find the exact length of the curve.\n$y = \\ln(\\sec(x))$, $0 \\leq x \\leq \\pi/4$

Explanation:

Step1: Find the derivative of (y)

The derivative of (y = \ln(\sec(x))) using the chain - rule. If (y=\ln(u)) and (u = \sec(x)), then (y^\prime=\frac{u^\prime}{u}). Since (u^\prime=\sec(x)\tan(x)), we have (y^\prime=\frac{\sec(x)\tan(x)}{\sec(x)}=\tan(x))

Step2: Use the arc - length formula

The arc - length formula for (y = f(x)) from (x = a) to (x = b) is (L=\int_{a}^{b}\sqrt{1+(y^\prime)^2}dx). Substitute (y^\prime=\tan(x)) into the formula: (L=\int_{0}^{\frac{\pi}{4}}\sqrt{1 + \tan^{2}(x)}dx) Using the trigonometric identity (1+\tan^{2}(x)=\sec^{2}(x)), the integral becomes (L=\int_{0}^{\frac{\pi}{4}}\sqrt{\sec^{2}(x)}dx=\int_{0}^{\frac{\pi}{4}}\sec(x)dx)

Step3: Evaluate the integral

The integral of (\sec(x)) is (\ln|\sec(x)+\tan(x)|). Evaluate (\left[\ln(\sec(x)+\tan(x))\right]_{0}^{\frac{\pi}{4}}) When (x=\frac{\pi}{4}), (\sec(\frac{\pi}{4})=\sqrt{2}), (\tan(\frac{\pi}{4}) = 1), so (\ln(\sqrt{2}+1)) When (x = 0), (\sec(0)=1), (\tan(0)=0), so (\ln(1 + 0)=0) Then (L=\ln(\sqrt{2}+1)-\ln(1)=\ln(1 + \sqrt{2}))

Answer:

(\ln(1+\sqrt{2}))