antiderivative?

Question

f ''(x) = 8x + 6 sin(x)

f'(0)= 4
f(0)= 3

What is f(5)?

( I am having trouble with the sin5/cos5 part)

Answer 1

Integrating f ''(x) = 8x + 6 sin(x) once gives
f '(x) = 8x^2/2 - 6 cos(x) + c.
Since f '(0) =4 we can work out the constant of integration c as
c = 4 - 0 +6cos(0) = 10.
Integrating again gives
f(x) = 4x^3/3 - 6 sin(x) +10x + d
Since f(0) = 3 we can work out d as
d = 3 - 0 + 6sin(0) -10(0) = 3.
So
f(x) = 4/3x^3 - 6sin(x) +10x +3
and f(5) = 4/3*5^3 - 6sin(5) +10(5) +3
= 4/3*125 +53 - 6sin(5).
I think it's OK to leave the answer as that.

Answer 2

f´(x) = 4x^2-6 cos x +C1
f(x) = 4/3x^3 -6sinx +C1 x +C2
f(0) = C2 =3
f´(0) =4=- 6+C1 so C1= 10
f(x) = 4/3 x^3-6sinx+10x+3
f(5) =500/3 -6sin 5 +50 +3 // sin (5) =-0.9589