To determine the intervals where the function \( f(x) = \log(\sin x) \) is monotonically increasing, we will follow these steps:
### Step 1: Find the derivative of \( f(x) \)
To find where \( f(x) \) is increasing, we need to compute its derivative \( f'(x) \).
\[
f'(x) = \frac{d}{dx}[\log(\sin x)]
\]
Using the chain rule, we have:
\[
f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x
\]
### Step 2: Set the derivative greater than zero
For the function to be monotonically increasing, we need:
\[
f'(x) > 0 \implies \cot x > 0
\]
### Step 3: Determine where \( \cot x > 0 \)
The cotangent function \( \cot x \) is positive in the intervals where sine and cosine have the same sign. Specifically, \( \cot x > 0 \) in the following intervals:
1. \( (0, \frac{\pi}{2}) \)
2. \( (\pi, \frac{3\pi}{2}) \)
3. \( (2\pi, \frac{5\pi}{2}) \)
4. And similar intervals for all multiples of \( \pi \).
### Step 4: Identify the relevant interval
Since we are interested in the interval \( (0, \frac{\pi}{2}) \), we can conclude that:
\[
f(x) \text{ is monotonically increasing in } (0, \frac{\pi}{2}).
\]
### Step 5: Verify the behavior of \( f(x) \) in other intervals
- In the interval \( (\frac{\pi}{2}, \pi) \), \( \cot x < 0 \), hence \( f(x) \) is decreasing.
- In the interval \( (\pi, \frac{3\pi}{2}) \), \( \cot x > 0 \), hence \( f(x) \) is increasing again.
- This pattern continues for other intervals.
### Conclusion
The function \( f(x) = \log(\sin x) \) is monotonically increasing in the interval:
\[
(0, \frac{\pi}{2}) \text{ and } (n\pi, n\pi + \frac{\pi}{2}) \text{ for any integer } n.
\]
