Finding the Anti-Derivative of x*cosh(x^2) using Hyperbolic Identities

Click For Summary

Homework Help Overview

The discussion revolves around finding the anti-derivative of the integral \(\int x \cosh(x^2) \, dx\). The subject area includes calculus, specifically integration techniques and hyperbolic functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use integration by parts and expresses difficulty in integrating \(\int \cosh(x^2) \, dx\). They mention the use of hyperbolic identities and Euler's formula as potential tools. Some participants suggest using u-substitution instead, specifically letting \(u = x^2\) and \(du = 2x \, dx\), while others note complications with the substitution process.

Discussion Status

Participants are exploring different methods to tackle the integral, with some suggesting u-substitution as a simpler approach. There is acknowledgment of the challenges faced with the original method proposed by the poster, and some have indicated they have resolved their confusion regarding the substitution process.

Contextual Notes

The original poster is new to the topic and seeks light explanations, indicating a potential gap in understanding hyperbolic functions and their integrals. There is also mention of external resources like Wolfram Alpha, which introduces additional concepts such as the error function that the poster is not yet familiar with.

Gondur
Messages
25
Reaction score
0

Homework Statement



Find the anti derivative of \int xcosh (x^2) dx

Homework Equations



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

The Attempt at a Solution



\int xcosh (x^2) dx

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

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

I then concentrated specifically on solving \int cosh (x^2) dx. 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 cosx+isinx=e^{ix} and its parallel sinhx+coshx=e^x 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.
 
Physics news on Phys.org
You don't need parts, all you need is u-substitution. u = x^2 and du = 2x dx
 
Panphobia said:
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 \sqrt {u}.

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

Sorry I got it.
 
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.
 
Panphobia said:
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
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
2K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K