How to Solve Trigonometric Integrals and Isolate y as a Function of x

[shocks90]

Im stuck with this

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

after doing I am left with

coth x/tan y= dy/dx

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

 

[Ackbach]

I would separate out this way:
$$\coth(x) \, dx= \tan(y) \, dy,$$
and integrate both sides.

 

[shocks90]

hmm...but the solution i get don't seem to be y as the function of x

i got this :

ln sinh x + C = -ln cos y

is this correct?

 

[Ackbach]

I would actually write $ \ln | \sinh(x)|+C= \ln | \sec(y)|$. Note that $-\ln| \cos(y)|=\ln|(\cos(y))^{-1}|= \ln| \sec(y)|$. So now you can simply solve for $y$:
\begin{align*}
e^{C} e^{\ln| \sinh(x)|}&=\sec(y) \\
e^{C} \sinh(x)&= \sec(y).
\end{align*}
Can you finish?

 

[shocks90]

Sadly I can't finish it. the more i look the more i get lost into it. :(

 

[Ackbach]

Sadly I can't finish it. the more i look the more i get lost into it. :(

Ok, next step:
$$e^{-C} \text{csch}(x)= \cos(y).$$
Can you finish now?

 

[Prove It]

Im stuck with this

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

after doing I am left with

coth x/tan y= dy/dx

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

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

\displaystyle \begin{align*} <br /> \cosh{(x)}\cos{(y)}\frac{dx}{dy} &amp;= \sinh{(x)}\sin{(y)} \\ \frac{\cosh{(x)}}{\sinh{(x)}} \,\frac{dx}{dy} &amp;= \frac{\sin{(y)}}{\cos{(y)}} \\ \int{\frac{\cosh{(x)}}{\sinh{(x)}} \, \frac{dx}{dy}\,dy} &amp;= \int{ \frac{\sin{(y)}}{\cos{(y)}} \, dy} \\ \int{ \frac{\cosh{(x)}}{\sinh{(x)}}\,dx} &amp;= -\int{ \frac{-\sin{(y)}}{\cos{(y)}}\,dy} \end{align*}

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

 

[shocks90]

i have a question, can y be be any function or it has to be y itself
for its asking the expression y of the function x?

 

[Prove It]

If it's an ordinary differential equation (as you have stated it is) then y has to be a function of x (or vice versa).

 

[shocks90]

Thanks...but i have yet to remove the trigos to get y alone.