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Question
Solution
\[\int \text{sin}^2 \left( 2x + 5 \right) \text{dx}\]
\[ = \int\left( \frac{1 - \cos \left( 4x + 10 \right)}{2} \right)dx \left[ \therefore \sin^2 A = \frac{1 - \cos2A}{2} \right]\]
\[ = \frac{1}{2}\int\left( 1 - \cos \left( 4x + 10 \right) \right)dx\]
\[ = \frac{1}{2}\left[ x - \frac{\sin \left( 4x + 10 \right)}{4} \right] + C\]
\[ = \frac{1}{2}x - \frac{\sin \left( 4x + 10 \right)}{8} + C\]
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