Revising for Exams: Understanding Euler Formula #8

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I'm doing revision for next semesters exams and I ran across this:

http://mathworld.wolfram.com/EulerFormula.html

Specifically formula no.8. I've seen it before but can't find it in my notes, I forgot what splitting

{\frac {d z}{d\theta}}

on either side of the equals sign was called. But I do remember being warned not to think that it is just division.
 
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That is "separating" the differential. What you have is a "separable" differential equation and you separated the variables.
 
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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