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