To prove that

$$\int_{0}^{\pi/4} \ln(\cos(x)) \, dx = \int_{0}^{\pi/4} \ln(\cos(\pi/4 - x)) \, dx,$$

let's use the substitution method and symmetry properties of trigonometric functions.

Step 1: Substitution in the second integral

Let's define $u = \frac{\pi}{4} - x$. Then,
$$du = -dx$$

When

• $x = 0 \implies u = \frac{\pi}{4}$,
• $x = \frac{\pi}{4} \implies u = 0$,

this reverses the limits of integration. The second integral becomes:
$$\int_{0}^{\pi/4} \ln(\cos(\pi/4 - x)) \, dx = \int_{\pi/4}^{0} \ln(\cos(u)) \, (-du)$$

Reversing the limits to restore the order of integration introduces a negative sign:
$$\int_{\pi/4}^{0} \ln(\cos(u)) \, (-du) = -\int_{\pi/4}^{0} \ln(\cos(u)) \, du = \int_{0}^{\pi/4} \ln(\cos(u)) \, du$$

then:

$$\int_{0}^{\pi/4} \ln(\cos(\pi/4 - x)) \, dx = \int_{0}^{\pi/4} \ln(\cos(u)) \, du$$

Step 2: Rename the variable

Since the variable of integration is a dummy variable, we can replace $u$ with $x$ in the final expression:
$$\int_{0}^{\pi/4} \ln(\cos(\pi/4 - x)) \, dx = \int_{0}^{\pi/4} \ln(\cos(x)) \, dx$$

Conclusion

We have shown that:
$$\int_{0}^{\pi/4} \ln(\cos(x)) \, dx = \int_{0}^{\pi/4} \ln(\cos(\pi/4 - x)) \, dx$$