text: ## Calculus (3rd Edition)
  
  $$\frac{1}{3} \cos (8-3 \theta)+C$$
  Given $$\int \sin (8-3 \theta) d \theta$$ Let $$u=8-3 \theta\ \ \ \Rightarrow \ \ \ du=-3d \theta$$ Then \begin{aligned} \int \sin (8-3 \theta) d \theta &=-\frac{1}{3} \int \sin u d u \\ &=-\frac{1}{3}(-\cos u+C) \\ &=\frac{1}{3} \cos u+C\\ &= \frac{1}{3} \cos (8-3 \theta)+C \end{aligned}