Find the arc length of the curve y = ln(cos x) between

Find the arc length of the curve $y = \ln(\cos x)$ between points with abscissas $x = 0$ and $x = \frac{\pi}{4}$.

We reduce the arc-length integrand to $\sec x$ via a standard trigonometric identity.

Step 1: Differentiate

\[\frac{dy}{dx} = \frac{-\sin x}{\cos x} = -\tan x\]

Step 2: Simplify using $1+\tan^2x=\sec^2x$

\[1+\left(\frac{dy}{dx}\right)^2 = 1+\tan^2x = \sec^2x\]

Since $\cos x>0$ on $\left[0,\frac\pi4\right]$, $\sec x>0$, so $\sqrt{\sec^2x}=\sec x$.

Step 3: Set up and evaluate the integral

\[L = \int_0^{\pi/4}\sec x\,dx = \left[\ln|\sec x+\tan x|\right]_0^{\pi/4}\]

At $x=\frac\pi4$: $\sec\frac\pi4=\sqrt2$, $\tan\frac\pi4=1$, giving $\ln(\sqrt2+1)$. At $x=0$: $\ln(1+0)=0$.

\[L = \ln(\sqrt2+1)\]

\[\boxed{L = \ln(\sqrt2+1) \text{ units}}\]

Step 4: Numerical value

\[L = \ln(\sqrt2+1) \approx 0.881 \text{ units}\]

Summary

\[\boxed{L = \ln(\sqrt2+1) \approx 0.881 \text{ units}}\]

Key insight: The identity $1+\tan^2x=\sec^2x$ is exactly what turns the arc-length square root into the bare secant function, whose antiderivative $\ln|\sec x+\tan x|$ is a standard (if not immediately obvious) result — here producing the compact constant $\ln(\sqrt2+1)$.