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Question
\[\int \sin^2\text{ b x dx}\]
Sum
Solution
\[\int \sin^2\text{ bx dx}\]
\[ = \int\left[ \frac{1 - \cos 2bx}{2} \right]dx \left[ \therefore \sin^2 x = \frac{1 - \cos 2x}{2} \right]\]
\[ = \frac{1}{2}\int\left( 1 - \cos 2bx \right)dx\]
\[ = \frac{1}{2}\left[ x - \frac{\sin 2bx}{2b} \right] + C\]
\[ = \frac{x}{2} - \frac{\sin 2bx}{4b} + C\]
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