Orthogonality and Inner Products: Understanding a Linear Algebra Proof

Dustinsfl
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How to start this proof?

<u-p, p> = 0
 
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What are \textbf{u} and \textbf{p} supposed to represent? That statement isn't true for two general vectors \textbf{u} and \textbf{p}.

What is the entire problem statement that you are given?
 
It has to do with inner product.
 
Is that equivalent to <u, p> - <p, p> = <u, p> - \left\|p\right\|2
 

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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