- #1

- 9

- 0

## Homework Statement

## Homework Equations

## The Attempt at a Solution

I evaluate the first integral and get [tex](1-cos(x^3))/x[/tex] then can't go further from that.

Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, the student is trying to solve a cosine integral but is having difficulty. He has tried to find the definition of the cosine integral and found that it is the one for Cin(x). He has also tried to use a u-substitution to make the integral look like the definition. He has also attempted to find the limits of the cosine integral and found them to be -∞ to +∞. He has concluded that he needs more help from the teacher or from others who are more knowledgeable in this subject.f

- #1

- 9

- 0

I evaluate the first integral and get [tex](1-cos(x^3))/x[/tex] then can't go further from that.

Last edited:

Calculus and Beyond Homework Help News on Phys.org

- #2

Science Advisor

Homework Helper

- 17,874

- 1,657

I'm reading ((1-cos(x)^3))/x is $$\frac{1-\cos^3 x}{x}$$ ... which is not correct.I evaluate the first integral and get ((1-cos(x)^3))/x then can't go further from that.

Please show your reasoning.

I'm thinking you may want to look at the class of functions like sinc.

- #3

- 9

- 0

Ops, I mean [tex](1-cos(x^3))/x[/tex]I'm reading ((1-cos(x)^3))/x is $$\frac{1-\cos^3 x}{x}$$ ... which is not correct.

Please show your reasoning.

I'm thinking you may want to look at the class of functions like sinc.

- #4

Science Advisor

Homework Helper

- 17,874

- 1,657

Cool ... well: $$\int_0^2\frac{1-\cos x^3}{x}\;dx = \int_0^2\frac{dx}{x}-\int_0^2\frac{\cos x^3}{x}\; dx$$ ... the first integral you can do right? So it must be the second one. Try graphing the integrand for clues.

Note: $$\frac{\cos x^3}{x} = \frac{x^2\cos x^3}{x^3}$$ ... suggests trying ##u=x^3## and investigating the properties of ##\cos(x)/x##.

Possibly you have been studying trigonometric integrals recently?

https://en.wikipedia.org/wiki/Trigonometric_integral#Cosine_integral

Note: $$\frac{\cos x^3}{x} = \frac{x^2\cos x^3}{x^3}$$ ... suggests trying ##u=x^3## and investigating the properties of ##\cos(x)/x##.

Possibly you have been studying trigonometric integrals recently?

https://en.wikipedia.org/wiki/Trigonometric_integral#Cosine_integral

Last edited:

- #5

- 9

- 0

I graphed the integrand and I'm still not getting it. What is the property, I can't find anything about it.

Note: $$\frac{\cos x^3}{x} = \frac{x^2\cos x^3}{x^3}$$ ... suggests trying ##u=x^3## and investigating the properties of ##\cos(x)/x##.

Possibly you have been studying trigonometric integrals recently?

https://en.wikipedia.org/wiki/Trigonometric_integral#Cosine_integral

- #6

Science Advisor

Homework Helper

- 17,874

- 1,657

Missed it? See "note" post #4.

- #7

- 9

- 0

Note: $$\frac{\cos x^3}{x} = \frac{x^2\cos x^3}{x^3}$$ ... suggests trying ##u=x^3## and investigating the properties of ##\cos(x)/x##.

Possibly you have been studying trigonometric integrals recently?

https://en.wikipedia.org/wiki/Trigonometric_integral#Cosine_integral

Aw man, I'm just taking a Calculus class, first time seeing Euler–Mascheroni constant. I'm guessing there is other way to do it besides that and I can't seem to find it. :( Is there any other hint or way?I gave you a link to the properties which you subsequently quoted above.

- #8

Science Advisor

Homework Helper

- 17,874

- 1,657

Compare the integrand in post #3 with the wiki page.

If this problem comes from an assignment or coursework, then it is testing you to see if you can recognise these special integrals and if you remember the lessons you had on how to deal with them. This means I cannot give you any more help. You have the needed parts now: you have to put them together.

- #9

- 9

- 0

The only thing I have learned is Fubini theorem and reversing the order, I don't know anything other than that. The teacher probably just gave a hard problem or chose a wrong problem from somewhere. Aw man, I have already spent hours trying to solve this problem. :( I tried comparing it with the wiki, I still can't get it.

Compare the integrand in post #3 with the wiki page.

If this problem comes from an assignment or coursework, then it is testing you to see if you can recognise these special integrals and if you remember the lessons you had on how to deal with them. This means I cannot give you any more help. You have the needed parts now: you have to put them together.

- #10

Science Advisor

Homework Helper

- 17,874

- 1,657

Use a u-substitution to make the integral you have look like the definition... don't forget to transform the limits too.

Your answer will include "Cin(?)" where some number goes in where the "?" is.

I have all-but told you the answer... there will be no more help.

- #11

- 9

- 0

Thanks for the hints. But, that's like giving an average man all the tools and expecting them to build a house with it. I never learned to build the house, if I did, I would have done it by now with all your hints. :( Oh well, I'll just have to run with it.

Use a u-substitution to make the integral you have look like the definition... don't forget to transform the limits too.

Your answer will include "Cin(?)" where some number goes in where the "?" is.

I have all-but told you the answer... there will be no more help.

- #12

- 9,568

- 774

I'm a bit late to this thread, but ##\int_0^2\frac{dx}{x}## is divergent.Cool ... well: $$\int_0^2\frac{1-\cos x^3}{x}\;dx = \int_0^2\frac{dx}{x}-\int_0^2\frac{\cos x^3}{x}\; dx$$ ... the first integral you can do right? So it must be the second one. Try graphing the integrand for clues.

- #13

- 9

- 0

Hmmm, does that make the whole thing divergent?I'm a bit late to this thread, but ##\int_0^2\frac{dx}{x}## is divergent.

- #14

- 9,568

- 774

I'm a bit late to this thread, but ##\int_0^2\frac{dx}{x}## is divergent.

Hmmm, does that make the whole thing divergent?

Not necessarily. But it does mean you can't work it by separating that term out. You need that numerator to "cancel out" the singularity at ##x=0## if the integral is to converge. I would think ##1-\cos(x^3)## has a high enough order zero to do that.

- #15

- 9

- 0

How would you solve that problem now since this comes into play?Not necessarily. But it does mean you can't work it by separating that term out. You need that numerator to "cancel out" the singularity at ##x=0## if the integral is to converge. I would think ##1-\cos(x^3)## has a high enough order zero to do that.

- #16

- 9,568

- 774

- #17

- 9

- 0

Yeah, that's why I think this problem was chosen by mistake. It's not your typical calculus problem.

Share:

- Replies
- 7

- Views
- 597

- Replies
- 12

- Views
- 429

- Replies
- 4

- Views
- 510

- Replies
- 23

- Views
- 312

- Replies
- 9

- Views
- 461

- Replies
- 3

- Views
- 157

- Replies
- 9

- Views
- 250

- Replies
- 2

- Views
- 204

- Replies
- 3

- Views
- 1K

- Replies
- 6

- Views
- 659