Free fall far away from Earth (integral substitution problem)

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[SOLVED] Free fall far away from Earth (integral substitution problem)

Homework Statement



Given:

v(x) = -v_1\sqrt{\left(\frac{R}{x} - \frac{R}{h}\right)}

Find the time t.

Homework Equations



Listed above where v_1 , R , h are all constant.

The Attempt at a Solution




v(x) = \frac{dx}{dt}

dt = \frac{dx}{\left[-v_1\sqrt{\left(\frac{R}{x} - \frac{R}{h}\right)}\right]}

t = -\frac{1}{v_1}\int\frac{dx}{\left[\sqrt{\left(\frac{R}{x} - \frac{R}{h}\right)}\right]}

I'm stuck on this integral but I'm convinced it's some kind of trigonometric substitution although i can't figure out which one. I've spent hours searching the internet for help and couldn't find anything so any help would be appreciated.

Perhaps if someone could at least point out which trigonometric substitution (if any at all) i should be trying to work with i'd be extremely grateful.

Thanks in advance.
 
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R/x - R/h =R(h-x)/hx.

The integrand becomes [1/(sqrt R)]*sqrt(hx)dx/sqrt(h-x). Put x = h*(sin^2 u).

It is now solved very easily. In the num, there is only sin^2u*du.
 
Thanks a lot that seemed to solve it for me ^_^
 
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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