Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 10

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

Since \begin{aligned} s &=\int_{0}^{\frac{\pi}{4}} \sqrt{1+f^{\prime}(x)^{2}}dx \\ =& \int_{0}^{\frac{\pi}{4}} \sqrt{1+(-\tan x)^{2}}dx \\ &=\int_{0} \frac{\pi}{4} \sqrt{\sec ^{2} x}dx \\ &=\int_{0}^{\frac{\pi}{4}} \sec x dx\\ &=\ln |\sec x+\tan x|\bigg|_{0}^{\frac{\pi}{4}} \\ &=\ln(\sqrt{2}+1) \end{aligned}