Calculus: Early Transcendentals, Chapter 4, 4.9, Section 4.9, Problem 63

a(t)=10sin(t)+3cos(t)
a(t)=v'(t)
v(t)=inta(t)dt
v(t)=int(10sin(t)+3cos(t))dt
v(t)=-10cos(t)+3sin(t)+c_1
v(t)=s'(t)
s(t)=intv(t)dt
s(t)=int(-10cos(t)+3sin(t)+c_1)dt
s(t)=-10sin(t)+3(-cos(t))+c_1t+c_2
s(t)=-10sin(t)-3cos(t)+c_1t+c_2
Let's find constants c_1 and c_2 , given s(0)=0 ans s(2pi)=12
s(0)=0=-10sin(0)-3cos(0)+c_1(0)+c_2
0=-3+c_2
c_2=3
s(2pi)=12=-10sin(2pi)-3cos(2pi)+c_1(2pi)+3
12=-10(0)-3(1)+2pic_1+3
12=2pic_1
c_1=6/pi
:. position of the particle is given by s(t)=-10sin(t)-3cos(t)+(6t)/pi+3