Changing Limits of x and y with Transformation to u and v

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rite I've got limits of 0<x<pi/2 and x-pi<y<x

with u=x-y, v=x+y

with a transformation from f(x,y) to f(u,v) which isn't really needed for what I am about to ask

i get he jacobian as 1/2
then get y=(v-u)/2 and x=(u+v)/2 all of which i know is right as i have the answers, this is purely revision
then when changing the limits i get every step until the last one which are
0<x<pi/2
0<(u+v)/2<pi/2
0<u+v<pi
-u<v<pi-u
0<u<pi<<<<<<<<<<<where does that come from
 
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similarly x-pi<y<x
(u+v)/2-pi<v-u<(u+v)/2
-2pi<-2u<0
-u<v<pi-u
 
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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