Hyperbolic Functions: Exploring coshz, sinhz, Derivatives & Integrals

In summary, The hyperbolic functions, coshz and sinhz, are defined as e^{z} + e^{-z} /2 and e^{z} - e^{-z} /2 respectively. They are related to the trigonometric functions cos and sin through the equations coshz = cos(iz) and sinhz = i*sin(iz). The derivatives of coshz and sinhz are sinh and cosh, respectively, and their integrals are arcsinh and arccosh. The identity cosh^2z - sinh^2z = 1 can be used to prove the integral of dx/sqrt[1+x^2] equals arcsinh x by substituting x = sinhz and
  • #36
Oblio said:
either way i don't see how i won't just get 1/1 = 1...

Right... you get the integral of dz, which is just z. z = arcsinh(x), since x = sinh(z).
 
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  • #37
learningphysics said:
Right... you get the integral of dz, which is just z. z = arcsinh(x), since x = sinh(z).

I had accidentally switched some z's for x's... that's a bad thing. lol

I get it.
The only thing that seems odd now, is
WHY can one just say x=sinhz? Was that shown in the question somehow already that I'm not seeing?
 
  • #38
Oblio said:
I had accidentally switched some z's for x's... that's a bad thing. lol

I get it.
The only thing that seems odd now, is
WHY can one just say x=sinhz? Was that shown in the question somehow already that I'm not seeing?

I agree with you that it is odd the order we did it...

The better way to approach is to let z = arcsinh(x) (so we're introducing a new variable z... x is already given.)... and then from there saying x = sinh(z). I think that makes more sense.
 
  • #39
So, am I correct in thinking though, that this is only true SINCE x =sinhx.
I mean, they made it a 'Hint', but that definition of x is completely necessary to solve it, isn't it?

Without the hint, it couldn't be done?
 
  • #40
I guess I mean, it should written that the statement is true WHEN x =sinhz. (made a typo above i see)
 
  • #41
agree or am i missing something still? lol
 
  • #42
Oblio said:
So, am I correct in thinking though, that this is only true SINCE x =sinhx.
I mean, they made it a 'Hint', but that definition of x is completely necessary to solve it, isn't it?

Without the hint, it couldn't be done?

No, that definition wasn't necessary...

We need to define z = arcsinh(x). we're not defining x... x is already given... we define z = arcsinh(x), and then we get x = sinh(z). but this isn't a definition... just a relationship... z is the variable being defined.

given any variable... say u...

I can simply define v = arcsinh(u). now I have defined v not u... but u = sinh(v), because of the way v has been defined.
 
  • #43
Oblio said:
I guess I mean, it should written that the statement is true WHEN x =sinhz. (made a typo above i see)

z is defined in such a way that x = sinhz... in other words we create a variable z that is defined as arcsinh(x).

so z is being defined, not x.
 
  • #44
Oblio said:
So, am I correct in thinking though, that this is only true SINCE x =sinhx.
I mean, they made it a 'Hint', but that definition of x is completely necessary to solve it, isn't it?

Without the hint, it couldn't be done?

Ok, switch x for z.
Was it necessary for the question to define z?
 
  • #45
Oblio said:
Ok, switch x for z.
Was it necessary for the question to define z?

The question didn't have to define it... the question suggested it as a hint to help you solve the problem...

Even if the question didn't define it... you could define a variable as arcsinh(x), to help you solve the problem.

The question is simply prove that arcsin(x) = integral(...)

The z = arcsinh(x) is only a hint to solve the problem... the question is complete even without that hint.
 

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