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Simple Integration Problem

  1. Feb 22, 2008 #1
    [SOLVED] Simple Integration Problem

    1. The problem statement, all variables and given/known data
    Integrate: [tex]\int\frac{dx}{3\sin x+2}[/tex]


    2. Relevant equations
    ....none


    3. The attempt at a solution

    I tried to do this by u-sub, and that was unsuccessful, then after that I tried a trig. sub with 3sinx=2tan(theta)^2, and that got very messy, really fast. So, I'm kinda stuck, any help would be appreciated.
     
    Last edited: Feb 22, 2008
  2. jcsd
  3. Feb 22, 2008 #2
    [tex]\int\frac{dx}{3\sin x+2}[/tex]

    Yes?
     
  4. Feb 22, 2008 #3
    Yea, sorry about that.
     
  5. Feb 22, 2008 #4
    OK, I updated it with the Latex.
     
  6. Feb 22, 2008 #5
    do you mean : [tex]dt=\frac{dx}{\sqrt{x^2-1}}[/tex] ?
     
  7. Feb 22, 2008 #6
    Grr. I'm stumped. Must think harder.
     
  8. Feb 22, 2008 #7
    Yea, I've been working on and off with this one for a few hours, and I even plugged it into Mathematica's Integrator, but I get some crazily complicated stuff. I'm really in need of help with it...
     
  9. Feb 22, 2008 #8
    Well, I have this so far ...

    [tex]t=\sin x[/tex]
    [tex]x=\sin^{-1}t[/tex]
    [tex]dx=\frac{dt}{\sqrt{1-t^2}}[/tex]

    [tex]\int\frac{dt}{(3t+2)\sqrt{1-t^2}}[/tex]

    Looking at Mathematica, I may be on the right track. I'm going to try a hyperbolic substitution now.
     
  10. Feb 22, 2008 #9
    I think I'm going to start over, I need to use the so called world's sneakiest substitution, lol. If I can remember how to use it :p Gotta look through my book.
     
  11. Feb 22, 2008 #10

    VietDao29

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    Homework Helper

    When seeing some integrals like this, i.e, those trigonometric functions, that clearly cannot use other normal u-substitution, then, we must think about substituting tan(x/2).

    So, let:

    • u = tan(x/2).

      ~~~> du = (1/2) (1 + tan2(x/2)) dx = [1/2 (1 + u2)] dx

      ~~~> dx = (2 du) / (1 + u2)
    • sin(x) = (2u) / (1 + u2)

    You know this formula, right? If not, here's a simple proof:

    [tex]\sin (x) = \sin (2 \times \frac{x}{2}) = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) = 2 \cos ^ 2 \left( \frac{x}{2} \right) \frac{\sin \left( \frac{x}{2} \right)}{\cos \left( \frac{x}{2} \right)}[/tex]

    [tex]= 2 \frac{1}{\frac{1}{\cos ^ 2 \left( \frac{x}{2} \right) }} \tan \left( \frac{x}{2} \right) = 2 \frac{\tan \left( \frac{x}{2} \right)}{1 + \tan ^ 2 \left( \frac{x}{2} \right) } = \frac{2u}{1 + u ^ 2}[/tex] (Q.E.D)

    So, your integral will become:

    [tex]\int \frac{\frac{2}{1 + u ^ 2}}{3 \times \frac{2u}{1 + u ^ 2} + 2} du[/tex]

    Can you go from here? :)
     
    Last edited: Feb 22, 2008
  12. Feb 22, 2008 #11
    Very nice VietDao :-] I was lookin through my book and next refresh I see your work. Very clean!
     
  13. Feb 22, 2008 #12
    Thanks a lot, that is the first time that I've seen that substitution used. How often is it usually employed? What situations does it work for?
     
  14. Feb 22, 2008 #13
    Try this
    [tex]x=2\,\arctan u\Rightarrow u=\tan \frac{x}{2},\,d\,x=\frac{2\,du}{1+u^2}[/tex]
    with
    [tex]\sin x=\frac{2\,\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}\Rightarrow \sin x=\frac{2\,u}{1+u^2}[/tex]
     
  15. Feb 22, 2008 #14
    Oupps! VietDao29 was faster :smile:
     
  16. Feb 22, 2008 #15

    VietDao29

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    Homework Helper

    Yah, thanx a lot.. :blushing: :blushing:

    Well, it works for some types like:

    [tex]\int \frac{\alpha}{\beta \sin (x) + \gamma } dx[/tex], [tex]\int \frac{\alpha}{\beta \cos (x) + \gamma } dx[/tex], [tex]\int \frac{\alpha}{\beta \sin (x) + \gamma \cos (x) + \delta } dx[/tex], ...

    where alpha, beta, gamma, and delta are all constants. :)

    There are some more, but the 3 listed above are the most common ones..
     
    Last edited: Feb 22, 2008
  17. Feb 22, 2008 #16
    When all hope is lost :-D
     
  18. Feb 22, 2008 #17
    Oh, that's interesting, it seems I've been deprived of any exposure to those types. Thanks to everyone for answering so quickly.
     
  19. Feb 22, 2008 #18

    Gib Z

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    It reduces any rational function of sin x and cos x into a rational function of t. In fact its not when all hope is lost, its the first thing you do for many trig integrals.
     
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