Changing the interval of integration

Amaelle
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Homework Statement
look at the image
Relevant Equations
Green theorem
Greetings Dear community!
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Here is the solutions using two different methods: the first method is the Green theorem and the second is the simple path integration method:


My question is why they integrate over [0.2pi] in the path integration method while they integrate within [0. pi] in the green method (I do agree with it)?


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thank you!
 

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They don’t seem to be using any parametrization at all when applying Stokes’ theorem.

The value of an integral does not depend on the parametrization. You could have picked a parameter taking values in ##[-200.5,\pi]## and you would get the same result (although a bit more annoying expressions for the differentials).
 
for the stock theroem they calculate the surface by integrating from [0 ,pi], this is why i don't understand why they didn't do the same for the path integral
 
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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