f ''(θ) = sin θ + cos θ, f(0) = 5, f '(0) = 3
When f'(0)=3, C=-2, so f'(θ)=cosθ-sinθ-2.
When f(0)=5, D=6, so f is -sinθ-cosθ-2x+6.
How is that wrong?
f'Ø) = -cosx + sinx + C
you differentiated instead of integrated.
Had you checked your answer by taking the derivative you would have seen your error
since f'(0) = 3
3 = -cos0 + sin0 + c
3 = -1 + 0 + c ----> c = 4
so f'(x) = -cosx + sinx + 4
f(x) = -sinx - cosx + 4x + d
and f(0) = 5
5 = -sin0 - cos0 + 0 + d
5 = 0 - 1 + d
d = 6
f(x) = -sinx - cosx + 4x + 6
Just realized I used x instead of Ø, no big deal to change
I suggest you take a better look at the derivatives of both sine and cosine
posted by Reiny