Is There a Faster Method for Solving Hyperbolic Function Problems?

Clara Chung
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Homework Statement


222.png


Homework Equations

The Attempt at a Solution


The attempt is in the picture. Is this the right method? Is there any faster method without cumbersome calculations?
 

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Mark44 said:
From this page, https://en.wikipedia.org/wiki/Inverse_hyperbolic_functions, I see that ##\sinh(\cosh^{-1}(x) = \sqrt{x^2 - 1}##, for |x| > 1.
Thank you. I get the answer.
x=sinh(-arccosh(x+2))
=-sinh(arccosh(x+2))
=-root(x^2+4x+3)
And the website is very helpful
 
@Clara Chung: The problem is, if you look at the graph it looks like there is a solution ##x=-\frac 3 4##. And you can check that works exactly in your last equation of your original post but not your root solution.
 
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LCKurtz said:
@Clara Chung: The problem is, if you look at the graph it looks like there is a solution ##x=-\frac 3 4##. And you can check that works exactly in your last equation of your original post but not your root solution.

So it is x^2=x^2+4x+3
X=-3/4
 
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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