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- Jan 26, 2012

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I would separate out this way:

$$\coth(x) \, dx= \tan(y) \, dy,$$

and integrate both sides.

$$\coth(x) \, dx= \tan(y) \, dy,$$

and integrate both sides.

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- Jan 26, 2012

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\begin{align*}

e^{C} e^{\ln| \sinh(x)|}&=\sec(y) \\

e^{C} \sinh(x)&= \sec(y).

\end{align*}

Can you finish?

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- Jan 26, 2012

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Ok, next step:Sadly I can't finish it. the more i look the more i get lost into it.

$$e^{-C} \text{csch}(x)= \cos(y).$$

Can you finish now?

It might be easier for you to integrate these functions if you keep them in terms of sines, cosines, shines and coshines...Im stuck with this

cosh x cos y dx/dy =sinh x sin y

after doing im left with

coth x/tan y= dy/dx

lost in trying to get y as a function of x due to integrating of trigo

[tex]\displaystyle \begin{align*}

\cosh{(x)}\cos{(y)}\frac{dx}{dy} &= \sinh{(x)}\sin{(y)} \\ \frac{\cosh{(x)}}{\sinh{(x)}} \,\frac{dx}{dy} &= \frac{\sin{(y)}}{\cos{(y)}} \\ \int{\frac{\cosh{(x)}}{\sinh{(x)}} \, \frac{dx}{dy}\,dy} &= \int{ \frac{\sin{(y)}}{\cos{(y)}} \, dy} \\ \int{ \frac{\cosh{(x)}}{\sinh{(x)}}\,dx} &= -\int{ \frac{-\sin{(y)}}{\cos{(y)}}\,dy} \end{align*}[/tex]

Each of these can easily be integrated using a substitution [tex]\displaystyle \begin{align*} u = \sinh{(x)} \implies du = \cosh{(x)}\,dx \end{align*}[/tex] and [tex]\displaystyle \begin{align*} v = \cos{(y)} \implies dv = -\sin{(y)}\,dy \end{align*}[/tex]

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