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Hyperbolic Integration

  • Thread starter Gondur
  • Start date
  • #1
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Homework Statement



Find the anti derivative of [tex] \int xcosh (x^2) dx[/tex]


Homework Equations



By parts formula and Hyperbolic Identities of sinh x and cosh x as well as others

The Attempt at a Solution



[tex] \int xcosh (x^2) dx[/tex]

The problem I'm having is integrating [tex] \int cosh (x^2) dx[/tex]

I tried setting variables [tex]u=x[/tex] and [tex]\frac{dv}{dx}= \int cosh (x^2) dx[/tex] with the assumption this could be solved using the by parts formula.

I then concentrated specifically on solving [tex] \int cosh (x^2) dx[/tex]. I haven't found a method that I know of that's appropriate given that the composite is (x^2) and not (cosh x)^2. Wolfram Alpha shows the solution with an error function - which I know nothing about yet.

I've touched up on Euler's formula [tex]cosx+isinx=e^{ix}[/tex] and its parallel [tex]sinhx+coshx=e^x[/tex] and I'm just about to learn its applications, maybe it should be used here. This area is new to me so light explanations are wise at this time.
 

Answers and Replies

  • #2
435
13
You don't need parts, all you need is u-substitution. u = x^2 and du = 2x dx
 
  • #3
24
0
You don't need parts, all you need is u-substitution. u = x^2 and du = 2x dx
I tried that but gave up because of the extraneous x which would mean substituting it for [tex]\sqrt {u}[/tex].

The x in the numerator cancels ut the out in x in the denominator.

Sorry I got it.
 
  • #4
435
13
look
u = x^2
du = 2x * dx

du/2 = x * dx

(1/2)∫cosh(u)du

Now from there its pretty easy as you can see.
 
  • #5
24
0
look
u = x^2
du = 2x * dx

du/2 = x * dx

(1/2)∫cosh(u)du

Now from there its pretty easy as you can see.

Yes the x variables cancel each other out. I figured it out
 

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