Calculus (3rd Edition)

$-\cos (\theta-8) +c$
Let $u=\theta-8$, then $du=d\theta$ and hence $$\int \sin(\theta-8) d\theta=\int \sin u du=-\cos u +c= -\cos (\theta-8) +c$$ Where we used the fact that the integral of sine is negative cosine.