# Homework Help: Antiderivative of cosh(x^2)

1. May 20, 2010

### WrittenStars

1. The problem statement, all variables and given/known data
Just trying to figure out the anti-derivative of cosh(x^2).

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

3. The attempt at a solution
I am baffled by this one :(. Any help and pointers would be much appreciated.

2. May 20, 2010

### WrittenStars

Sorry but isn't the chain rule for differentiating, not antidifferentiating?

3. May 21, 2010

### mg0stisha

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. May 21, 2010

### WrittenStars

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. May 21, 2010

### mg0stisha

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. May 21, 2010

### WrittenStars

Good question, dont 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:
$= \int_0^1 (1- x/3)cosh(x^2)dx= \int_{x=0}^1 cosh(x^2)dx- \frac{1}{3}\int_0^1 xcosh(x^2)dx$

From here I cant figure out how to differentiate cosh(x^2). I hope I made a mistake and there is an easier way!

7. May 21, 2010

### mg0stisha

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

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. May 21, 2010

### thepatient

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. May 21, 2010

### thepatient

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. May 21, 2010

### thepatient

Oh it's a double integral... I should read ahead before posting. :\

11. May 21, 2010

### gabbagabbahey

When you sketch the integration region for

$$\int_0^1\int_{3y}^3\cosh(x^2)dxdy$$

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 $dx$, you should find that $y$ ranges from zero to $\frac{x}{3}$, not x/3 to one.

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

12. May 21, 2010

### mg0stisha

And there's where my lack of knowledge on multiple integrals comes in! Good luck on the rest of the problem!

13. May 21, 2010

### thepatient

And from there is really easy. :]