Answer
$\sin \left( {\tan y} \right) - \frac{1}{3}{\sin ^3}\left( {\tan y} \right) + C$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^2}y{{\cos }^3}\left( {\tan y} \right)} dy \cr
& {\text{Split }}{\cos ^3}\left( {\tan y} \right){\text{ as }}{\cos ^2}\left( {\tan y} \right)\cos \left( {\tan y} \right) \cr
& = \int {{{\sec }^2}y{{\cos }^2}\left( {\tan y} \right)\cos \left( {\tan y} \right)} dy \cr
& {\text{Rewrite }}{\sin ^2}x{\text{ using }}{\sin ^2}x + {\cos ^2}x = 1 \cr
& = \int {{{\sec }^2}y\left[ {1 - {{\sin }^2}\left( {\tan y} \right)} \right]\cos \left( {\tan y} \right)} dy \cr
& = \int {\left[ {1 - {{\sin }^2}\left( {\tan y} \right)} \right]\cos \left( {\tan y} \right){{\sec }^2}y} dy \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = \sin \left( {\tan y} \right),{\text{ }}du = \cos \left( {\tan y} \right){\sec ^2}ydy \cr
& {\text{Substituting}} \cr
& = \int {\left( {1 - {u^2}} \right)} du \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{1}{{n + 1}}{u^{n + 1}} + C \cr
& = u - \frac{1}{3}{u^3} + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}\sin \left( {\tan y} \right){\text{ for }}u \cr
& = \sin \left( {\tan y} \right) - \frac{1}{3}{\sin ^3}\left( {\tan y} \right) + C \cr} $$
