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Calculus Double Integrals

  • Thread starter kapitalpro
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  • #1

Homework Statement


xnP3QCF.png


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.
 
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Answers and Replies

  • #2
Simon Bridge
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I evaluate the first integral and get ((1-cos(x)^3))/x then can't go further from that.
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.
 
  • #3
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.
Ops, I mean [tex](1-cos(x^3))/x[/tex]
 
  • #4
Simon Bridge
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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
 
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  • #5
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
I graphed the integrand and I'm still not getting it. What is the property, I can't find anything about it.
 
  • #6
Simon Bridge
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I gave you a link to the properties which you subsequently quoted above.
Missed it? See "note" post #4.
 
  • #7
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
I gave you a link to the properties which you subsequently quoted above.
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?
 
  • #8
Simon Bridge
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You need to be more observant ... hint: look at post #3: there is more than one kind of cosine integral.
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
You need to be more observant ... hint: look at post #3: there is more than one kind of cosine integral.
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.
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.
 
  • #10
Simon Bridge
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The definition you want is the one for Cin(x).
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
The definition you want is the one for Cin(x).
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.
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.
 
  • #12
LCKurtz
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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.
I'm a bit late to this thread, but ##\int_0^2\frac{dx}{x}## is divergent.
 
  • #13
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?
 
  • #14
LCKurtz
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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
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.
How would you solve that problem now since this comes into play?
 
  • #16
LCKurtz
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I think this is not an elementary integral and I wouldn't expect to see it in a typical calculus class. I haven't worked it all out, but it looks like Simon's suggestion about a u substitution and the Cin function would be the way to go. I don't think you can find a "simple" antiderivative solution.
 
  • #17
I think this is not an elementary integral and I wouldn't expect to see it in a typical calculus class. I haven't worked it all out, but it looks like Simon's suggestion about a u substitution and the Cin function would be the way to go. I don't think you can find a "simple" antiderivative solution.
Yeah, that's why I think this problem was chosen by mistake. It's not your typical calculus problem.
 

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