Compute the arc length of the curve y= ln cos x between the points with the abscissas `x=0, x= (pi)/(4)`

Answer

To compute the arc length of the curve \( y = \ln(\cos x) \) between the points \( x = 0 \) and \( x = \frac{\pi}{4} \), we will use the formula for arc length given by:

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

### Step 1: Find \(\frac{dy}{dx}\)

Given \( y = \ln(\cos x) \), we need to differentiate it with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}(\ln(\cos x)) = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \]

### Step 2: Substitute \(\frac{dy}{dx}\) into the arc length formula

Now we substitute \(\frac{dy}{dx}\) into the arc length formula:

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

This simplifies to:

\[ L = \int_0^{\frac{\pi}{4}} \sqrt{1 + \tan^2 x} \, dx \]

### Step 3: Use the identity \(1 + \tan^2 x = \sec^2 x\)

Using the trigonometric identity \(1 + \tan^2 x = \sec^2 x\):

\[ L = \int_0^{\frac{\pi}{4}} \sqrt{\sec^2 x} \, dx = \int_0^{\frac{\pi}{4}} \sec x \, dx \]

### Step 4: Compute the integral of \(\sec x\)

The integral of \(\sec x\) can be computed as follows:

\[ \int \sec x \, dx = \ln |\sec x + \tan x| + C \]

Thus, we have:

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

### Step 5: Evaluate the definite integral

Now we evaluate the limits:

1. At \( x = \frac{\pi}{4} \):
\[ \sec\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \tan\left(\frac{\pi}{4}\right) = 1 \]
Therefore,
\[ \sec\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right) = \sqrt{2} + 1 \]

2. At \( x = 0 \):
\[ \sec(0) = 1, \quad \tan(0) = 0 \]
Therefore,
\[ \sec(0) + \tan(0) = 1 + 0 = 1 \]

Putting it all together:

\[ L = \ln(\sqrt{2} + 1) - \ln(1) = \ln(\sqrt{2} + 1) \]

### Final Answer

Thus, the arc length of the curve \( y = \ln(\cos x) \) between \( x = 0 \) and \( x = \frac{\pi}{4} \) is:

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