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Simple Trigonometric Integral

  1. Jul 2, 2013 #1
    Hi, I am just wondering what the antiderivative of this integral was. It looks easy to me, but I no matter what I did just could not evaluate it. Can someone help me?:
    $$\int \frac{\tan(x)}{x}\;dx$$
     
  2. jcsd
  3. Jul 2, 2013 #2

    MarneMath

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    Have you tried integration by parts? It can be a bit tedious but it isn't to bad.
     
  4. Jul 4, 2013 #3
    Actually, that integral isn't so simple. It cannot be stated as a finite combination of elementary functions.
     
    Last edited: Jul 4, 2013
  5. Jul 4, 2013 #4
    In fact, that is true for all trig functions over x.
     
  6. Jul 5, 2013 #5
    How about using the Fourier series for tan?
     
  7. Jul 6, 2013 #6
    It depends on the application. If this is part of a larger problem it may be more convenient to define a new function in terms of the integral rather than write an infinite series repeatedly.
    [tex]f'(x)=\frac{\tan{x}}{x}[/tex]
    [tex]f(x)=\int_{0}^{x}\frac{\tan{t}}{t}dt[/tex]
    One problem that may occur with infinite series is the radius of convergence. While tan(x)/x may have a value at certain values of x, the infinite series may not converge.
     
    Last edited: Jul 6, 2013
  8. Jul 6, 2013 #7
    Use integration by parts twice.
     
  9. Jul 6, 2013 #8
    Please demonstrate.
     
  10. Jul 6, 2013 #9

    MarneMath

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    Bleh, let dv = 1/x u = tanx for the first integration by parts.

    Then with the second integral you get let dv = 1/x and dv = sec^2(x)dx. You'll notice that after you do integration by parts twice, that you'll have the same integral on both sides, so combine like terms and come out with ln(x)tan(x)
     
  11. Jul 6, 2013 #10
    You are mistaken.
    [tex]\frac{d}{dx}(\log{x}\tan{x})=\log{x}\sec^{2}{x}+\frac{\tan{x}}{x}[/tex]

    Here is where you went wrong.
    [tex]\int\frac{\tan{x}}{x}dx=\log{|x|}\tan{x}-\int\log{|x|}\sec^{2}{x}dx[/tex]
     
    Last edited: Jul 6, 2013
  12. Jul 9, 2013 #11
    What would this look like?
     
  13. Jul 9, 2013 #12
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