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Let f(x)=∫sec(x) dx2cos(2x)−1f(x) = \displaystyle\int \dfrac{\sec(x) \, dx}{2\cos(2x)-1} f(x)=∫2cos(2x)−1sec(x)dx
Given that f(0)=1f(0) = 1f(0)=1, find the value of f(π12)f\left( \dfrac{ \pi }{12} \right)f(12π).
Give your answer to 3 decimal places.
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