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Trigonometric Integrals

  1. Aug 23, 2008 #1
    1. The problem statement, all variables and given/known data
    ∫ cosx+sin 2X/sinX


    2. Relevant equations
    sin^2 x =1/2 (1-cos2x), cos^2=1/2 (1+cos2x)
    if cosine is odd, u sin x, cos^2 x=1-sin^x) , if sine is odd u=cosx sin^2x =1-cos^2x)


    3. The attempt at a solution
    i'm not sure how to start this one because i've never came across a function like this before
     
  2. jcsd
  3. Aug 23, 2008 #2

    HallsofIvy

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    First, clarify. Do you mean cos(x)+ (sin(2x)/sin(x)) or do you mean (cos(x)+ sin(2x))/sin(x)?

    Along with formulas for sin2(x) and cos2(x) you also need sin(2x)= 2 sin(x)cos(x).
     
  4. Aug 23, 2008 #3

    tiny-tim

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    Hi afcwestwarrior! :smile:

    Hint: one of the standard trigonometric identities …

    sin2X = 2 sinX cosX :smile:
     
  5. Aug 23, 2008 #4
    I mean (cos(x)+sin(2x))/ sin(x)

    so it would be (cos(x) + 2 sin(x)cos(x))/sin(x)
     
  6. Aug 23, 2008 #5
    Thanks guys
     
  7. Aug 23, 2008 #6
    so would it be like this
    cos(x)/sin(x) - 2 sin(x) cos(x)/ sin (x)= cos(x)/sin(x) - 2cos(x)
     
  8. Aug 23, 2008 #7
    isn't cosX/sinx= to something, i forgot
     
  9. Aug 23, 2008 #8

    tiny-tim

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    That's right! :smile:

    And both those are easy to integrate (hint: one's a ln).
     
  10. Aug 23, 2008 #9
    I already found the answer thanks once again.
     
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