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Integration with Trig Substitution

  1. Mar 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Ok, so I was doing a problem on the electric field strength of a continuous charge distribution and I arrived at this seemingly easy integral

    [tex] \int \frac{1}{({l^2+a^2})^\frac{3}{2}} dl[/tex]
    sorry the latex is lagging badly, you can see the correct integral by clicking on it. it is 1/l squared + a squared with the denominator to the power of 3/2.

    2. Relevant equations

    3. The attempt at a solution
    I do not know how to solve this.... at all, and I am pretty sure that this integral is correct.
    Last edited: Mar 11, 2008
  2. jcsd
  3. Mar 11, 2008 #2
    Recheck your latex and include which variable you're integrating wrt.
  4. Mar 11, 2008 #3
    This requires nothing more than a trig substitution.

    [tex]dl=a\sec^{2}\theta d\theta[/tex]

    [tex]*\cos^2 x=\frac{1}{2}(1+\cos{2x})[/tex]
    Last edited: Mar 11, 2008
  5. Mar 11, 2008 #4
    omggggg........ this is killing me! wth invented integration

    The final answer was supposed to be trigo free. Can you hint a little more, Im a little confused =x

    btw it is supposed to be a definite integral from infinity to negative infinity.

    oh yes, I am able to convert [tex]cos\theta[/tex] and [tex]sin\theta[/tex] to [tex]\frac{a}{(l^2+a^2)^\frac{1}{2}}[/tex] or[tex] \frac{l}{(l^2+a^2)^\frac{1}{2}}[/tex]


    that [tex]\theta[/tex] refers to the angle that [tex](l^2+a^2)^{0.5}[/tex] makes to the vertical.

    By the way, this is a question of fidning the electric field strength of an infinitely long charged wire.
    Last edited: Mar 11, 2008
  6. Mar 11, 2008 #5
    You don't have to use trig sub., but I find it the easiest. Here is another thread with the exact problem you are doing except with different letters.

    bob suggests to factor out the constant ... if you need more help post

  7. Mar 11, 2008 #6

    I guess I should start on my calculus 1 course first before attempting intro physics.

    Something i take comfort in though, I managed to come up with the electric field strength after I worked with the correct solution in the page your provided.
    Last edited: Mar 11, 2008
  8. Mar 11, 2008 #7
    You haven't taken Calculus? Does this Intro course only require like a wimpy Intro or Elements to Calculus course?
  9. Mar 11, 2008 #8
    nope, Im actualyl studying for my exemption test where I will get exempted rfom first course modules if I pass the tests.

    I did study some calculus in high school which was required from the syllabus, either I am rusty, or they did not cover enough ground.
  10. Mar 11, 2008 #9
    What courses are in the "first course modules"?

    Referring to Calculus: Trust me, it's worth taking the course. If you need more convincing, post in the Academic sub-forum. What you learn in a College Calculus course is far more in-depth than what you learned in HS.
  11. Mar 11, 2008 #10
    six modules in total:

    Cal 1, linear algebra, diff equations

    mechanics, Waves, EM
  12. Mar 11, 2008 #11
    ... You want to skip all those courses?
  13. Mar 11, 2008 #12
    If I could... Why not? I still have 5 months left,

    Im done with mechanics which was easy, halfway through EM, guess i should start on Cal 1 soon.

    oh yes, diff equations, I guess I will need them badly when doing electric potentials.
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