Rotation of Axis to Eliminate Cross-Product Term

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Homework Statement



The question si the make a rotation of axis to elmininate the cross-product term in the work i provided in this pic: http://img295.imageshack.us/img295/9579/untitledtu4.png

Homework Equations





The Attempt at a Solution



I've drawn of the example in the pic I've given a link.

my questions:

how do i know that 2 theta = pi/3 and theta is pi/6

where does the sqrt(3)/2 and 1/2 come from?

i know there is some kind of 30/60/90 triangle going on here i think.
can someone please help me understand where the numbers are coming from.

if you expand the pic its MUCH easier to view.

thankyou.
 
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a response in your method =p

http://img96.imageshack.us/img96/1745/sresponsewm5.jpg
 
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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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