Answer
$s(t) = -sin~t-3~cos~t+3t+3$
Work Step by Step
$s(t)$ is the position of the particle at time $t$
$\frac{ds}{dt} = v(t)$ is the velocity of the particle at time $t$
$\frac{dv}{dt} = a(t)$ is the acceleration of the particle at time $t$
$a(t) = sin~t+3~cos~t$
$v(t) = \int (sin~t+3~cos~t)~dt = -cos~t+3~sin~t+C_1$
Note that $v(0) = 2$
We can find the value of $C_1$:
$v(0) = -cos~(0)+3~sin~(0)+C_1 = 2$
$-1+0+C_1 = 2$
$C_1 = 3$
Then:
$v(t) = -cos~t+3~sin~t+3$
$s(t) = \int (-cos~t+3~sin~t+3)~dt = -sin~t-3~cos~t+3t+C_2$
Note that $s(0) = 0$
We can find the value of $C_2$:
$s(0) = -sin~(0)-3~cos~(0)+3(0)+C_2 = 0$
$0-3+0+C_2 = 0$
$C_2 = 3$
Therefore:
$s(t) = -sin~t-3~cos~t+3t+3$
