Please help me find f(x) of f '(x) = 2t - 3sint f(0) = 5

I got f(t) = t² - 3cost + C which is correct but when I'm trying to find C I keep getting 8 when C is suppose to be 2. Please show me how you get to.
the answer is f(t) = t² - 3cost + 2

f(t) = t² + 3*cos(t) + C

Plugging in initial condition:

5 = 3 + C

C = 2

f(t) = t² + 3*cos(t) + 2