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Integration question, please help!

  1. Aug 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following integral:

    2. Relevant equations

    [tex]\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx[/tex]

    3. The attempt at a solution

    I tried to split is into ln(1) times ln(tanx) but didnt work.
     
  2. jcsd
  3. Aug 8, 2012 #2
    Because you can't split it up like that perhaps?

    Use this identity of integrals: Substituting x=a-t, dx=-dt:

    [tex]\int_{0}^{a}f(a-t)\,dt=-\int_{a}^{0}f(x)\,dx=\int_{0}^{a}f(x)\,dx[/tex]
     
  4. Aug 8, 2012 #3
    can u please show me how to do it, i tried it this way but i messed it up.
     
  5. Aug 8, 2012 #4
    That is a pretty difficult integral. If you really wrote it down correctly, the best approach is likely to evaluate it numerically.
     
  6. Aug 8, 2012 #5
    I'll give you a headstart, but you should solve it yourself.
    Applying the result I gave above to your problem, we get that your integral is equal to
    [tex]\int_{0}^{\pi/4}\log(1+\tan(\pi/4-x))dx[/tex]
    Now, note that [itex]\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}[/itex] so that
    [tex]\tan(\pi/4-x)=\frac{\sin(\pi/4-x)}{\cos(\pi/4-x)}=\frac{\sin(\pi/4)\cos(x)-\cos(\pi/4)\sin(x)}{\cos(\pi/4)\cos(x)+\sin(\pi/4)\sin(x)}=\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}[/tex]
    This is easily seen to be equal to
    [tex]\frac{1-\tan(x)}{1+\tan(x)}[/tex]
    Now, please take it from there.
     
  7. Aug 8, 2012 #6
    The indefinite integral is in terms of the logarithmic integral function but the definite integral has a solution.
     
  8. Aug 8, 2012 #7
    Thank u a lot.

    great forum.
     
  9. Aug 8, 2012 #8
    What does that mean? Do you have the answer??
     
  10. Aug 8, 2012 #9
    I don't have the answer but thanks to u, i solved it, the answer is pi/4 ln(2)
     
  11. Aug 8, 2012 #10
    guys, im learning harder integration so can any one give me an example of an extremely difficult integral?
     
    Last edited: Aug 8, 2012
  12. Aug 8, 2012 #11
    [tex]\int_{0}^{\infty}\frac{\cos(x)\,dx}{1+x^2}[/tex]
    Good luck!
     
  13. Aug 8, 2012 #12
    wow thats a hard one
    do you have so really big, difficult and intimidating integrals that looks almost impossible to solve
     
  14. Aug 8, 2012 #13
    Discussion here is not really appropriate, I shall send some to you via PM.
     
  15. Aug 8, 2012 #14
    thank you
    im new here so excuse me please :)
     
  16. Aug 9, 2012 #15

    gabbagabbahey

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    Homework Helper
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    Is it? You might want to sketch the integrand and see if your result is reasonable for the area under the curve ;0).
     
  17. Aug 10, 2012 #16
    Your really close, you probably made a simple arithmetic mistake in your work. The graph of that is very close to a triangle and the area you found wouldn't fit in the bounds of a triangle. That was fun integral though!
     
  18. Aug 10, 2012 #17

    Ray Vickson

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    Wrong. If we let [itex] J = \int_0^{\pi/4} \, \ln(1 + \tan(x))\, dx,[/itex] your claim is that [itex]J = \pi/4\ln(2).[/itex] If you mean [itex]J = (\pi/4) \ln(2) \doteq 0.5443965,[/itex] that is wrong. If you mean [itex] J = \pi/(4 \ln(2)) \doteq 1.1330900,[/itex] that is also wrong. The correct answer is
    [tex] J = \frac{i}{2}\text{dilog}\left(\frac{1+i}{2}\right) - \frac{i}{2}\text{dilog}\left(\frac{1-i}{2}\right) + \frac{\pi \ln(2)}{4} - C,[/tex]
    where C is Catalan's constant
    [tex] C = \sum_{i=0}^{\infty} \frac{(-1)^i}{(2i+1)^2} [/tex]
    and
    [tex] \text{dilog}(w) = \int_1^w \frac{\ln(t)}{1-t} \, dt[/tex]
    and [itex]i = \sqrt{-1}.[/itex] The numerical value of the correct answer is
    [itex]J \doteq 0.27219826.[/itex] (All this courtesy of Maple11).


    RGV
     
  19. Aug 10, 2012 #18
    J≐0.27219826 looks like [tex] \frac{\pi}{8}\ln{2}[/tex] which is why I think it likely he made a simple arithmetic mistake.
     
  20. Aug 10, 2012 #19
    He probably made the same mistake I did when I worked it out and got the same answer as him :blushing: but I did eventually get[tex]\frac{\pi}{8}\ln2[/tex]
     
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