## 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}
The text passage contains a detailed solution to an integral problem from Thomas' Calculus 13th Edition, showing the steps to evaluate the integral of \(\sin 3x \sin 3x\) from \(-\pi\) to \(\pi\).
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
## 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}
user
The text passage contains a detailed solution to an integral problem from Thomas' Calculus 13th Edition, showing the steps to evaluate the integral of \(\sin 3x \sin 3x\) from \(-\pi\) to \(\pi\).

TYPE: SINGLE_QUESTION_WITH_ANSWER