Answer:
y=tanx+C
Step-by-step explanation:
[tex] \frac{dy}{dx} = {sec}^{2} \: x \\ \\ dy = {sec}^{2} \: x \: dx \\ integrating \: both \: sides \\ \\ \int \:1\: dy = \int {sec}^{2} \: x \: dx \\ \\ \huge \red{ \boxed{ \therefore \: y = tan \: x + c}} \\ [/tex]