To show that
\[
\int_{0}^{\frac{\pi}{2}} \left( \sin x - \cos x \right) \log \left( \sin x + \cos x \right) \, dx = 0,
\]
we can use the property of definite integrals that states:
\[
\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx.
\]
In our case, we have \( a = 0 \) and \( b = \frac{\pi}{2} \), so we can rewrite the integral as follows:
1. **Define the integral:**
Let
\[
I = \int_{0}^{\frac{\pi}{2}} \left( \sin x - \cos x \right) \log \left( \sin x + \cos x \right) \, dx.
\]
2. **Apply the property of definite integrals:**
Using the property mentioned, we can find \( I \) as follows:
\[
I = \int_{0}^{\frac{\pi}{2}} \left( \sin \left( \frac{\pi}{2} - x \right) - \cos \left( \frac{\pi}{2} - x \right) \right) \log \left( \sin \left( \frac{\pi}{2} - x \right) + \cos \left( \frac{\pi}{2} - x \right) \right) \, dx.
\]
3. **Simplify the expressions:**
We know that:
\[
\sin \left( \frac{\pi}{2} - x \right) = \cos x \quad \text{and} \quad \cos \left( \frac{\pi}{2} - x \right) = \sin x.
\]
Therefore, we can rewrite the integral:
\[
I = \int_{0}^{\frac{\pi}{2}} \left( \cos x - \sin x \right) \log \left( \cos x + \sin x \right) \, dx.
\]
4. **Combine the two integrals:**
Now we have two expressions for \( I \):
\[
I = \int_{0}^{\frac{\pi}{2}} \left( \sin x - \cos x \right) \log \left( \sin x + \cos x \right) \, dx,
\]
and
\[
I = \int_{0}^{\frac{\pi}{2}} \left( \cos x - \sin x \right) \log \left( \cos x + \sin x \right) \, dx.
\]
5. **Add both expressions:**
Adding both expressions for \( I \):
\[
2I = \int_{0}^{\frac{\pi}{2}} \left( \left( \sin x - \cos x \right) + \left( \cos x - \sin x \right) \right) \log \left( \sin x + \cos x \right) \, dx.
\]
Notice that the terms \(\sin x\) and \(-\sin x\) cancel out, as do \(\cos x\) and \(-\cos x\):
\[
2I = \int_{0}^{\frac{\pi}{2}} 0 \cdot \log \left( \sin x + \cos x \right) \, dx = 0.
\]
6. **Conclusion:**
Therefore, we have:
\[
2I = 0 \implies I = 0.
\]
Thus, we conclude that:
\[
\int_{0}^{\frac{\pi}{2}} \left( \sin x - \cos x \right) \log \left( \sin x + \cos x \right) \, dx = 0.
\]
