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Figuring out the solution of the following Integral

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data

    Hello,I'm having problems figuring out the solution of the following Integral.Any hints would be much appreciated!


    There it is:


    I= sin(x)/sqrt{4sin(x)^2+cos(x)^2} dx
     
    Last edited: Sep 15, 2007
  2. jcsd
  3. Sep 15, 2007 #2
    [tex]I=\int\frac{sin(x) dx}{\sqrt{4sin^2(x)+cos^2(x)}}[/tex]

    O.O wow thats...a toughie, no substitution seems to work...you could try tan(x/2)=y but you'd get a polynomial of degree 4 on the square root...

    It would be so much nicer if that 4 was gone...

    ok you could try getting rid of those sin^2 x by using those formulas
    [tex]sin^2 (x) = \frac{1-cos 2x}{2}[/tex]
    [tex]cos^2 (x) = \frac{1+cos 2x}{2}[/tex]

    that might lead you somewhere decent but you'll have some problems with sin(x) on the top but square root of acos2x + or - something...
     
  4. Sep 15, 2007 #3
    Yeah,I've tried everything you've listed above,but had no success lol.
    Anyway,thank you !
     
  5. Sep 15, 2007 #4

    Gib Z

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    Not expressible in elementary functions or in a closed form.
     
  6. Sep 15, 2007 #5

    dynamicsolo

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    It's always a bad sign when it doesn't show up in Gradshteyn & Ryzhik...

    Are you sure that you're asked for the indefinite integral? The closest entry I found was [3.676 #1] in G&R, 5th edition:

    rewrite [tex]4sin^2(x)+cos^2(x)[/tex] as [tex]1+3sin^2(x)[/tex] and use

    definite integral from 0 to pi/2 of [tex]\int\frac{sin(x) dx}{\sqrt{1+p^2sin^2(x)}}[/tex]

    = (1/p) arctan p , with p^2 = 3 .

    It looks like this may be one of those integrals that only has nice solutions for *certain* definite limits, but not in general...
     
  7. Sep 15, 2007 #6
    so yea I tried it and my integral has complex numbers, so i'm going to go recheck my work.
     
  8. Sep 16, 2007 #7

    HallsofIvy

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    [tex]I=\int\frac{sin(x) dx}{\sqrt{4sin^2(x)+cos^2(x)}}= \int\frac{sin(x)dx}{\sqrt{4- 5cos^2(x)}}[/tex]

    Let u= cos(x) so du= sin(x)dx and the integral becomes
    [tex]\int\frac{du}{\sqrt{4- 5u^2}}= \frac{1}{2}\int\frac{du}{\sqrt{1- \frac{5}{4}u^2}}[/tex]
    Letting [itex]sin(\theta)= \sqrt{5}/2 u[/itex] should convert it to a rational integral.
     
  9. Sep 16, 2007 #8

    Gib Z

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    Thats some really nice work there Halls!! I wish i'd seen that =] Just One tiny mistake though, u=cos x so du = - sin x dx, so the actual integral is the one you gave, just with a negative sign stuck out the front.
     
  10. Sep 16, 2007 #9

    dynamicsolo

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    I am likewise chagrined to have missed that; I was focusing on the sine-squared term, rather than the cosine-squared one.

    One other little thing, though: 4 (sin x)^2 + (cos x)^2 =
    4 [1 - (cos x)^2] + (cos x)^2 =
    4 - 4 (cos x)^2 + (cos x)^ 2 =
    4 - 3 (cos x)^2 , no?
     
  11. Sep 16, 2007 #10

    HallsofIvy

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    +, - what does it really matter!!!:redface::cry:
     
  12. Sep 16, 2007 #11

    dynamicsolo

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    Yeah, signs are *really* only important if you go to pick something up and the nuclei turn out to have negative charge...
     
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