Partial Derivative Matrix Proof

critter
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x=rcos(θ), y=rsin(θ) Do these formulas look familiar? They give the relationship between two coordinate systems in the plane. Evaluate:
|x'r x'θ|

|y'r y'θ|

I know that the x primes are cos(θ) and -rsin(θ), and the y primes are sin(θ) and rcos(θ), respectively. I am not sure what to do with the matrix. The original functions do look familiar, but I do not remember what they mean.
 
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The notation stands for the determinant.

<br /> \left|\begin{matrix} a&amp; b \\ c &amp; d \end{matrix} \right|= ad - bc<br /> [/itex]
 
As for the meaning: these are polar coordinates.
 
Thanks!
 
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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