The integral \( I = \int_0^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} \, dx \) is to be evaluated. Employing symmetry via the substitution \( x = \frac{\pi}{2} - t \) yields \( dx = -dt \), \( \sin\left(\frac{\pi}{2} - t\right) = \cos t \), and \( \cos\left(\frac{\pi}{2} - t\right) = \sin t \). The integral transforms to \( I = \int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2} - t}{\cos t + \sin t} \, dt \). Rewriting this, we get \( I = \int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2}}{\sin x + \cos x} \, dx - \int_0^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} \, dx \), which simplifies to \( I = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx - I \). Solving for \( I \) gives \( 2I = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx \). Given that \( \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx = \sqrt{2} \), we have \( 2I = \frac{\pi}{2} \times \sqrt{2} \), resulting in \( I = \frac{\pi \sqrt{2}}{4} \).