Answer
$f(\theta)=-\sin\theta-\cos\theta+5\theta+4$
Work Step by Step
$ f''(t)=\sin\theta+\cos\theta$
Using the antiderivatives table,
$f'(\theta)=-\cos\theta+\sin\theta+C$
Given that $f'(0)=4$ it follows that
$4=-1+0+C$
$C=5,$
$f'(\theta)=-\cos\theta+\sin\theta+5$
Using the antiderivatives table,
$f(\theta)=-\sin\theta-\cos\theta+5\theta+D$
Given that $f(0)=3$ it follows that
$3=-0-1+0+D$
$D=4,$
$f(\theta)=-\sin\theta-\cos\theta+5\theta+4$