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

[WrittenStars]

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!

 

[WrittenStars]

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

 

[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 e^x2, which doesn't have a simple antiderivative.

 

[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!

 

[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?

 

[WrittenStars]

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:
[math]= \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[/math]

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!

 

[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).

 

[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?

 

[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

 

[thepatient]

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

 

[gabbagabbahey]

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

\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

 

[mg0stisha]

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

 

[thepatient]

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

And from there is really easy. :]