Can anyone prove the following theorem?

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Can anyone prove the following theorem?
<br /> \int_{0}^{z} \frac{dy}{dx} dz = \int_{0}^{y} \frac{dz}{dx} dy<br />
 
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What is the relationship between y and z and how are they dependent on x?
 
RSGuy said:
Can anyone prove the following theorem?
<br /> \int_{0}^{z} \frac{dy}{dx} dz = \int_{0}^{y} \frac{dz}{dx} dy<br />

HINT: dz = \frac {dz}{dy} dy
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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