What is the Integral of the Square Root of Tangent of X?

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please let me know if i have any mistakes up to this point! it would suck to have been doing it wrong after so much work :-[

http://img149.imageshack.us/img149/7546/tanxdu0.jpg
 
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everything looks right. and your final integrals = the one right before you did the partial fractions.
 
next step? integration by parts? advice ... I've gone blind.

http://img99.imageshack.us/img99/6840/partialfractionsvg7.jpg
 
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k i guess ima do it that way ... damn this integral is so ugly!

edit: nvm! w00t :-]
 
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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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