Hyperbolic substitition question:

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SUMMARY

The integral of the function \(\int \sqrt{1+v^{2}} \, dv\) can be solved using hyperbolic substitution, specifically involving the inverse hyperbolic sine function, arcsinh(v). The solution is given as \(\frac{1}{2} v \sqrt{1+v^{2}} + \frac{1}{2} \text{arcsinh}(v)\). The discussion emphasizes that arcsinh can be expressed in terms of logarithmic functions, specifically \(\log_e(v + \sqrt{v^2+1})\), and suggests that using trigonometric substitutions like \(u = \tan(x)\) can also be beneficial.

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  • Understanding of integral calculus and hyperbolic functions.
  • Familiarity with inverse hyperbolic functions, particularly arcsinh.
  • Knowledge of trigonometric substitutions in integration.
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  • Study the properties and applications of inverse hyperbolic functions, focusing on arcsinh.
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  • Explore the derivation and applications of the logarithmic form of hyperbolic functions.
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Homework Statement



[tex]\int \!\sqrt {1+{v}^{2}}{dv}[/tex]

Homework Equations



Maple tells me that I have to throw in an arcsinh into the solution some how.

The Attempt at a Solution



I've tried substituting with tan(x) but that got me no where and from the solution I'm given:

[tex]1/2\,v\sqrt {1+{v}^{2}}+1/2\,{\it arcsinh} \left( v \right)[/tex]

I'm not sure how you get the first term and I know that arcsinh(v) is the integral of 1/sqrt(1+v^2)
 
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If the answer involves inverse sinh, why not put sinh (or cosh or tanh) into the equation?
 
If you don't exactly like arcsinh, just rewrite sinh in terms of the exponential function and find its inverse. This shows you that arcsinh is just a fancier way of writing: [tex]\log_e (x + \sqrt{x^2+1})[/tex]

Edit: P.S. u= tan x was a good idea :) Go along with it.
 

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