To solve the problem, we will analyze the function \( f(x) = x \cos^{-1}(-\sin |x|) \) for \( x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
### Step 1: Simplify the Function
We start with the function:
\[
f(x) = x \cos^{-1}(-\sin |x|)
\]
Using the property of the inverse cosine function, we know:
\[
\cos^{-1}(-x) = \pi - \cos^{-1}(x)
\]
Thus, we can rewrite \( f(x) \) as:
\[
f(x) = x \left( \pi - \cos^{-1}(\sin |x|) \right)
\]
This simplifies to:
\[
f(x) = x \pi - x \cos^{-1}(\sin |x|)
\]
### Step 2: Analyze \( \cos^{-1}(\sin |x|) \)
Next, we need to analyze \( \cos^{-1}(\sin |x|) \). Since \( |x| \) is non-negative in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we can consider two cases based on the sign of \( x \).
1. **Case 1: \( x \in \left(-\frac{\pi}{2}, 0\right) \)**
Here, \( |x| = -x \), so:
\[
f(x) = x \pi - x \cos^{-1}(\sin(-x)) = x \pi - x \cos^{-1}(-\sin x)
\]
Using the property \( \cos^{-1}(-x) = \pi - \cos^{-1}(x) \):
\[
f(x) = x \pi - x \left( \pi - \cos^{-1}(\sin x) \right) = x \cos^{-1}(\sin x)
\]
2. **Case 2: \( x \in \left(0, \frac{\pi}{2}\right) \)**
Here, \( |x| = x \), so:
\[
f(x) = x \cos^{-1}(\sin x)
\]
### Step 3: Combine Cases
Thus, we can express \( f(x) \) as:
\[
f(x) =
\begin{cases}
x \cos^{-1}(\sin x) & \text{if } x > 0 \\
x \cos^{-1}(\sin(-x)) & \text{if } x < 0
\end{cases}
\]
### Step 4: Differentiate the Function
Now, we will differentiate \( f(x) \).
1. **For \( x \in \left(-\frac{\pi}{2}, 0\right) \)**:
\[
f'(x) = \cos^{-1}(\sin x) + x \cdot \frac{d}{dx}(\cos^{-1}(\sin x))
\]
Using the chain rule:
\[
\frac{d}{dx}(\cos^{-1}(\sin x)) = -\frac{\cos x}{\sqrt{1 - \sin^2 x}} = -\frac{\cos x}{\cos x} = -1
\]
Thus,
\[
f'(x) = \cos^{-1}(\sin x) - x
\]
2. **For \( x \in \left(0, \frac{\pi}{2}\right) \)**:
\[
f'(x) = \cos^{-1}(\sin x) + x \cdot \frac{d}{dx}(\cos^{-1}(\sin x)) = \cos^{-1}(\sin x) - x
\]
### Step 5: Analyze Increasing/Decreasing Behavior
- **For \( x < 0 \)**: \( f'(x) < 0 \) when \( \cos^{-1}(\sin x) < x \) (decreasing).
- **For \( x > 0 \)**: \( f'(x) > 0 \) when \( \cos^{-1}(\sin x) > x \) (increasing).
### Conclusion
The function \( f(x) \) is:
- Decreasing on \( \left(-\frac{\pi}{2}, 0\right) \)
- Increasing on \( \left(0, \frac{\pi}{2}\right) \)
