Step 1: Understanding the Integral
The integral to evaluate is:
\[ I = \int_0^{\frac{\pi}{2}} \frac{\ln(\cos x)}{1 + \sin^2 x} \, dx \]
This integral involves a logarithmic function and a trigonometric identity, which can be approached using symmetry and properties of definite integrals.

Step 2: Utilizing Symmetry
To analyze this integral, we can use the substitution \( x = \frac{\pi}{2} - t \):
- The limits change: when \( x = 0 \), \( t = \frac{\pi}{2} \) and when \( x = \frac{\pi}{2} \), \( t = 0 \).
- The integral transforms as follows:
\[ I = \int_{\frac{\pi}{2}}^0 \frac{\ln(\cos(\frac{\pi}{2} - t))}{1 + \sin^2(\frac{\pi}{2} - t)} (-dt) \]
This simplifies to:
\[ I = \int_0^{\frac{\pi}{2}} \frac{\ln(\sin t)}{1 + \cos^2 t} \, dt \]

Step 3: Summing Integrals
Now, we can sum the two representations of the integral:
\[ 2I = \int_0^{\frac{\pi}{2}} \left( \frac{\ln(\cos x)}{1 + \sin^2 x} + \frac{\ln(\sin x)}{1 + \cos^2 x} \right) dx \]
This can be combined into a single integral:
\[ 2I = \int_0^{\frac{\pi}{2}} \frac{\ln(\sin x \cos x)}{1 + \sin^2 x + \cos^2 x} \, dx \]
Since \( \sin^2 x + \cos^2 x = 1 \):
\[ 2I = \int_0^{\frac{\pi}{2}} \frac{\ln(\frac{1}{2} \sin(2x))}{2} \, dx \]

Step 4: Final Result
Evaluating this integral using properties of logarithms and symmetry yields:
\[ I = -\frac{\pi}{4} \ln(2) \]
Thus, the integral \( \int_0^{\frac{\pi}{2}} \frac{\ln(\cos x)}{1 + \sin^2 x} \, dx \) evaluates to:
\[ I = -\frac{\pi}{4} \ln(2) \]