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Trigo Integral

  1. May 20, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data

    [tex]\int{\frac{xsec^2x}{tanx+\sqrt{3}}}dx[/tex]

    3. The attempt at a solution

    I let [tex]tanx+\sqrt{3}=u[/tex]

    Then, [tex]\frac{du}{dx}=sec^2x , dx=\frac{du}{sec^2x}[/tex]

    Substituting into the integral: [tex]\int{\frac{x}{u}du[/tex]

    But now I don't know what to do. That little x is really screwing me over on this... Maybe a whole other approach needs to be taken?
     
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  3. May 20, 2009 #2

    Cyosis

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    From your substitution it follows that [itex]x=\arctan(u-\sqrt{3})[/itex]. That said I don't think this integral has a solution within the elementary functions. It will involve some kind of poly logarithm.
     
  4. May 20, 2009 #3

    Mentallic

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    If this integral has no elementary solution, I think I may give up and walk away from it.

    Just out of curiosity, what would [tex]\int{arctanx}dx[/tex] be? I'm sure it's not elementary but the anti-derivative must've been found/solved for such a simple result?
     
  5. May 20, 2009 #4

    Mark44

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    This one can be integrated using integration by parts.
    u = tan-1x, dv = dx
    du = dx/(1 + x2), v = x

    This results in
    [tex]\int tan^{-1}x dx = xtan^{-1}x - \int xdx/(1 + x^2)[/tex]
    The latter integral can be done with an ordinary substitution, u = 1 + x2.
     
  6. May 20, 2009 #5

    Cyosis

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    That one is simpler than you think, write it as [itex] \int 1*\arctan x dx[/itex], then use partial integration.
     
  7. May 21, 2009 #6

    Mentallic

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    I think I might start reading up on integration by parts, since I couldn't follow where the v came from and the dv=dx, v=x?

    If it comes down to not being able to solve an integral with elementary functions, are there any methods available to solve one as such? Other than taking guesses and differentiating to be sure :smile:
     
  8. May 21, 2009 #7

    HallsofIvy

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    That's exactly where the v came from: if dv= dx, then, integrating both sides, v= x!

    If it really is true that a given integral cannot be done with "elementary functions", then what in the world would you "guess"? It not a matter of certain methods not working, it is that the answer itself cannot be written in terms of elementary functions. In that case you either define a new function to be its integral or try to convert to a such an integral for which the integral has already been defined.

    For example,
    [tex]\int e^{-x^2}dx[/tex]
    cannot be done "in terms of elementary functions" so the non-elementary function Erf(x) (the "error" function because that integral shows up in probability and calculating "random errors") is defined as that integral.
    [tex]\int e^{-x^2}dx= Erf(x)+ C[/tex]

    But if you had, now
    [tex]\int e^{-(2x-3)^2} dx[/tex]
    you could make the substitution u= 2x- 3 so that du= 2dx and dx= (1/2)du so
    [tex]\int e^{-(2x-3)^2} dx= \frac{1}{2}\int e^{-u^2}du= \frac{1}{2}Erf(u)+ C= \frac{1}{2}Erf(2x-3)+ C[/tex].
     
  9. May 21, 2009 #8

    Mentallic

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    Not sure what the v is doing there in the first place. Oh thanks, I could've guessed that if dv=dx then v=x but I never quite trusted any algebraic manipulations of dx.


    Ahh then I was misunderstanding 'elementary functions'. I suppose I confused it with elementary methods of integrating.
    This is quite interesting! Has it been proven that [tex]\int{e^{-x^2}}dx[/tex] cannot be expressed with elementary functions, or is it just that the integral (still possible to express with elementary functions) has yet to be found?
     
  10. May 21, 2009 #9

    Cyosis

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    It has been proven that it does not exist. As for integration by parts, if you find differentials tricky you can also write it like this.

    [tex]\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x) dx[/tex].

    Now write [itex]\int \arctan x dx =\int 1*\arctan x dx[/itex]. Take [itex]f(x)=\arctan x, \;\;\; g'(x)=1[/itex]. Now try to apply integration by parts.
     
  11. May 21, 2009 #10

    Mark44

    Staff: Mentor

    Integration by parts is the integration counterpart to the product rule of differentiation.
    d/dx(f(x)*g(x)) = df(x)/dx*g(x) + f(x)*dg(x)/dx
    Multiply by dx to get d(f(x) * g(x)) = df(x) * g(x) + f(x)*dg(x)

    Antidifferentiate the above to get
    [itex]\int d(f(x) * g(x)) = \int df(x) * g(x) + \int f(x)*dg(x)[/itex]

    or
    [itex]f(x)*g(x) = \int df(x) * g(x) + \int f(x)*dg(x)[/itex]
    or finally,
    [itex]\int df(x) * g(x) = f(x)*g(x) - \int f(x)*dg(x)[/itex]
     
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