Meden Agan's latest activity
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Do you mean I should substitute ##x = \sin^2 (\theta)## into ##\dfrac{(x - 1)\left(x - 1 + x \sqrt{9 - 16x}\right)}{1-2x}##? If yes, I...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Mhm... I've discovered something which puzzles me, but I have no idea how to prove it. From post #1, we have the double integral $$I =...
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with
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I have tried to solve this by my favorite method: "remove what disturbs the most" and set ## y^2=9-16x ,## which gave me some symmetries... -
MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Typo. The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{u} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0...
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MMeden Agan posted the thread Prove that the integral is equal to ##\pi^2/8## in Calculus and Beyond Homework Help.Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx =...
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MMeden Agan reacted to PeroK's post in the thread Projectile motion with ##N## bounces on the ground with Like.
For ##N > 1## the two functions intersect at a point greater than ##\sqrt 2 -1##. You do know that is the positive zero of the...
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MMeden Agan reacted to haruspex's post in the thread Projectile motion with ##N## bounces on the ground with Like.
Your original approach strongly suggested that limiting value, it just wasn't a valid proof.