Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 2

$$\int_0^{\pi/4}\sqrt{1+\sec^4}dx.$$

To find the arc length $s$ of the curve $y= \tan x$ between $x=0$ and $x=\pi/4$, we fist calculate $y'=\sec^2 x$. Then we have the integral, $$s=\int_0^{\pi/4}\sqrt{1+(y')^2}dx=\int_0^{\pi/4}\sqrt{1+\sec^4}dx.$$