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Tricky Integral (need help with substitution)

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]
    {\int_{}^{}}{ \frac{ds}{{({s}^{2}+{d}^{2})}^{\frac{3}{2}}}}
    [/tex]

    [itex]s[/itex] [itex]\equiv[/itex] variable

    [itex]d[/itex] [itex]\equiv[/itex] constant

    2. Relevant equations

    u-substitution techniques for integration.

    3. The attempt at a solution

    This integral is particularly tricky as I have already made several attempts using conventional u-substitution, however this integral is not coming out right.

    Below is my best attempt,

    If I split the denominator and multiply the top and bottom by [itex]s[/itex],

    [tex]
    {\frac{sds}{{s{({s}^{2}+{d}^{2})}^{\frac{1}{2}}}{{({s}^{2}+{d}^{2})}^{1}}}}
    [/tex]

    And let,

    [tex]
    u = {s}^{2}+{d}^{2}
    [/tex]

    [tex]
    du = 2sds
    [/tex]

    With the substitution yields,

    [tex]
    {\frac{1}{2}}{\int_{}^{}}{\frac{du}{{{u}^{\frac{3}{2}}}{\sqrt{{u}-{d}^{2}}}}}
    [/tex]

    However, this seems to get me no where.

    Any help is appreciated.

    Thanks,

    -PFStudent

    P.S.: I do realize I can look this up in the integraion tables, however I would like to know how to do this on my own without using a table.
     
    Last edited: Sep 19, 2007
  2. jcsd
  3. Sep 19, 2007 #2

    Gib Z

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    Homework Helper

    Let s= d tan x, so ds = d sec^2 x dx. Then the integral becomes [tex]\int \frac{d \sec^2 x}{ ( d^2 \tan^2 x + d^2)^{3/2}} dx. Shouldn't be too hard from there.
     
  4. Sep 19, 2007 #3

    dynamicsolo

    User Avatar
    Homework Helper

    As a bit of advice, when you see a term in your integrand that looks like

    [tex]{{({s}^{2}+{d}^{2})}^{\frac{1}{2}}}[/tex] ,

    [tex]{{({s}^{2}-{d}^{2})}^{\frac{1}{2}}}[/tex] , or

    [tex]{{({d}^{2}-{s}^{2})}^{\frac{1}{2}}}[/tex] ,

    it is an immediate candidate for the "trigonometic substitution" method when there is no (s ds) term outside the radical.
     
  5. Sep 19, 2007 #4
    try letting s = d*tan(s)
     
  6. Sep 19, 2007 #5

    dextercioby

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    Science Advisor
    Homework Helper

    Make the substitution [itex]s=d\sinh t [/itex] and you'll get 1/d^2 times

    [tex]\int \frac{dt}{\cosh^2 t}=\tanh t +C [/tex]
     
  7. Sep 19, 2007 #6
    Hey,

    Thanks for the help guys.

    Yea, I forgot that I could use the technique of Trigonometric substitution for Integrals.

    Yea, so what I did was the following,

    [tex]
    {\int_{}^{}}{ \frac{ds}{{({s}^{2}+{d}^{2})}^{\frac{3}{2}}}}
    [/tex]

    [tex]
    {\int_{}^{}}{\frac{ds}{\left[{\left({\frac{{d}^{2}}{1}}{\frac {1}{{d}^{2}}}\right)}{\left({s}^{2}+{d}^{2}\right)\right]^{\frac{3}{2}}}}}
    [/tex]

    [tex]
    {\int_{}^{}}{\frac{ds}{{d^3}{\left({\left(\frac{s}{d}\right)}^{2}+{1}^{}\right)}^{\frac{3}{2}}}}
    [/tex]

    Letting,

    [tex]
    {tan}{\theta} = {\frac{s}{d}}
    [/tex]

    and considering the trigonometric identity,

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

    [tex]
    {\int_{}^{}}{\frac{ds}{{d^3}{\left({sec}^{2}{\theta}\right)}^{\frac{3}{2}}}}
    [/tex]

    Rewriting,

    [tex]
    {s} = {d}{tan}{\theta}
    [/tex]

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

    [tex]
    {\int_{}^{}}{\frac{\left({d}{sec}^{2}{\theta}{d{\theta}}\right)}{{d^3}{\left({sec}^{2}{\theta}\right)}^{\frac{3}{2}}}}
    [/tex]

    Simplifying and factoring out all constants,

    [tex]
    {\frac{1}{{d}^{2}}}{\int_{}^{}}{{cos}{\theta}{d\theta}}
    [/tex]

    Integrating and noting that,

    [tex]
    {sin}{\theta} = {\frac{s}{\sqrt{{s}^{2}+{d}^{2}}}}
    [/tex]

    Yields,

    [tex]
    {\frac{1}{{d}^{2}}}\left[{\frac{s}{\sqrt{{s}^{2}+{d}^{2}}}}+{C_{1}}\right]
    [/tex]

    Reducing to,

    [tex]
    {\frac{s}{{{d}^{2}}{\sqrt{{s}^{2}+{d}^{2}}}}+{C}
    [/tex]

    Thanks for the help Gib Z, dynamicsolo, JonF. Also, I did not know I could do the subsititution with hyperbolic trigonometric functions, thanks dextercioby.

    -PFStudent
     
    Last edited: Sep 19, 2007
  8. Sep 19, 2007 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You may have noticed that dextercioby always prefers hyperbolic trig substitutions!
     
  9. Sep 19, 2007 #8

    dynamicsolo

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    Homework Helper

    Yes, I was going to say that, while the hyperbolic trig substitutions are certainly elegant, a lot of people in first-year calculus are not even shown those functions. Indeed, outside of physics and some branches of engineering, hardly anyone even uses them, so they are usually discussed or even introduced only in courses for students majoring in those fields...
     
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