Answer
$$s\left( t \right) = - 2\cos t + t + 2$$
Work Step by Step
$$\eqalign{
& a\left( t \right) = 2\cos t;\,\,\,v\left( 0 \right) = 1,\,\,\,\,s\left( 0 \right) = 0 \cr
& v'\left( t \right) = a\left( t \right),{\text{ then}} \cr
& v\left( t \right) = \int {a\left( t \right)dt} \cr
& v\left( t \right) = \int {2\cos tdt} \cr
& v\left( t \right) = 2\sin t + C \cr
& {\text{Use the initial condition }}v\left( 0 \right) = 1 \cr
& 1 = 2\sin \left( 0 \right) + C \cr
& C = 1 \cr
& {\text{Thus}}{\text{, }} \cr
& v\left( t \right) = 2\sin t + 1 \cr
& s'\left( t \right) = v\left( t \right),{\text{ then}} \cr
& s\left( t \right) = \int {v\left( t \right)dt} \cr
& s\left( t \right) = \int {\left( {2\sin t + 1} \right)dt} \cr
& s\left( t \right) = - 2\cos t + t + C \cr
& {\text{Use the initial condition }}s\left( 0 \right) = 0 \cr
& 0 = - 2\cos \left( 0 \right) + 0 + C \cr
& C = 2 \cr
& {\text{Thus}}{\text{, }} \cr
& s\left( t \right) = - 2\cos t + t + 2 \cr} $$