Find f.
f ''(θ) = sin(θ) + cos(θ), f(0) = 5, f '(0) = 4

[tex] \text{I will use x instead of}\ \Theta.\\\\f''(x)=\sin x+\cos x\\f'(0)=4\\f(0)=5\\\\f'(x)=\int f''(x)dx\\\\f'(x)=\int(\sin x+\cos x)dx=\int \sin x\ dx+\int \cos x\ dx=-\cos x+\sin x+C\\\\f'(0)=4\to-\cos0+\sin0+C=4\\\\-1+0+C=4\ \ \ \ |+1\\\\C=5\\\\\text{therefore}\ f'(x)=\sin x-\cos x+5\\\\f(x)=\int f'(x)\ dx\\\\f(x)=\int(\sin x-\cos x+5)dx=\int\sin x\ dx-\int\cos x\ dx+\int5\ dx\\\\=-\cos x-\sin x+5x+C\\\\f(0)=5\to -\cos0-\sin0+5\cdot0+C=5\\\\-1-0+0+C=5\ \ \ |+1\\\\C=6 [/tex]

[tex] \text{therefore}\ f(x)=-\cos x-\sin x+5x+6\\\\\\\text{Answer:}\ f(\Theta)=-\cos\Theta-\sin\Theta+5\Theta+6 [/tex]