To find the length of the curve $y = \ln(\cos(x))$ for $0 \leq x \leq \frac{\pi}{4}$, we use the formula for the arc length of a curve:

$L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$

Step 1: Compute $\frac{dy}{dx}$

Given $y = \ln(\cos(x))$, we apply the chain rule to differentiate:

$\frac{dy}{dx} = \frac{1}{\cos(x)} \cdot (-\sin(x)) = -\tan(x)$

Step 2: Set up the arc length formula

The arc length formula becomes:

$L = \int_0^{\frac{\pi}{4}} \sqrt{1 + (-\tan(x))^2} \, dx$

Since $\tan^2(x) + 1 = \sec^2(x)$, we simplify the integrand:

$L = \int_0^{\frac{\pi}{4}} \sqrt{\sec^2(x)} \, dx = \int_0^{\frac{\pi}{4}} \sec(x) \, dx$

Step 3: Integrate

The integral of $\sec(x)$ is known:

$\int \sec(x) \, dx = \ln|\sec(x) + \tan(x)|$

Thus, the arc length is:

$L = \left[ \ln|\sec(x) + \tan(x)| \right]_0^{\frac{\pi}{4}}$

Step 4: Evaluate the bounds

At $x = \frac{\pi}{4}$:

$\sec\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \tan\left(\frac{\pi}{4}\right) = 1$

So,

$\ln|\sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right)| = \ln(\sqrt{2} + 1)$

At $x = 0$:

$\sec(0) = 1, \quad \tan(0) = 0$

So,

$\ln|\sec(0) + \tan(0)| = \ln(1) = 0$

Step 5: Final answer

The length of the curve is:

$L = \ln(\sqrt{2} + 1)$

This is the exact length of the curve.

---

Would you like more details or have any further questions? Here are 5 related questions to explore further:

1. What is the geometric interpretation of the curve $y = \ln(\cos(x))$?
2. How do we derive the formula for arc length in calculus?
3. What are some real-world applications of curve length calculations?
4. How does the chain rule work when differentiating logarithmic functions?
5. Can we approximate the length of a curve using numerical methods?

Tip: In problems involving arc length, always remember to simplify the square root expressions as much as possible before integrating!