I can't seem to follow the manipulation of the equation

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Hello!

In the picture below, i can't seem to understand how they get from the first line, to the second line.

Fal3T


(https://imgur.com/a/Fal3T)

When i write it up as dr/dT on the leftside, and separate the r's and the T's, i get dr/((1/2)*r-(1/16)*r^5)=dT which is not the same as in the picture.

Where do i go wrong?
 
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How is it not the same thing? It is exactly the same thing just factorising out ##r/2## from the denominator.
 
Kasper Larssen said:
Hello!

In the picture below, i can't seem to understand how they get from the first line, to the second line.

Fal3T


(https://imgur.com/a/Fal3T)

When i write it up as dr/dT on the leftside, and separate the r's and the T's, i get dr/((1/2)*r-(1/16)*r^5)=dT which is not the same as in the picture.

Where do i go wrong?

You forgot Algebra 101: just multiply your numerator and denominator by 2.
 
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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