How Are Cos(x) and Cosh(x) Related in Mathematical Identities?

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Homework Statement





Homework Equations


from the identities found on the internet:

cos(x)=\frac{(e^{ix}+e^{-ix})}{2}

and

cosh(x)=\frac{(e^{x}+e^{-x})}{2}



The Attempt at a Solution



Assuming for the definition of cosh(x), if we take x as being equal to (ix), then surely this shows that cosh(x)=cos(x)? Can someone explain why this is wrong please? because i can't see it
 
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It shows cosh(ix)=cos(x), not cosh(x)=cos(x). There's nothing wrong with that.
 
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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