Solving for Particular Integral in Differential Equations: Arctan Definition

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Homework Statement


I'm trying to find a particular integral to the differential equation on the attatched image, but I come up with two slightly different answers depending on the method I use. I suspect the problem (if it's not just me being very thick and making a silly algebra error) is the way arctan is defined but I can't see what I should do differently in either solution.

The Attempt at a Solution


Attatched.

Thanks
 

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?? You give just the answers and they look identical to me!
 
One has a plus thi one has a minus thi for some reason.
 
I think you mean phi, \phi
I haven't read the question, I probably wouldn't understand it, but perhaps Halls was giving a hint that its equal to zero?
 
My mistake sorry, meant phi. I think phi's only going to be zero in a special case, for most values of a,b,c,w it's going to be non-zero.

Does anyone have any other suggestions? The second version is the one that works in the differential equation but I still can't see why the first one doesn't.
Thanks
 
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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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