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Homework Help: Integration of sin terms

  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data
    (This is a part of the entire problem. I'm just struggling with going to the next step since it involves solving this integral.)


    $$ \int \frac {1}{\sin \theta \sqrt {R^2\sin ^2 \theta - a^2} } d\theta $$

    2. Relevant equations

    R and a are simply constants. Only $$ \theta $$ is a variable.

    3. The attempt at a solution

    I have tried performing substitutions (i.e. u = \sqrt $$ {R^2\sin ^2\theta - a^2} $$ and in another attempt u = $$ \sin \theta $$) but this seems to get me stuck in the same situation (or worsening the situation). I have also tried integration by parts, but it just seems very messy when I do it this way, and I feel like it may be the wrong approach. I might be completely missing it, but is there a particularly good substitution or method I should approach this problem with? None of the ones I've tried so far seem to work, so any suggestions would be greatly appreciated!
  2. jcsd
  3. Oct 23, 2015 #2


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    That's not going to be an easy integral but I would start with the substitution [itex]u= R^2sin^2(\theta)[/itex]. Then [itex]du= 2R^2 sin(\theta)cos(\theta) d\theta[/itex]. Now multiply both numerator and denominator by [itex]2R^2 sin(\theta)cos(\theta)d\theta[/itex]. We can write the integral as [tex]\frac{1}{2R^2}\int \frac{2R^2sin(\theta)cos(\theta)d\theta}{sin^2(\theta)cos(\theta)\sqrt{R^2sin^2(\theta)- a^2}}[/tex]
    Now [itex]sin^2(\theta)= u/R^2[/itex] and [itex]cos(\theta)= \sqrt{1- cos^2(\theta)}= \sqrt{1- \frac{u}{R^2}}[/itex] so that becomes
    [tex]\frac{1}{2}\int \frac{du}{u\sqrt{1- \frac{u^2}{R^2}}\sqrt{u^2- a^2}}[/tex]
  4. Oct 23, 2015 #3

    Simon Bridge

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    I'd look for a trig identity.
  5. Oct 23, 2015 #4

    Thank you for the response. I actually got a result very similar to yours but this still doesn't seem like an easily solvable integrable. Am I missing something? Integration by parts doesn't exactly work and and applying a trig substitution would bring me back to where I was earlier.
  6. Oct 23, 2015 #5
    Besides applying:

    $$ cos 2\theta = 1 - 2sin^2\theta $$

    is there any other obvious choice for a better simplification? This still seems to not make the integral much better.
  7. Oct 24, 2015 #6
    It seems to me that starting with a ##u = \cos \theta ## change of variable, followed by a ## v = \tanh^{-1} u ## would greatly simplify your integral. With a last hyperbolic change of variable, and a discussion over the sign of a constant depending of ##a## and ##R##, you could solve your problem.
  8. Oct 24, 2015 #7

    Simon Bridge

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    Do you have any reason to think that this will turn out to be an easy integral to solve, if only you knew the trick?
  9. Oct 24, 2015 #8
    Except for the first change of variable which is tricky, the next steps come quite naturally.
  10. Oct 25, 2015 #9
    After looking into your approach, I resorted to using WolframAlpha to help figure out this integral. I was expecting a nicer solution since it is based on a real problem, and this definitely seems off from what I was expecting. Nonetheless, thank you for the suggestion.

    Attached Files:

  11. Oct 25, 2015 #10


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    I'd start by pulling ##a## out of the integral, leaving you with
    $$ \frac 1a \int \frac{1}{\sin\theta \sqrt{b^2\sin^2 \theta-1}} \,d\theta$$ where ##b = R/a##. Then use the substitution ##u = \csc \theta##.

    Are you sure the integral you derived is correct?
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