Probably easy proof for you guys

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Homework Statement



Prove if \vec{x}\times\vec{v}=\vec{x}\times \vec{u}, then \vec{v}=\vec{u}
where \vec{x}\neq \vec{0}

2. The attempt at a solution

Use the definition of cross product and tried to reduce it.
 
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That is false as stated. Let x = <-2, 3, 1>, u = <1,2,3>, v = < 3,-1,2>. Then x cross u and x cross v are both <7,7,-7>. So your first step in solving the problem is to state it correctly.
 
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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