text: ## Thomas' Calculus 13th Edition
  
  $$\pi$$
  \eqalign{ & \int_{ - \pi }^\pi {\sin 3x} \sin 3xdx \cr & {\text{Use the identity }}\sin mx\sin xnx = \frac{1}{2}\left[ {\cos \left( {m - n} \right)x - \cos \left( {m + n} \right)x} \right]\,\,\left( {{\text{see page 463}}} \right) \cr & \int_{ - \pi }^\pi {\sin 3x} \sin 3xdx = \int_{ - \pi }^\pi {\frac{1}{2}\left[ {\cos \left( {3 - 3} \right)x - \cos \left( {3 + 3} \right)x} \right]dx} \cr & = \int_{ - \pi }^\pi {\frac{1}{2}\left[ {1 - \cos 6x} \right]dx} \cr & = \frac{1}{2}\int_{ - \pi }^\pi {\left( {1 - \cos 6x} \right)dx} \cr & {\text{integrating}} \cr & = \frac{1}{2}\left( {x - \frac{1}{6}\sin 6x} \right)_{ - \pi }^\pi \cr & = \frac{1}{2}\left( {\pi - \frac{1}{6}\sin 6\pi } \right) - \frac{1}{2}\left( { - \pi - \frac{1}{6}\sin 6\left( { - \pi } \right)} \right) \cr & {\text{simplifying}} \cr & = \frac{1}{2}\left( {\pi - 0} \right) - \frac{1}{2}\left( { - \pi - 0} \right) \cr & = \frac{1}{2}\pi + \frac{1}{2}\pi \cr & = \pi \cr}
explanation: The document contains a detailed solution to a calculus problem involving the evaluation of a definite integral. The problem is to evaluate the integral \(\int_{ - \pi }^\pi {\sin 3x} \sin 3x \, dx\). The solution uses a trigonometric identity and step-by-step integration to arrive at the final answer, which is \(\pi\).
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: ## Thomas' Calculus 13th Edition
  
  $$\pi$$
  \eqalign{ & \int_{ - \pi }^\pi {\sin 3x} \sin 3xdx \cr & {\text{Use the identity }}\sin mx\sin xnx = \frac{1}{2}\left[ {\cos \left( {m - n} \right)x - \cos \left( {m + n} \right)x} \right]\,\,\left( {{\text{see page 463}}} \right) \cr & \int_{ - \pi }^\pi {\sin 3x} \sin 3xdx = \int_{ - \pi }^\pi {\frac{1}{2}\left[ {\cos \left( {3 - 3} \right)x - \cos \left( {3 + 3} \right)x} \right]dx} \cr & = \int_{ - \pi }^\pi {\frac{1}{2}\left[ {1 - \cos 6x} \right]dx} \cr & = \frac{1}{2}\int_{ - \pi }^\pi {\left( {1 - \cos 6x} \right)dx} \cr & {\text{integrating}} \cr & = \frac{1}{2}\left( {x - \frac{1}{6}\sin 6x} \right)_{ - \pi }^\pi \cr & = \frac{1}{2}\left( {\pi - \frac{1}{6}\sin 6\pi } \right) - \frac{1}{2}\left( { - \pi - \frac{1}{6}\sin 6\left( { - \pi } \right)} \right) \cr & {\text{simplifying}} \cr & = \frac{1}{2}\left( {\pi - 0} \right) - \frac{1}{2}\left( { - \pi - 0} \right) \cr & = \frac{1}{2}\pi + \frac{1}{2}\pi \cr & = \pi \cr}
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a detailed solution to a calculus problem involving the evaluation of a definite integral. The problem is to evaluate the integral \(\int_{ - \pi }^\pi {\sin 3x} \sin 3x \, dx\). The solution uses a trigonometric identity and step-by-step integration to arrive at the final answer, which is \(\pi\).
  
  Label: question-and-answer