How to Correctly Use Substitution for Polar Coordinates in Integrals?

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The Attempt at a Solution


http://img484.imageshack.us/my.php?image=picture155jy1.jpg
i get an extra cos² thi term! WHY!
am i doing the substitution completely wrong?? or i forgot/left something out which i can not seem to see!

Thank you
 
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4th line, dw_1= -w\sin \phi d\phi unfortunately :(
 
oops, i meant to write dw1 = cos thi dw.
nevermind, i got it.
I have to use Jacobian matrix instead of what i did.
 
Last edited:

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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