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Determine a parameter from two exact integrals

Find the exact value of pp such that ∫01dx4−x2=∫01/(2p)dx1−p2x2\displaystyle\int_0^1\frac{dx}{4-x^2}=\int_0^{1/(2p)}\frac{dx}{\sqrt{1-p^2x^2}}.

CJC · Discussion · Question 3

Use the endpoint px = 1/2 in the inverse-sine integral.

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