Arc Length for Hyperbolic Sin

[JCMateri]

I am having trouble with the arc length for hyperbolic sine. Can anyone help?

$$L=\int_{0}^{X}\sqrt{1+[\frac{dsinh(x)}{dx}]^2}dx=\int_{0}^{X}\sqrt{1+cosh^2(x)}dx$$

I'm having trouble evaluating the final integral.

 

[Mark44]

Wolfram Alpha reports that $\int \sqrt{1 + \cosh^2(x)}dx = \sqrt{\cosh^2(x)}\tanh(x) + C$

I suspect that it might be using a (hyperbolic) trig substitution to arrive at that result.

 

[JCMateri]

I don't see how that can be right. Isn't the RHS = sinh(x)+C and its derivative does nor equal \sqrt{1+cosh^2(x)}

 

[JCMateri]

Sorry,but my tex material keeps disappearing when I save.

 

[haruspex]

I am having trouble with the arc length for hyperbolic sine. Can anyone help?

$$L=\int_{0}^{X}\sqrt{1+[\frac{dsinh(x)}{dx}]^2}dx=\int_{0}^{X}\sqrt{1+cosh^2(x)}dx$$

I'm having trouble evaluating the final integral.

According to AI Overview

• Evaluation:The resulting integral
  
  
  
  $\int\sqrt{1+cosh^2(x)}dx$
  does not have a simple elementary closed-form solution and often requires numerical methods for precise calculation.

 

[JCMateri]

Thanks. I suspected that.

 

[Mark44]

I'm getting a result that's different from the one I posted earlier. I was pretty sure I asked it to integrate $\sqrt{1 + \cosh^2(x)}dx$, but it's possible that what I entered wasn't that.

Here's what it shows now:

The part on the right in parentheses is the elliptic integral of the second kind with parameter $m = k^2$.