Solve Change of Variable: Triangular to Square/Rectangle

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[SOLVED] Change of Variable

In case of the triangular area with vertices: (-1 1) (0 0) and (1 1), how would you do a change of variable to make that look like a square or rectangular region in terms of u an v?

x = (1/2)(u+v) y = (1/2)(u-v)

so

u = x+y v = x-y
 
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No (continuous) change of variable will make a 3 sided region into a 4 sided region!
 
oh. well what i did was to do the change of variable and then do the intervals as if i had a whole quadrilateral region and then i halved the answer. It seems to be right but i am not sure however.
 
Choosing u= x+y, v= x- y as coordinates will make two sides of the triangle, y= -x and y= x correspond to u= 0 and v= 0 respectively. The third side of the triangle, y= 1, becomes u+ v= 2.

I have no idea what you mean by "intervals" or what answer you "halved". Apparently this was part of a larger problem you haven't mentioned.
 
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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