## 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}
The text passage contains a calculus problem involving an integral and its solution. The problem and its solution are presented step-by-step.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## 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}
user
The text passage contains a calculus problem involving an integral and its solution. The problem and its solution are presented step-by-step.

TYPE: SINGLE_QUESTION_WITH_ANSWER