Answer
$${\sin ^{ - 1}}\left( {\tan y} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sec }^2}ydy}}{{\sqrt {1 - {{\tan }^2}y} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = \tan y,{\text{ so that }}du = {\sec ^2}ydy \cr
& {\text{then}} \cr
& \int {\frac{{{{\sec }^2}ydy}}{{\sqrt {1 - {{\tan }^2}y} }}} = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = {\sin ^{ - 1}}\left( {\frac{u}{1}} \right) + C \cr
& {\text{write in terms of }}y;{\text{ replace }}\tan y{\text{ for }}u \cr
& = {\sin ^{ - 1}}\left( {\tan y} \right) + C \cr} $$
