Answer
$$f\left( x \right) = \sin 2x + 1$$
Work Step by Step
$$\eqalign{
& f'\left( x \right) = 2\cos 2x;{\text{ }}f\left( 0 \right) = 1 \cr
& {\text{Calculating the general solution}} \cr
& f\left( x \right) = \int {f'\left( x \right)} dx \cr
& f\left( x \right) = \int {2\cos 2x} dx \cr
& f\left( x \right) = \sin 2x + C \cr
& {\text{Calculating the particular solution for }}f\left( 0 \right) = 1 \cr
& 1 = \sin 2\left( 0 \right) + C \cr
& C = 1 \cr
& {\text{The particular solution is}} \cr
& f\left( x \right) = \sin 2x + 1 \cr
& {\text{Graphing general solutions for }}C = - 4,{\text{ 1, 1 and the particular}} \cr
& {\text{solution }}f\left( x \right) = \sin 2x + 1 \cr} $$
