1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration question

  1. Jun 27, 2010 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations


    3. The attempt at a solution

    [tex]

    \mbox{Let }x = R\\tan\theta \Rightarrow dx = R\\sec^{2}d\theta\\
    x^{2} = R^{2}\\tan^{2}\theta\\
    \Rightarrow R^{2} + x^{2} = R^{2}\\( 1 + tan^{2}\theta)\\
    \Rightarrow R^{2} + x^{2} = R^{2}\\sec^{2}\\
    \Rightarrow (R^{2} + x^{2})^{3/2} = R^{3}\\sec^{3}\theta\\
    \\
    Therefore, \int\frac{dx}{(R^{2}+x^{2})^{3/2}}\\
    =\int\frac{R\\sec^{2}\theta}{R^{3}\\sec^{3}\theta}\\d\theta\\
    =\frac{1}{R^{2}}\\ \int^{\infty}_{-\infty}cos\theta\\ d\theta\\
    =\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty} \int^{N}_{-N}cos\theta\\ d\theta\\
    =\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(sin\theta)\mid^{N}_{-N}\\
    =\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(sin N + sin N)
    =\frac{2}{R}\stackrel{lim}{n\rightarrow\infty}sin N
    [/tex]

    and then??????????????????????
     
    Last edited: Jun 27, 2010
  2. jcsd
  3. Jun 27, 2010 #2
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations


    3. The attempt at a solution

    [tex]\mbox{Let }x = R\\tan\theta \Rightarrow dx = R\\sec^{2}d\theta[/tex]

    [tex]x^{2} = R^{2}\\tan^{2}\theta[/tex]

    [tex]\Rightarrow R^{2} + x^{2} = R^{2}\\( 1 + tan^{2}\theta)[/tex]

    [tex]\Rightarrow R^{2} + x^{2} = R^{2}\\sec^{2}\theta[/tex]

    [tex]\Rightarrow (R^{2} + x^{2})^{3/2} = R^{3}\\sec^{3}\theta[/tex]

    [tex]\mbox{Therefore, }\int\frac{dx}{(R^{2}+x^{2})^{3/2}}[/tex]

    [tex]=\int\frac{R\\sec^{2}\theta}{R^{3}\\sec^{3}\theta}\\d\theta[/tex]

    [tex]=\frac{1}{R^{2}}\\ \int^{\infty}_{-\infty}cos\theta\\ d\theta[/tex]

    [tex]=\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty} \int^{N}_{-N}cos\theta\\ d\theta[/tex]

    [tex]=\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(sin\theta)\mid^{N}_{-N}[/tex]

    [tex]=\frac{1}{R^{2}}\stackrel{lim}{n\rightarrow\infty}(sin N + sin N)[/tex]

    [tex]=\frac{2}{R^{2}}\stackrel{lim}{n\rightarrow\infty}sin N[/tex]

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



    2. Relevant equations



    3. The attempt at a solution
     
  4. Jun 27, 2010 #3

    Char. Limit

    User Avatar
    Gold Member

    Why did you suddenly switch from indefinite to definite (across the entire real axis, no less) integrals partway through? I would assume that you continue to use the indefinite integral.
     
  5. Jun 27, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Fixed your LaTex.
    You need to change the limits of the integral when you change variables.
     
  6. Jun 27, 2010 #5
    I thought that an indefinite integral is just a definite integral with limits that are infinity.
    Isn't that true?
     
  7. Jun 27, 2010 #6
    Nope, that is unfortunately not true at all.
     
  8. Jun 27, 2010 #7
    But [tex]\theta[/tex] from [tex]\infty[/tex] to [tex]-\infty[/tex] is the same as [tex]x[/tex] from [tex]\infty[/tex] to [tex]-\infty[/tex]. Isn't it???
     
  9. Jun 27, 2010 #8
    What is an indefinite integral, then?
     
  10. Jun 27, 2010 #9

    zcd

    User Avatar

    [tex]\lim_{\theta\to\infty}{\tan(\theta)}[/tex] doesn't exist. You have a better chance reversing the substitution before taking the limits.
     
  11. Jun 27, 2010 #10
    An indefinite integral simply produces the antiderivative of the function being integrated.
    [tex]\int f(x) dx = F(x) + c[/tex]
    where differentiating F(x) yields the function f(x).

    Meanwhile, a definite integral of a function over a specified interval is equal to the difference between the values of the antiderivative at the endpoints of the interval.
    [tex]\int_{a}^{b}f(x) dx = F(b) - F(a)[/tex]

    One can view the antiderivative as the basis with which we employ the fundamental theorem of calculus to compute definite integrals.
     
  12. Jun 27, 2010 #11

    Mark44

    Staff: Mentor

    As already noted, that's not true. You might be thinking of an improper integral.
     
  13. Jun 27, 2010 #12

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integration question
  1. Integral Questions (Replies: 2)

  2. Integration question (Replies: 2)

  3. Integral Question (Replies: 5)

  4. Integral Question (Replies: 8)

  5. Integration Question (Replies: 4)

Loading...