Solve Integration Homework: Prove Integral in Attachment

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Homework Statement


Please prove the integral in the attachment

Homework Equations


The Attempt at a Solution


Tried many ways to solve it. Integration by parts, trig sub etc.
The closest one is the substitution $U=\frac{r^2}{2}-\frac{R^2/2}+x^2$, which gives some arcsin. I change it into arctan, but it is not the arctan mathematica shows. I guess it has something to do with the $\sqrt{R^2 - x^2}$, since it will blow up at R. May need delta function to handle it.
 

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Welcome to PF;
Please prove the integral in the attachment
Nice try :)
This is your homework - we can help you where you get stuck but you have to do the work.

Please give it a go and show your working as far as you get.
Try to describe how you are thinking about the problem as you go.
Don't worry about looking stupid - we've all done this before: you are among understanding, and friendly, people here.
 
You tried many ways- please show us at least some. If possible one where you feel you came closest to solving the problem. That will make it easier to give suggestions.
 
Tried many ways to solve it. Integration by parts, trig sub etc.
The closest one is the substitution ##U=\frac{r^2}{2}-\frac{R^2}{2}+x^2##, which gives some ##\arcsin##. I change it into ##\arctan##, but it is not the ##\arctan## Mathematica shows. I guess it has something to do with the ##\sqrt{R^2 - x^2}##, since it will blow up at ##R##. May need delta function to handle it.
Well your integral (off the supplied attachment) seems to be: $$\int_0^b \frac{x.dx}{\sqrt{a^2+x^2}\sqrt{b^2-x^2}}=\frac{a\arctan(b/a)}{\sqrt{a^2}}$$ ... since there is no ##r## or ##R## in the integrand, it is difficult to see your reasoning behind the substitution. From context - I'm guessing R=b and r=a?

Considering the arctan in the solution - perhaps a substitution like ##x=a\tan\theta## and a table of trig identities will be the way to go?
Did you try that?

However, HallofIvy is right - it is better for you to go step-by-step through the one you felt got the closest.
So you made the substitution: what was the next step?

(I know: it's a pain typing out the math. Use the "quote" button off the bottom of this post and see how I was able to type the math out ... you can copy and paste then make minor adjustments.)
 
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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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