find the exact length of the curve y=ln(sec(x)) between x=0 and x=pi/4

Question

Find the exact length of the curve y=ln(sec(x)) between 0 and pi/4.

First do I find dy/dr?

Answer 1

Arc length is $\int_{a}^{b} \sqrt{1+(y')^2}\text{d}x$ so $\frac{\text{d}}{\text{d}x}\ln \sec x=\frac{\sec x \tan x}{\sec x}=\tan x$ so the arc length from 0 to $\frac{\pi}{4}$ is $\int_{0}^{\frac{\pi}{4}} \sqrt{1+\tan^2 x}\text{d}x=\int_{0}^{\frac{\pi}{4}} \sec x\text{d}x=$ $$\ln\left | \sec x +\tan x \right |\rvert_{0}^{\frac{\pi}{4}}=\ln (\sqrt 2+1)$$