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}
explanation: The document contains a calculus problem and its solution. It starts with the given integral \(\int \sin (8-3 \theta) d \theta\) and proceeds to solve it using substitution, resulting in the final answer \(\frac{1}{3} \cos (8-3 \theta) + C\).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## 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}
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a calculus problem and its solution. It starts with the given integral \(\int \sin (8-3 \theta) d \theta\) and proceeds to solve it using substitution, resulting in the final answer \(\frac{1}{3} \cos (8-3 \theta) + C\).
  
  Label: question-and-answer