Answer
$$\frac{1}{5}\tan 5x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^2}5x} dx \cr
& {\text{set }}u = 5x{\text{ then }}\frac{{du}}{{dx}} = 5,\,\,\,\,\,\,\,\,\frac{{du}}{5} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{{\sec }^2}5x} dx = \int {{{\sec }^2}u} \left( {\frac{{du}}{5}} \right) \cr
& = \frac{1}{5}\int {{{\sec }^2}u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {{{\sec }^2}x} dx = \tan x + C \cr
& = \frac{1}{5}\tan u + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{5}\tan 5x + C \cr} $$
