What is the antiderivative of cosh(x^2)?

In summary, the conversation is about trying to figure out the anti-derivative of cosh(x^2) for a double integral problem. The initial attempt at using the chain rule for antidifferentiating is discussed, but it is determined that there is no elementary antiderivative for cosh(x^2). The conversation then turns to discussing possible alternative methods, including using the definition of cosh(x) and expressing the antiderivative as an infinite sum using the Maclaurin series. The conversation also touches on the issue of changing the order of integration in a double integral problem and provides a helpful tip for solving the problem.
  • #1
WrittenStars
16
0

Homework Statement


Just trying to figure out the anti-derivative of cosh(x^2).


Homework Equations


I knowthe antiderivative cannot be expressed as an elementary function but I am pretty clueless of getting the antiderivative though!


The Attempt at a Solution


I am baffled by this one :(. Any help and pointers would be much appreciated.

thanks in advance!
 
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  • #2
Sorry but isn't the chain rule for differentiating, not antidifferentiating?
 
  • #3
I don't think there is an elementary antiderivative for cosh(x2). What's this for?

EDIT: I know there isn't an elementary antiderivative because it would involve integrating ex2, which doesn't have a simple antiderivative.
 
  • #4
i am trying to solve a double integral question but I can't figure out how to anti-differentiate cosh(x^2).

it's quite frustrating!
 
  • #5
Hmm. Were there prior steps to this problem where you could've made a mistake, giving you the wrong integral to deal with at this point?
 
  • #6
Good question, don't think I've done anything wrong to that point.

Basically I am trying to evaluate the following double integral, by changing the order of integration first.

cosh(x^2).dx.dy for 3y < or equal to: x < or equal to 3
and 0 < or equal to: y < or equal to 1

I changed the order of integration to get:

cosh(x^2).dy.dx for x/3 < or equal to: y < or equal to 1
and 0 < or equal to: x < or equal to 3

Then:
\(\displaystyle = \int_0^1 (1- x/3)cosh(x^2)dx= \int_{x=0}^1 cosh(x^2)dx- \frac{1}{3}\int_0^1 xcosh(x^2)dx\)

From here I can't figure out how to differentiate cosh(x^2). I hope I made a mistake and there is an easier way!
 
  • #7
[tex]= \int_0^1 (1- x/3)cosh(x^2)dx= \int_{x=0}^1 cosh(x^2)dx- \frac{1}{3}\int_0^1 xcosh(x^2)dx[/tex]

I can't see a problem in that line of math, but I also haven't encountered multiple integrals in my studies yet, perhaps someone else will be able to help there.

The second integral you have there is easy enough (by u-substitution), but the first one is quite troubling.

EDIT: Just looking at your LaTeX (use tex and /tex in your brackets for the proper coding).
 
  • #8
Oh yea, antiderivative. I thought I saw derivative. XD

I guess.. the only way to express the antiderivative of cosh(x^2) would be expressing it as a infinite sum using the maclaurin series?
 
  • #9
Or you can also use the definition of cosh(x). Which is 1/2 (e^x+e^-x). But that won't help since you have cosh(x^2), not cosh(x), so therefore you'll have a sum of two infinite maclaurin series for e^x^2. XD
 
  • #10
Oh it's a double integral... I should read ahead before posting. :\
 
  • #11
WrittenStars said:
Basically I am trying to evaluate the following double integral, by changing the order of integration first.

cosh(x^2).dx.dy for 3y < or equal to: x < or equal to 3
and 0 < or equal to: y < or equal to 1

I changed the order of integration to get:

cosh(x^2).dy.dx for x/3 < or equal to: y < or equal to 1
and 0 < or equal to: x < or equal to 3

When you sketch the integration region for

[tex]\int_0^1\int_{3y}^3\cosh(x^2)dxdy[/tex]

you should get a triangle with its base along the x-axis and vertices (0,0), (1,0) and (1,1). When you cut that region into vertical slices of thickness [itex]dx[/itex], you should find that [itex]y[/itex] ranges from zero to [itex]\frac{x}{3}[/itex], not x/3 to one.

[tex]\int_0^1\int_{3y}^3\cosh(x^2)dxdy=\int_0^3\int_0^{x/3}\cosh(x^2)dydx[/tex]
 
  • #12
And there's where my lack of knowledge on multiple integrals comes in! Good luck on the rest of the problem!
 
  • #13
gabbagabbahey said:
When you sketch the integration region for

[tex]\int_0^1\int_{3y}^3\cosh(x^2)dxdy[/tex]

you should get a triangle with its base along the x-axis and vertices (0,0), (1,0) and (1,1). When you cut that region into vertical slices of thickness [itex]dx[/itex], you should find that [itex]y[/itex] ranges from zero to [itex]\frac{x}{3}[/itex], not x/3 to one.

[tex]\int_0^1\int_{3y}^3\cosh(x^2)dxdy=\int_0^3\int_0^{x/3}\cosh(x^2)dydx[/tex]

And from there is really easy. :]
 

1. What is the definition of an antiderivative of cosh(x^2)?

An antiderivative of cosh(x^2) is a function whose derivative is equal to cosh(x^2). In other words, it is the inverse operation of finding the derivative of cosh(x^2).

2. How do you find the antiderivative of cosh(x^2)?

There is no general formula for finding the antiderivative of cosh(x^2). It involves using various techniques and methods, such as integration by parts or substitution, depending on the complexity of the function.

3. What is the difference between an antiderivative and an indefinite integral?

An antiderivative of a function f(x) is a specific function whose derivative is equal to f(x). On the other hand, an indefinite integral is a family of functions that differ only by a constant, all of which have the same derivative.

4. Can the antiderivative of cosh(x^2) be expressed in terms of elementary functions?

No, the antiderivative of cosh(x^2) cannot be expressed in terms of elementary functions. It is considered a special function and is denoted by F(x).

5. Why is the antiderivative of cosh(x^2) important in mathematics and science?

The antiderivative of cosh(x^2) is important in various areas of mathematics and science, such as in the study of differential equations, probability, and statistics. It also has applications in fields such as physics, engineering, and economics.

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