cos(x) \):
\[ \sin(x) \cos(x) = \frac{1}{2} \sin(2x) \]
Thus, we can write:
\[ I = \frac{1}{4} \int \sin(2x) \, dx \]

6. **Integrate**:
Now we integrate:
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C \]
Therefore,
\[ I = \frac{1}{4} \left(-\frac{1}{2} \cos(2x) + C\right) = -\frac{1}{8} \cos(2x) + C \]

7. **Final Result**:
The final result of the integral is:
\[ I = -\frac{1}{8} \cos(2x) + C \]