Answer
$$ \frac{1}{3} \cos (8-3 \theta)+C$$
Work Step by Step
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}