To solve the integral
\[
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x \, dx}{1 + 2 \lfloor \sin^{-1}(\sin x) \rfloor}
\]
we start by simplifying the expression inside the integral.
### Step 1: Simplifying \(\sin^{-1}(\sin x)\)
The function \(\sin^{-1}(\sin x)\) returns \(x\) for \(x\) in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\). Therefore, we can write:
\[
\lfloor \sin^{-1}(\sin x) \rfloor = \lfloor x \rfloor
\]
### Step 2: Determine the value of \(\lfloor x \rfloor\)
The greatest integer function \(\lfloor x \rfloor\) takes different values depending on the interval of \(x\):
- For \(x \in [-\frac{\pi}{2}, -1)\), \(\lfloor x \rfloor = -2\)
- For \(x \in [-1, 0)\), \(\lfloor x \rfloor = -1\)
- For \(x \in [0, 1)\), \(\lfloor x \rfloor = 0\)
- For \(x \in [1, \frac{\pi}{2}]\), \(\lfloor x \rfloor = 1\)
### Step 3: Break the integral into segments
We can split the integral into four parts based on the intervals identified:
\[
I = \int_{-\frac{\pi}{2}}^{-1} \frac{\cos x \, dx}{1 + 2(-2)} + \int_{-1}^{0} \frac{\cos x \, dx}{1 + 2(-1)} + \int_{0}^{1} \frac{\cos x \, dx}{1 + 2(0)} + \int_{1}^{\frac{\pi}{2}} \frac{\cos x \, dx}{1 + 2(1)}
\]
This simplifies to:
\[
I = \int_{-\frac{\pi}{2}}^{-1} \frac{\cos x \, dx}{-3} + \int_{-1}^{0} \frac{\cos x \, dx}{-1} + \int_{0}^{1} \cos x \, dx + \int_{1}^{\frac{\pi}{2}} \frac{\cos x \, dx}{3}
\]
### Step 4: Evaluate each integral
1. **First Integral:**
\[
\int_{-\frac{\pi}{2}}^{-1} \frac{\cos x \, dx}{-3} = -\frac{1}{3} \left[ \sin x \right]_{-\frac{\pi}{2}}^{-1} = -\frac{1}{3} (\sin(-1) - \sin(-\frac{\pi}{2})) = -\frac{1}{3} (\sin(-1) + 1)
\]
2. **Second Integral:**
\[
\int_{-1}^{0} \frac{\cos x \, dx}{-1} = -\left[ \sin x \right]_{-1}^{0} = -(\sin(0) - \sin(-1)) = \sin(-1)
\]
3. **Third Integral:**
\[
\int_{0}^{1} \cos x \, dx = [\sin x]_{0}^{1} = \sin(1) - \sin(0) = \sin(1)
\]
4. **Fourth Integral:**
\[
\int_{1}^{\frac{\pi}{2}} \frac{\cos x \, dx}{3} = \frac{1}{3} \left[ \sin x \right]_{1}^{\frac{\pi}{2}} = \frac{1}{3} (\sin(\frac{\pi}{2}) - \sin(1)) = \frac{1}{3} (1 - \sin(1))
\]
### Step 5: Combine all parts
Now, we combine all the parts:
\[
I = -\frac{1}{3} (\sin(-1) + 1) + \sin(-1) + \sin(1) + \frac{1}{3} (1 - \sin(1))
\]
### Step 6: Simplify the expression
Combining the terms, we notice that \(\sin(-1) = -\sin(1)\):
\[
I = -\frac{1}{3} (-\sin(1) + 1) + (-\sin(1)) + \sin(1) + \frac{1}{3} (1 - \sin(1))
\]
This simplifies to:
\[
I = 0
\]
### Final Result
Thus, the value of the integral \(I\) is:
\[
\boxed{0}
\]
