Integration of hyperbolic functions

[Gauss's Prodigy]

Hello,

 

I was wondering if someone could help me to interpret the following integral calculus problem:

 

(Please open and study the attached file)

 

Many thanksIntegral Problem.pdf

[imatfaal]

G'sP - welcome to the forum and you've chosen a name that's gonna take some living up to

 

Your problem is beyond my ken - but I did have one vague thought; is it acceptable to cancel out the constants of integration, are they necessarily identical?

[Gauss's Prodigy]

G'sP - welcome to the forum and you've chosen a name that's gonna take some living up to

 

Your problem is beyond my ken - but I did have one vague thought; is it acceptable to cancel out the constants of integration, are they necessarily identical?

 

Thank you for your response imatfaal,

 

I can appreciate what you said about the constants of integration. But I would have imagined the constants of integration to be identical since they are both originating from the same function that was integrated. The solution of the integral is simply written in two different forms, both of which seem to be mathematically correct. Therefore I believe the constants should theoretically be equal.

 

But even if we were to agree that the constants of integration may not be identical; lets say that they can take any value, so long as C minus C is not equal to 0.5, the hyperbolic identity (coshx)^2 - (sinhx)^2 = 1 will still NOT be true, as it will no longer equal 1.

Edited May 3, 2012 by Gauss's Prodigy

[Xittenn]

The great thing about constants of integration is they can be anything you want them to be.

 

[math]

\begin{aligned}

\frac{\cosh^2 0}{2} & \neq \frac{\sinh^2 0}{2} \\

\frac{1}{2} & \neq 0 \\

\end{aligned}

[/math]

 

 

The question then is with what constants of integration do these expressions become algebraically equivalent.

 

[math]

\begin{aligned}

\frac{\cosh^2 x}{2} & = \frac{\sinh^2 x}{2} + \frac{1}{2} \\

\frac{\cosh^2 x}{2} - \frac{\sinh^2 x}{2} & = \frac{1}{2} \\

cosh^2 x - sinh^2 x & = 1 \\

\end{aligned}

[/math]

Edited May 3, 2012 by Xittenn

[Gauss's Prodigy]

The great thing about constants of integration is they can be anything you want them to be.

 

[math]\begin{aligned}\frac{\cosh^2 0}{2} & \neq \frac{\sinh^2 0}{2} \\\frac{1}{2} & \neq 0 \\\end{aligned}[/math]

 

 

The question then is with what constants of integration do these expressions become algebraically equivalent.

 

[math]\begin{aligned}\frac{\cosh^2 x}{2} & = \frac{\sinh^2 x}{2} + \frac{1}{2} \\\frac{\cosh^2 x}{2} - \frac{\sinh^2 x}{2} & = \frac{1}{2} \\cosh^2 x - sinh^2 x & = 1 \\\end{aligned}[/math]

 

Thank you for sharing your view Xittenn.

 

You say that C can assume any value we chose, what would happen if both of the constants of integration was equal to 0?

 

Would you agree that the hyperbolic identity would not be fulfilled?

 

Thanks

[Xittenn]

The hyperbolic identity would be irrelevant. You can not make two equations that are not equal to each other, equal to each other. With each constant set to 0 the left hand is equal to 0.5 and the right hand is equal to 0, and this is an inequality so you can not start making statements about the identity. When dealing with integration the proper approach is to combine the constants to form a +c in no special way just drop the constant on the left and keep the constant on the right, and then solve for the constant. The idea here is that the integral differs within itself by a constant, in a case like this you don't even know if the constant is positive or negative, the constant is a placeholder.

Edited May 3, 2012 by Xittenn

[Gauss's Prodigy]

The hyperbolic identity would be irrelevant. You can not make two equations that are not equal to each other, equal to each other. With each constant set to 0 the left hand is equal to 0.5 and the right hand is equal to 0, and this is an inequality so you can not start making statements about the identity. When dealing with integration the proper approach is to combine the constants to form a +c in no special way just drop the constant on the left and keep the constant on the right, and then solve for the constant. The idea here is that the integral differs within itself by a constant, in a case like this you don't even know if the constant is positive or negative, the constant is a placeholder.

 

I understand your point, thanks