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Hardest integral ever

  1. Apr 7, 2013 #1
    aaaa
     

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    Last edited: Apr 7, 2013
  2. jcsd
  3. Apr 7, 2013 #2

    arildno

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    You may start with the substitution u=ln(x). What do you get then?
     
  4. Apr 7, 2013 #3
    believe me Ive tried almost everything.. it doesnt give me something special but maybe you can do something
     
  5. Apr 7, 2013 #4

    arildno

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    I won't give you anything at all, as long as you refuse to do the guided exercises I give you ( I happen to know the way ahead).
    So:
    Post the result of that substitution.
     
  6. Apr 7, 2013 #5
    .......I think the u=lnx substitution is quite obvious, I mean the 1/x factor is right there.
    After you use the substituion, the 1/x factor will go away and you can factorize that junk in the root sign, I think you can take it from there.
     
  7. Apr 7, 2013 #6

    arildno

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    Factorization under the square root won't help you a bit, but completing the square beneath it will.
     
  8. Apr 7, 2013 #7
    thank you all but ive tried it all...
     
  9. Apr 7, 2013 #8

    arildno

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    No, you haven't.
    And, you won't get any further help, because you are too damn lazy and refuse to follow up the exercises I give you. What do wish for? A solution handout you don't have to work for? You're not getting it.
     
  10. Apr 7, 2013 #9
    what execises? Im new here and I have no idea what your talking about. believe me I know a thing or two in math and this integral i couldnt solve so dont call me laze because Ive tried for 5 hours befor I decided to ask for help. you can see that it is my first post
     
  11. Apr 7, 2013 #10

    SteamKing

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    That's funny. I looked at your attachment and I didn't see any integral at all. All I saw was y = mess.
     
  12. Apr 7, 2013 #11
    you need to integrate that mes... :)
     
  13. Apr 7, 2013 #12

    arildno

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    And, for the last time:
    I gave you an exercise in POSTING your result when you make that u-substitution. Do so.
     
  14. Apr 7, 2013 #13
    ok sorry for the miss understanding
     

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  15. Apr 7, 2013 #14

    arildno

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    OK, then we're on the right track! :smile:

    Now, the first learning point is to compare THAT expression with the previous one. Do you agree that the new one is strictly simpler than what you began with? I assume you do (if not, disagree!).

    Now, why is this important?
    Simply because you should NOT, ever, believe that a single substition will make the rest trivial, or at least, easier than your first substition.

    By the u-substition, you have made definite progress relative to your starting point, and THAT is why it is a good substitution! We have removed SOME nasties from the original, now we'll tackle the next one trying to frighten us.

    Now, the really ugly thing confronting us is that square root expression, with a second degree polynomial inside it.

    Square roots can only be removed by having a perfect square beneath it. But, we haven't got that.

    Can we manipulate the stuff underneath the root sign to get a perfect square? That's our next job!

    1. Option 1: We could FACTORIZE easily enough what's beneath the root, but that doesn't give us a required SQUARE, does it? So, we'll drop factorization.

    2. option 2: What other manipulation than factorization can we do with a second-degree polynomial? We can "complete" the square.

    Let's see where THAT takes us, so your next exercise is to write what is beneath the root sign in the shape (u+a)^2+b, where "a" and "b" will be numbers.

    Post your result, so that we get a progressing solution in this thread.
     
    Last edited: Apr 7, 2013
  16. Apr 7, 2013 #15
    thats what i got :)
     

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  17. Apr 7, 2013 #16

    arildno

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    Good!

    Well, how to proceed next?

    Note that essentially, what we've got now is an expression of the shape:
    [tex]\sqrt{v^{2}+w^{2}}, v=u+1, w=\sqrt{3}[/tex]
    That is, we have a sum of squares under our root sign, rather than a single square which is what we REALLY want!
    ----
    Isn't, however, that expression awfully alike the expression for the length of the hypotenuse in a right-angled triangle?

    And the functions of sine, cosine are tightly related to trigonometry..
    -----

    Thus, it would make sense to study trigonometric relations a bit, to see if they include an expression where a sum of squares can transform itself into a single square (our dearest wish!)
    ---

    Generally, we have the identity:
    [tex]\sin^{2}(v)+\cos^{2}(v)=1[/tex]
    Here, we DO have a sum of squares, but it is distinctly different from the one WE've got, because in that expression, only ONE of the addends is a variable, the other being the pesky number 3.

    How can we improve on that nastiness?
    If we divide through our identity with the squared cosine, we get:
    [tex]\tan^{2}(v)+1=\frac{1}{\cos^{2}(v)}[/tex]
    (Remember THAT identity in the future, it is extremely useful!)

    If we now multiply that identity throughout with [itex](\sqrt{3})^{2})[/itex], we get:
    [tex](\sqrt{3}\tan(v))^{2}+3=(\frac{\sqrt{3}}{\cos(v)})^{2}[/tex]
    Here, we have precisely what we need to transform the sum of squares (LHS) beneath our root sign, into a single square on the RHS!
    ----------------------
    Thus, our next substitution should be [itex]u+1=\sqrt{3}\tan(v)[/itex]

    Make that substitution, remember also to use it on the "u" outside the rootside, and on the differential element "du"
     
    Last edited: Apr 7, 2013
  18. Apr 7, 2013 #17
    thats what i got :)
     
  19. Apr 7, 2013 #18

    arildno

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    Well, that was good preparatory work on your part!
    Now, write down what you then get, and I'll provide you with a really sleazy substitution that will help you further on! :smile:
     
  20. Apr 7, 2013 #19
    aaaa
     

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  21. Apr 7, 2013 #20

    arildno

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    Correct!
    Now, we multiply that cosine factor into the parenthesis, and we get the integral:
    [tex]\int\frac{dv}{\sqrt{3}\sin(v)-\cos(v)}[/tex]
    Agreed?
    We have managed to simplify our original integral into one with a linear combination of trig functions in the denominator.

    The major sleaze now, is to use the half-angle formulas for the trig functions in a creative way.

    We set:
    [tex]s=\tan(\frac{v}{2})[/tex]
    (This is called the Weierstrass substitution)

    Note that:
    [tex]\frac{ds}{dv}=\frac{1}{2}\frac{1}{\cos^{2}(\frac{v}{2})}=\frac{1}{2}(1+\tan^{2}(\frac{v}{2}))=\frac{1}{2}(1+s^{2}), dv=\frac{2ds}{1+s^{2}}[/tex]

    Furthermore, we have, by the half-angle formulas,
    [tex]\sin(v)=2\sin(\frac{v}{2})\cos(\frac{v}{2}), cos(v)=\cos^{2}(\frac{v}{2})-\sin^{2}(\frac{v}{2})[/tex]
    and, of course (for use of numerator transformation: [tex]1=\cos^{2}(\frac{v}{2})+\sin^{2}(\frac{v}{2})[/tex]

    Now, inserting all these relations into your integral, and dividing through with the square of cosines (remembering that s equals the "tan of v half", you get:
    [tex]\int\frac{1+s^{2}}{2\sqrt{3}s-(1-s^{2})}\frac{2ds}{1+s^{2}}=\int\frac{2ds}{s^{2}+2\sqrt{3}s-1}[/tex]

    Verify these steps, and see if the road ahead looks a bit brighter! :smile:
     
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