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Trigonometric Integral

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data
    solve the integral.


    2. Relevant equations
    integral (5pi/6 to pi) (cosx)^4 / sqrt(1-sinx) dx


    3. The attempt at a solution
    Following the solution manual:

    integral (5pi/6 to pi) ( (cosx)^4 / sqrt(1-sinx)) * sqrt(1+sinx)/sqrt(1+sinx)

    however i am not sure how to do the algebra here.

    it should convert to integral (5pi/6 to pi) ( (cosx)^4 / sqrt(1-sin^2x)) * sqrt(1+sinx) dx
     
  2. jcsd
  3. Feb 23, 2013 #2

    LCKurtz

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    When you say "it should convert" do you mean you don't see how do get that last step or you don't know what to do next? You do know that ##1-\sin^2 x = \cos^2 x##, right? You could change all the even powers of ##\cos x## to sines and do a u substitution.
     
  4. Feb 23, 2013 #3
    solved.

    how about..

    interal (sin2x)^4 / (sqrt(1-cos2x) dx
     
  5. Feb 23, 2013 #4

    LCKurtz

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    Those problems are so similar that whatever you did for the first one (which you haven't shown us) should work on the second one.
     
  6. Feb 23, 2013 #5
    i can't remember what i did and i lost the scrap piece of paper. i think i multiplied by the conjugate but im stuck.
     
  7. Feb 23, 2013 #6

    SammyS

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    That should work.

    What are you stuck with?
     
  8. Feb 23, 2013 #7
    after multiplying by the conjugate and simplifying i get: integral (sin2x)^3 * sqrt(1+cos2x) dx
     
  9. Feb 23, 2013 #8

    SammyS

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    OK. So you have
    [itex]\displaystyle \int (\sin(2x))^3\sqrt{1+\cos(2x)\,}\,dx\ .[/itex]​

    What did LCKurtz have you do in post #2 ?
    This time all the arguments are 2x, rather than x, but so what?
     
  10. Feb 23, 2013 #9
    there are no even power cosines.
     
  11. Feb 23, 2013 #10
    Got it.
    Take u=2x and then you have to use the fact that sin^3x=sin^2x*sinx...
     
    Last edited: Feb 23, 2013
  12. Feb 24, 2013 #11

    SammyS

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    There certainly is an even power of cosine.

    cos3(2x)=cos2(2x)∙cos(2x).

    There's a cos2 for you.
     
  13. Feb 24, 2013 #12
    what about the square root with 1+cos2x, though?

    the problem is (sin2x)^3 not (cos2x)^3.
     
  14. Feb 24, 2013 #13
    in my solutions manual, in has the following as part of the solution. how is this correct algebra?

    (-2/3) + (2/3)(3/2)^(3/2) --> = sqrt(3/2) - 2/3

    what happened to the 3 in the exponent (3/2)??
     
  15. Feb 24, 2013 #14
    (2/3) is the same as 1/(3/2) = (3/2)^-1.
     
  16. Feb 24, 2013 #15

    SammyS

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    My mistake !

    (sin2x)^3 = (sin2x)2(sin2x)

         = (1-cos2(2x))sin(2x)

    That all suggests to me the substitution, u = cos(2x) .

    Added in Edit:

    Perhaps better yet is u = 1 + cos(2x). Do you see why?
     
  17. Feb 24, 2013 #16

    LCKurtz

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    @whatlifeforme: What I don't get about this thread is that in post #3 you claimed you have solved your original problem. Then you posted an almost identical problem just changing the sines to cosines and the x's to 2x's. So why are you going on and on with questions about every little step when you have already solved an essentially identical problem? What's going on here??
     
  18. Feb 24, 2013 #17
    Then you do another u substitution and then an integration by parts from what I remember doing.
     
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