Proof for cross product a x b= -b x a

senorclean
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This problem involves vectors and it's a fairly basic proof but I can't seem to wrap my head around it.

I tried to just separate the vectors into their components and cross them but then I just get another set of coordinates that doesn't seem factorable. I really can't see any way to manipulate it.

I thought about just doing the cross product of -(b x a) and showing that the coordinates would be the negative version of a x b but I'm not sure if that's even proving anything. I hate proofs. Any guidance would be greatly appreciated.
 
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Yes,doing the cross product of a x b and showing it is equal (component wise) to -b x a is enough.
 
ok then. gracias
 
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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