Which one of the followng is correct in respect of the function
`f(x)=x sin x + cos x + 1/2 cos^(2)x` ?

Options

A. It is increasing in the interval `(0, (pi)/(2))`
B. It remain constant in the interval `(0, (pi)/(2))`
C. It is decreasing in the interval `(0, (pi)/(2))`
D. It is decreasing in the interval `((pi)/(4), (pi)/(2))`

Explanation

To determine the behavior of the function \( f(x) = x \sin x + \cos x + \frac{1}{2} \cos^2 x \), we need to find its derivative and analyze the intervals of increase and decrease.

### Step-by-Step Solution:

1. **Differentiate the Function**:
We start by differentiating \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x \sin x) + \frac{d}{dx}(\cos x) + \frac{d}{dx}\left(\frac{1}{2} \cos^2 x\right) \]

Using the product rule for \( x \sin x \):
\[ \frac{d}{dx}(x \sin x) = x \cos x + \sin x \]

The derivative of \( \cos x \) is:
\[ \frac{d}{dx}(\cos x) = -\sin x \]

For \( \frac{1}{2} \cos^2 x \), we use the chain rule:
\[ \frac{d}{dx}\left(\frac{1}{2} \cos^2 x\right) = \cos x \cdot (-\sin x) = -\sin x \cos x \]

Now, combining these results:
\[ f'(x) = (x \cos x + \sin x) - \sin x - \sin x \cos x \]
Simplifying this gives:
\[ f'(x) = x \cos x - \sin x \cos x \]
Factoring out \( \cos x \):
\[ f'(x) = \cos x (x - \sin x) \]

2. **Determine Critical Points**:
To find where \( f'(x) = 0 \):
\[ \cos x (x - \sin x) = 0 \]
This gives us two cases:
- \( \cos x = 0 \) which occurs at \( x = \frac{\pi}{2} + n\pi \) (not relevant for our interval)
- \( x - \sin x = 0 \) which we will analyze further.

3. **Analyze the Interval \( [0, \frac{\pi}{2}] \)**:
We need to check the sign of \( f'(x) \) in the interval \( [0, \frac{\pi}{2}] \):
- In this interval, \( \cos x \) is positive.
- We need to analyze \( x - \sin x \):
- At \( x = 0 \): \( 0 - \sin(0) = 0 \)
- At \( x = \frac{\pi}{2} \): \( \frac{\pi}{2} - \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - 1 \) (which is positive since \( \frac{\pi}{2} \approx 1.57 > 1 \))

Since \( x - \sin x \) is increasing in \( [0, \frac{\pi}{2}] \) and starts from \( 0 \), it remains positive throughout the interval.

4. **Conclusion**:
Since \( f'(x) > 0 \) in the interval \( [0, \frac{\pi}{2}] \), we conclude that \( f(x) \) is increasing in this interval.

### Final Answer:
The function \( f(x) \) is increasing in the interval \( [0, \frac{\pi}{2}] \).