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Question
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
Sum
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Solution
\[\int \sin^3 \left( 2x + 1 \right)dx\]
\[ = \frac{1}{4}\int\left[ 3 \sin \left( 2x + 1 \right) - \sin \left( 3\left( 2x + 1 \right) \right) \right]dx \left[ \therefore \sin \left( 3\theta \right) = 3 \sin\theta - 4 \sin^3 \theta \Rightarrow \sin^3 \theta = \frac{1}{4}\left( 3\sin \theta - \sin \left( 3\theta \right) \right) \right] \]
\[ = \frac{3}{4}\int\sin \left( 2x + 1 \right)dx - \frac{1}{4}\int\sin \left( 6x + 3 \right)dx\]
\[ = \frac{3}{4}\left[ - \frac{\cos \left( 2x + 1 \right)}{2} \right] - \frac{1}{4}\left[ - \frac{\cos \left( 6x + 3 \right)}{6} \right] + C\]
\[ = \frac{- 3}{8}\cos \left( 2x + 1 \right) + \frac{1}{24} \cos \left( 6x + 3 \right) + C\]
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