Cant understand this solution in certain points

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on a circle with radius 1 there are two points one is static the other is
distributed evenly on the circle (i don't know the proper english term)
we sign X to be the hypotenuse on the circle

find the density function of the variable X
?

solution(i partialy understood it)the questions come after the photo

n3t35v.jpg

they say that we take x to be an angle from o til pi/2
we scan only the northern hemisphere because its symmetric.

what is the curv graph represents
what is t?
what is the meaning of y<=t?
why if they say that x should go from o til pi/2?
in the probability function they put x fro 0 till arcsin t/2 where t goes from
0<t<=2
?

after this step they did the derivative to find the density function
i understand that step
 
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Hi nhrock3! :smile:

(have a pi: π and a ≤ :wink:)
nhrock3 said:
on a circle with radius 1 there are two points one is static the other is
distributed evenly on the circle (i don't know the proper english term)
we sign X to be the hypotenuse on the circle

find the density function of the variable X

they say that we take x to be an angle from o til pi/2
we scan only the northern hemisphere because its symmetric.

what is the curv graph represents
what is t?
what is the meaning of y<=t?
why if they say that x should go from o til pi/2?
in the probability function they put x fro 0 till arcsin t/2 where t goes from
0<t<=2
?

after this step they did the derivative to find the density function
i understand that step

(we usually say "uniformly", but "evenly" is also ok :smile: and we usually say "around a circle")

No, I don't think x is the hypotenuse.

It's a little difficult to tell, but I think x and y are angles

(and what is t … is it a chord? :confused:)

Let's call the fixed point A, the movable point X, and the centre O. And let B be the other end of the diameter AOB.

Then I think x is the angle between the two points, which equals AOX, and y is the angle ABX.

y is always x/2 (if you don't know that, you need to look at a geometry book).

Carry on from there. :smile:
 
you are correct its a hord not a hipottenuse
but you didnt answer my questions?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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