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Simple integral

  1. Feb 23, 2010 #1
    Here's an easy one, I know the answer but can't get there...

    1. The problem statement, all variables and given/known data
    [tex]\int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx[/tex]

    2. Relevant equations
    na


    3. The attempt at a solution
    I know this approach is wrong, but why? Or am I not finishing somehow?

    [tex]\int\frac{1}{x^3}dx+\int\frac{1}{z^3}dx[/tex]

    [tex]\int{x^{-3}}dx+\int{z^{-3}}dx[/tex]

    [tex]\frac{x^{-2}}{-2}+xz^{-3}[/tex]


    The correct answer should be:
    [tex]\frac{x}{z^2(x^2+z^2)^{1/2}}[/tex]
     
    Last edited: Feb 23, 2010
  2. jcsd
  3. Feb 23, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    For one

    (x+y)n≠xn+yn

    and secondly

    1/(a+b) ≠ 1/a + 1/b


    What you will need to do is make trig substitution. Try x=ztanθ.
     
  4. Feb 23, 2010 #3
    That doesn't seem right- how would that help? Seems like substitution would work here...

    This seems closer...

    [tex]
    \int(x^2+z^2)^\frac{-3}{2}}dx
    [/tex]

    [tex]
    \frac{(z^2+x^2)^\frac{-1}{2}}{\frac{-1}{2}}
    [/tex]

    I kind of see the

    [tex]\frac{x}{z^2}[/tex]

    term, but why no

    [tex]\frac{x^3}{3}[/tex]
     
  5. Feb 23, 2010 #4

    Mentallic

    User Avatar
    Homework Helper

    Because [tex]\frac{d}{dx}\left(\frac{(z^2+x^2)^\frac{-1}{2}}{\frac{-1}{2}}\right)=(x^2+z^2)^\frac{-3}{2}}.2x[/tex]


    Why don't you at least try it and see how it might turn out?
    Think a little here: what trig substitutions do you know? Mainly along the lines of [itex]sin^2x+cos^2x=1[/itex]
     
  6. Feb 23, 2010 #5

    Mark44

    Staff: Mentor

    Somebody (it might have been Einstein) once said, "Make things as simple as possible, but no simpler."

    You are making things more simple than is possible. In your first post you made two errors (already pointed out by rock.freak667) that a calculus student should not be making.
    [tex]\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}[/tex]
    [tex]\frac{1}{a + b} \neq \frac{1}{a} + \frac{1}{b}[/tex]

    Errors like these show that you are either forgetting what you learned in your algebra course, or that you didn't learn some pretty basic concepts very well. If you don't have a firm grasp on the basic algebra and trig skills, you will almost certainly have a very difficult time even following the example problems in calculus, let alone being able to solve the problems on your own.

    The math classes you have taken don't stand alone; each one forms part of the foundation for the classes that follow. If you have a difficult time in one of these classes, it will make being successful in the next class even harder.

    The third mistake was in your attempt to integrate using an ordinary substitution. Integration using this technique can be described as the integration counterpart of the chain rule in differentiation. Unless you have a solid understanding of the chain rule, you will have a difficult time using this basic technique of antidifferentiation (integration).

    If you'll take some advice, I would advise you to identify any algebra and trig areas where you might not be as strong as you should be, and go back to your algebra or trig textbooks and review these sections and work as many problems as necessary to get yourself back up to where you need to be.
     
  7. Feb 23, 2010 #6
    Unfortunately I took 4+ years of university calculus over two decades ago. I've gone well beyond this but obviously haven't used these skills in a very long time, which is why I'm having trouble getting started. I'm not looking for anyone to do my homework for me, I just can't remember very much at all. I guess I'll just ask someone who is willing to show me.
     
  8. Feb 23, 2010 #7

    Mark44

    Staff: Mentor

    With your background, it shouldn't take too long to review the basics of algebra and trig. Once you have learned something, even though you have forgotten much of it, you can relearn it more easily than you did the first time.

    Lots of people are willing to help here at Physics Forum, including myself. rock.freak667 and Mentallic gave some good advice in this thread.
     
  9. Feb 23, 2010 #8
    [tex]x^2=z^2tan^2\theta[/tex]

    [tex]dx=zsec^2\theta{d\theta}[/tex]

    [tex]
    \int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx
    [/tex]

    [tex]\int\frac{zsec^2\theta}{(z^2+z^2tan^2\theta)^\frac{3}{2}}d\theta[/tex]

    [tex]\int\frac{zsec^2\theta}{(z^2(1+tan^2\theta)^\frac{3}{2}}d\theta[/tex]

    [tex]\int\frac{zsec^2\theta}{z^2(sec^2\theta)^\frac{3}{2}}d\theta[/tex]

    [tex]\int\frac{sec^2\theta}{z(sec^2\theta)^\frac{3}{2}}d\theta[/tex]

    [tex]\int\frac{1}{z(sec^2\theta)^\frac{1}{2}}d\theta[/tex]

    [tex]\int\frac{1}{z(sec\theta)}d\theta[/tex]

    Ok, where did I mess up this time...
     
    Last edited: Feb 23, 2010
  10. Feb 23, 2010 #9

    Mark44

    Staff: Mentor

    In the line above, you have (z2)3/2. That's z3.
    With the correction noted above, and moving the z factor outside the integral (which is done with respect to x only), you have the following. What's a simple identity you can use to replace sec(theta) in the denominator?
    [tex]\frac{1}{z^2}\int\frac{1}{(sec\theta)}d\theta[/tex]

    The resulting integral is very easy to do.
     
  11. Feb 23, 2010 #10
    [tex]\frac{1}{z^2}\int\cos\theta d\theta[/tex]

    [tex]\frac{1}{z^2}\sin\theta +c[/tex]

    ???
     
  12. Feb 23, 2010 #11

    Mark44

    Staff: Mentor

    Yes. Now all you need to do is to undo your trig substitution. You had x = z*tan(theta), and you want to replace sin(theta) by something involving x and z.

    When I worked this problem, to make the trig substitution, I drew a right triangle with the adjacent side labeled z, the opposite side labeled x, and the hypotenuse labeled sqrt(x^2 + z^2). One acute angle was labeled theta. From this triangle you can get sin(theta).
     
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