Calculus (3rd Edition)

$$y= \tan x +1$$
Given$$\frac{d y}{d x}=\sec^2 x, \quad y(\pi/4)=2$$ Since :\begin{align*} y&= \int (\sec^2 x) dx\\ &= \tan x +c \end{align*} at $x=\pi/4,\ \ \ y= 2$, then $c= 1$, hence $$y= \tan x +1$$