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Tough Integral

  1. Aug 4, 2005 #1
    I'm working on a problem that I've almost completed except for a single integral that I can't seem to figure out. I can plug it into mathematica and get the answer, but I'd really like to know how to work it out. So if anyone can walk me through it, I'd appreciate it.

    [tex]\int\frac {k(\sin^2\theta-\cos^2\theta-k\sin^4\theta)} {(1-k\sin^2\theta)^{\frac {3}{2}}} d\theta [/tex]

    Thanks in advance!
  2. jcsd
  3. Aug 4, 2005 #2

    James R

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    Mathematica says the solution is:

    [tex]-\frac{k \sin 2\theta}{\sqrt{4 - 2k + 2k \cos 2\theta}}[/tex]

    That doesn't give you a method, though.
  4. Aug 4, 2005 #3
    It may or may not help to simplify it to something like

    [tex] k\int {\frac{{\cos {2\theta } - k\sin ^4 \theta }}
    {{\left( {1 - k\sin ^2 \theta } \right)^{\frac{3}{2}} }}d\theta } [/tex]
    [tex] k\int {\frac{{\sin ^2 \theta \left( {1 - k\sin ^2 \theta } \right) - \cos ^2 \theta }}{{\left( {1 - k\sin ^2 \theta } \right)^{\frac{3}{2}} }}d\theta } [/tex]

    *Well...it still feels like I'm missing something obvious ~
    Last edited: Aug 4, 2005
  5. Aug 4, 2005 #4
    This looks like a prime candidate for contour integration on the complex plane...
  6. Aug 4, 2005 #5
    Ahh--no wonder my CalcII couldn't do it! :tongue2:

    And I thought it was a basic CalcII integral! :grumpy:
    Last edited: Aug 5, 2005
  7. Aug 4, 2005 #6
    Thing is, contour integrals deal with definite integration. The integral stated has no limits...
  8. Aug 4, 2005 #7

    Hmm, don't think I've learnt about contour integration yet. I'll look that up. I didn't mention it before, but the integral I'm working with actually does have limits from 0 to [tex]\phi[/tex] if that helps.
  9. Aug 4, 2005 #8


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    Now, just one piece here:
    By the identity [tex]\sin^{2}\theta=\frac{1-\cos(2\theta)}{2}[/tex]
    the denominator of the integrand may be rewritten as:
    since [tex]4^{\frac{3}{2}}=8[/tex]

    Perhaps I'll try to do something further later on.
    Last edited: Aug 4, 2005
  10. Aug 4, 2005 #9
    Does phi have an upper limit? You've got to be careful with contour integration with the values of functions, what with functions such as log z being multi-valued at certain points (leading to lovely things such as branch points, Reimann surfaces, that kind of jazz).

    I suggest Boas (Mathematical Methods in the Physical Sciences) for an introduction to Contour Integration (look at Cauchy Residue Theorem - an integral and beautiful part of solving things like this).
  11. Aug 5, 2005 #10
    Integrals of type [tex]\int R(\sin x, \cos x)dx[/tex], where R is rational function are solved by following substitution [tex]\tan \frac{x}{2}=t[/tex], thus, [tex]\sin x=\frac{2t}{1+t^2},\quad \cos x=\frac{1-t^2}{1+t^2},\quad dx=2\frac{dt}{1+t^2}[/tex], and the previous integral is reduced to integral of a rational function which I hope you know how to solve.
  12. Aug 5, 2005 #11
    Hmm, that seems to make sense, but it gets me to an integral I'm not having much more luck with. The only thing I can think of is some other subsitution that I can't figure out.

    [tex] - 2k\int {\frac{{t^8 - 4t^6 + 2(8k - 5)t^4 - 4t^2 + 1}}{{(1 + t^2 )^3 (1 - \frac{{4kt^2 }}{{(1 + t^2 )^2 }})^\frac {3}{2} }}} [/tex]

    Making [tex] u=1 - \frac{{4kt^2 }}{{(1 + t^2 )^2 }} [/tex] almost seems to work, but not quite. Any advice on what to do from here?
    Last edited: Aug 5, 2005
  13. Aug 5, 2005 #12
    I've misread your integral, it has [tex]\frac{3}{2}[/tex] exponent in denominator so it is irational not rational function. I guess I'll have to take paper and pencil. If I get to some solution, I will post it.
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