Solve the given trigonometry problem

chwala
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Homework Statement
##7 \sinh x + 3 \cosh x = 9##
Relevant Equations
hyperbolic trig. equations
My question is on the highlighted part (circled in red);

Why is it wrong to pre-multiply each term by ##e^x##? to realize ,

##5e^{2x} -2-9e^x=0## as opposed to factorising by ##e^{-x} ## ?

The other steps to required solution ##x=\ln 2## is quite clear and straightforward to me.



1716546593601.png
 
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What makes you think you cannot?
 
...because of this next step:

1716547615185.png


I think i get it now...to find the solution for ##x##, we can solve it as i had indicated but for the integration bit; we have to make use of all the transforms...
 
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That solves an entirely different question than the one you asked. The one you asked about asked for the solutions of a particular equality. What that thing is is integrating 1 divided by one of the sides of the equality.
 
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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