To investigate the behavior of the functions for monotonicity in the given intervals, we will follow these steps:
### (i) For the function \( f(x) = -\sin^3 x + 3\sin^2 x + 5 \) in the interval \( x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \):
1. **Differentiate the function**:
\[
f'(x) = \frac{d}{dx}(-\sin^3 x + 3\sin^2 x + 5)
\]
Using the chain rule:
\[
f'(x) = -3\sin^2 x \cos x + 6\sin x \cos x
\]
Simplifying:
\[
f'(x) = 3\sin x \cos x (2 - \sin x)
\]
2. **Find critical points**:
Set \( f'(x) = 0 \):
\[
3\sin x \cos x (2 - \sin x) = 0
\]
This gives us:
- \( \sin x = 0 \) which occurs at \( x = 0 \)
- \( 2 - \sin x = 0 \) which gives \( \sin x = 2 \) (not possible since \( \sin x \) ranges from -1 to 1)
3. **Test intervals**:
We will test the sign of \( f'(x) \) in the intervals \( [-\frac{\pi}{2}, 0] \) and \( [0, \frac{\pi}{2}] \):
- For \( x \in [-\frac{\pi}{2}, 0] \): Choose \( x = -\frac{\pi}{4} \):
\[
f'(-\frac{\pi}{4}) = 3\sin(-\frac{\pi}{4})\cos(-\frac{\pi}{4})(2 - \sin(-\frac{\pi}{4})) > 0
\]
Thus, \( f'(x) > 0 \) in this interval, so \( f(x) \) is increasing.
- For \( x \in [0, \frac{\pi}{2}] \): Choose \( x = \frac{\pi}{4} \):
\[
f'(\frac{\pi}{4}) = 3\sin(\frac{\pi}{4})\cos(\frac{\pi}{4})(2 - \sin(\frac{\pi}{4})) > 0
\]
Thus, \( f'(x) > 0 \) in this interval, so \( f(x) \) is also increasing.
4. **Conclusion**:
The function \( f(x) \) is increasing on the entire interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
### (ii) For the function \( f(x) = \sec x - \csc x \) in the interval \( x \in (0, \frac{\pi}{2}) \):
1. **Differentiate the function**:
\[
f'(x) = \frac{d}{dx}(\sec x - \csc x)
\]
Using the derivatives of secant and cosecant:
\[
f'(x) = \sec x \tan x + \csc x \cot x
\]
2. **Find critical points**:
Set \( f'(x) = 0 \):
\[
\sec x \tan x + \csc x \cot x = 0
\]
This equation does not have simple roots, so we will analyze the sign of \( f'(x) \).
3. **Test the sign of \( f'(x) \)**:
Since both \( \sec x \) and \( \csc x \) are positive in the interval \( (0, \frac{\pi}{2}) \):
- \( \sec x \tan x > 0 \)
- \( \csc x \cot x > 0 \)
Therefore, \( f'(x) > 0 \) for all \( x \in (0, \frac{\pi}{2}) \).
4. **Conclusion**:
The function \( f(x) \) is increasing on the interval \( (0, \frac{\pi}{2}) \).
### Summary:
- \( f(x) = -\sin^3 x + 3\sin^2 x + 5 \) is increasing on \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).
- \( f(x) = \sec x - \csc x \) is increasing on \( (0, \frac{\pi}{2}) \).
