Another integration

  1. [tex]\int \frac{1}{x^2 +4} [/tex] How to integrate this without knowing the derivatives of inverse function?
     
    Last edited: Dec 28, 2005
  2. jcsd
  3. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

    If you don't know about an anti-derivative of the integrand, or are unable to transform your integrand in such a manner that an anti-derivative becomes apparent, then the fundamental theorem of calculus is of minor use to you in that particular case in the evaluation of your integral.
     
  4. I think he/she is asking whether you need to know inverse trig derivatives/integrals to do this problem. You don't need to know the derivative of inverse trigonometric functions to do this problem, however you need to know about trig substitutions, if you can do trig substitutions then this integral can be done in two steps, one is a regular substitution that gets it in a form such as 1/(u^2 +1) and another trig substitution after that.
     
  5. You can use some basic algebra and the method of partial fractions..
    If you facter (x^2 + 4) you get (x + 2i)(x - 2i) where i is the complex number, sqrt(-1).
    From there you can decompose 1/(x^2 + 4) into two fractions:

    1/(x^2 + 4) = A/(x + 2i) + B/(x - 2i)

    And you'll end up with a complex logarithm as the answer-- which will fit the definition of arctan, if I'm not mistaken. (Or something like that)

    Either way plugging numbers in will still get you the same results as the real answer. :)
     
  6. Could you explain further?
     
  7. I'll assume you can do the first step, then use the trig substitution x = tan(θ). Remember after you have your answer you have to get it back in terms of x, your answer will be in terms of theta.

    Edit: forgot a comma and a then, it might make it confusing.
     
    Last edited: Dec 28, 2005
  8. VietDao29

    VietDao29 1,422
    Homework Helper

    Are you sure the substitution is [itex]x = \tan \theta[/itex]?
    @frozen7:
    You should look it up in your textbook or try here
    So, what's the substitution? Is it [itex]x = \tan \theta[/itex]? Or what is it?
     
  9. in the integral, you're missing the dx.

    draw a triangle, it should help you see which values to substitute.
     
    Last edited: Dec 28, 2005
  10. If you know integral of inverse function it's easy.
    At first [tex]\int \frac 1 {x^2+1} dx = \arctan(x)[/tex]

    Now put [tex]u=2x[/tex] and you'll get

    [tex] \frac 1 8 \int \frac 1 {u^2+1} du = 1/8 \arctan(2x)[/tex]
     
    Last edited: Dec 28, 2005
  11. maverick, the substitution is not u=2x but u=x/2

    [tex]du = \frac{dx}{2}[/tex]
    [tex]\frac{1}{x^2+4} = \frac{1}{4(\frac{x^2}{4} + 1)} = \frac{1}{4( ( \frac{x}{2} )^2 + 1)}[/tex]

    marlon
     
    Last edited: Dec 28, 2005
  12. oh my bad! Thank you! I'm a bit euphoric because my favorite female speed skater Tomomi Okazaki was chosen for Trino Olympic :smile:

    Then correction:

    [tex] \frac 1 2 \int \frac 1 {u^2+1} du = \arctan(x/2)[/tex]

    hm, this looks smarter.
     
    Last edited: Dec 28, 2005
  13. I don't wanna be whinning but the right hand side of your equation is not correct either. You are forgetting something :)

    marlon
     
  14. ooops!!! lol

    [tex]1/2 \arctan(x/2)[/tex]

    maybe i need sleep. It's 3:30am.
     
  15. Aha, that happens to me too.

    Here in Western Europe it's 19.30 pm and i am looking foreward to attend the SAW2 premiere tonight.

    OOHH YES, THERE WILL BE BLOOD

    Sleep well :)

    marlon
     
    Last edited: Dec 28, 2005
  16. Got up now..but it's still 7:40am..I'll sleep again.. :smile: New Years day my family will pay first visit to a nearby shrine and my father's grave, and from Jan.2 I'll be "Home alone" because my wife and daughters will be in my wife's home town for nearly a week. Still sleepy... nite... (looks like I'm hijacking this thread..:biggrin: this is a users' forum......) zzzzz...
     
  17. What's up with all of the arctans? Why not be frank with the function and do what you gotta do..... use partial fractions!?
     
  18. Because
    [tex]\int \frac 1 {u^2+a^2} du = \frac 1 {a} \arctan( \frac u {a} ) + C[/tex]
     
    Last edited: Dec 29, 2005
  19. Because than you would be needing complex functions. That's against the rules if the calculus must remain "real"...

    [tex]\frac{1}{x^2+a^2} = \frac{1}{x+ia}\frac{1}{x-ia} [/tex]

    marlon
     
  20. Here is the solution i was hinting at, without knowing the derivative of inverse tangent...
    [tex]\int \frac{1}{x^2+4} dx \ u = x/2 \ du =1/2 dx
    [/tex]
    [tex]1/2 \int \frac{1}{u^2+1} du \ u =\tan(\theta) \ du = \sec^2(\theta) d\theta
    [/tex]
    [tex]= 1/2 \int \frac{\sec^2(\theta)}{\tan^2(\theta) +1} d\theta = 1/2 \int \frac{\sec^2(\theta)}{\sec^2(\theta)} d\theta = 1/2 \int 1 \ d\theta
    [/tex]
    [tex]= 1/2(\theta + C)[/tex]
    Now we have to bring back in terms of x so we use the way we defined the substitutions to help us..
    [tex] \theta = \arctan(u) = \arctan(x/2)[/tex]
    therefore....
    [tex]\int \frac{1}{x^2+4} dx = 1/2\arctan(x/2) +C [/tex]
     
    Last edited: Dec 29, 2005
  21. It will-- you could then convert back to arctan using the definition of the function from Oiler's formula.... the final answer will always be real wherever arctan is defined as real.. so why not?
     
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