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Homework Help: Asking For Detailed Integrals Solutions

  1. Jun 25, 2009 #1
    1. The problem statement, all variables and given/known data

    I would like a detailed solution of following Integrals:

    [tex]\int^{z_{2}}_{z_{1}} 1/ \left(s^{2}+z^{2}\right)^{1/2} dz[/tex]

    2. Relevant equations

    [tex]1+tan^{2} \theta = sec^{2} \theta[/tex]

    3. The attempt at a solution

    [tex]\int 1/ \left(s^{2}+z^{2}\right)^{1/2} dz = \int 1/ \left[s\left(1+ \left(z/s \right)^{2}\right)^{1/2} \right] dz[/tex]


    [tex]z = s \tan \theta \rightarrow dz = s \sec^{2} \theta d \theta[/tex]
    [tex]\theta = tg^{-1} \left(z/s\right) [/tex]


    [tex]\int^{z_{2}}_{z_{1}} 1/ \left[s\left(1+ \left(z/s \right)^{2}\right)^{1/2} \right] dz = \int^{\theta_{2}}_{\theta_{1}} s \sec^{2} \theta/ \left[s\left(1+tan^{2} \theta \right)^{1/2} \right] d\theta[/tex]

    [tex]\int^{\theta_{2}}_{\theta_{1}} s \sec^{2}\theta / \left[s \sec \theta \right] d\theta = \int^{tg^{-1} \left(z_{2}/s\right)}_{tg^{-1} \left(z_{1}/s\right)} sec \theta d \theta[/tex]

    However this not give me the answer i need here (i checked final solution):

    [tex]\int^{z_{1}}_{z_{2}} 1/ \left(s^{2}+z^{2}\right)^{1/2} dz = \left[ln\left(z^{2}+\sqrt{z^{2}+s^{2}\right)}\right]\right|^{z_{2}}_{z_{1}}[/tex]

    Please, help!
    Last edited: Jun 25, 2009
  2. jcsd
  3. Jun 25, 2009 #2
    Another one:

    [tex]\int u/ \left(R^{2}+r^{2}+2Rru\right)^{1/2} du[/tex]
    Last edited: Jun 26, 2009
  4. Jun 25, 2009 #3


    User Avatar
    Homework Helper

    Are you sure that's not a correct answer? Just because your solution looks different doesn't mean it's not the same!

    For instance, note that the hyperbolic trig function sinh and cosh satisfy

    [tex]1 + \sinh^2x = \cosh^2x[/tex],

    so you could just as well make the substitution [itex]z = s\sinh\theta[/itex]. Your answer would then be in terms of an inverse hyperbolic function - which can be expressed in terms of a logarithm. I would suspect that's why your answer doesn't match what you looked up.

    Edit: this refers to the integral in the first post.

    Also, if you want to do fractions, you use: \frac{numerator}{denominator}
    Just be careful with the sqiggly brackets when doing complex fractions, as missing just one of those tends to completely ruin the output (my latex pdf compiler won't even compile anything if that happens)
  5. Jun 25, 2009 #4


    User Avatar

    you can simplify s sec^2 theta/ s sec theta into
    \int^{\theta_{2}}_{\theta_{1}} \sec \theta \right] d\theta
    which equals ln|sec(theta) + tan(theta)|, and then reverse substitute the z.
  6. Jun 25, 2009 #5


    Staff: Mentor

    Since you are integrating with respect to z, and assuming that R, r, and u are not functions of z, what you have there is the integral of a constant. It doesn't matter that the constant is complicated by having a radical in the denominator.

    In other words,
    \int \frac{dz}{K} = \frac{1}{K}z + C

    On the other hand, if this is what you got after doing a substitution, then you didn't go far enough, since you have both u and dz. When you do a substitution, your new integral should not have any of the old variables in it, including the differential.
  7. Jun 26, 2009 #6
    Sorry, i made a typing mistake here, the integral is:

    [tex]\int u/ \left(R^{2}+r^{2}+2Rru\right)^{1/2} du[/tex]
    Last edited: Jun 26, 2009
  8. Jun 26, 2009 #7
    I cannot deal with arctg z in the argument of sec ( ). Its necessary to my purpose find a solution without arctg or sec ( ). However Mute's suggestion would be usefull, i ll try this one and made another attempt.
  9. Jun 26, 2009 #8


    Staff: Mentor

    I would start with this substitution: w = R2 + r2 + 2Rru. Another one I would try if the first one didn't get me anywhere is w = (R2 + r2 + 2Rru)^(1/2)
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