Answer
$${\text{part a) }}\frac{1}{2}\sin 2x + C{\text{ part b)}}\frac{1}{2}\tan {x^2} + C$$
Work Step by Step
$$\eqalign{
& {\text{part a}} \cr
& \cr
& \int {\cos 2} xdx \cr
& {\text{substitute }}u = 2x,{\text{ }}du = 2dx \cr
& \int {\cos 2} xdx = \int {\cos u} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int {\cos u} du \cr
& {\text{find antiderivative}} \cr
& = \frac{1}{2}\sin u + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{2}\sin 2x + C \cr
& \cr
& {\text{part b}} \cr
& \int {x{{\sec }^2}{x^2}} dx \cr
& {\text{substitute }}u = {x^2},{\text{ }}du = 2xdx \cr
& \int {x{{\sec }^2}{x^2}} dx = \int {{u^2}} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int {{{\sec }^2}u} du \cr
& {\text{find antiderivative}} \cr
& = \frac{1}{2}\tan u + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{2}\tan {x^2} + C \cr} $$
