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Help integrating sin^2(x-pi/6)

  1. Oct 6, 2013 #1
    Hi there everyone,

    sin^2(x-pi/6) dx

    I have the following integral to solve but am unsure where I should start, I first thought about integrating by parts as I thought you could split it into [Sin(x-pi/6)][Sin(x-pi/6)]. But couldn't seem to figure that out. I was wondering if you could use a trig identity but again am unsure which one.

    Any suggestions?
     
  2. jcsd
  3. Oct 6, 2013 #2

    arildno

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    Use the double angle formula. This is a very handy formula to reduce the exponent in a trigfunction appearing in your problem.

    Integration by parts works nicely as well, if you are careful with your notation.
     
  4. Oct 6, 2013 #3
    which double angle formula would I use, we still have a Sin^2 to deal with
     
  5. Oct 6, 2013 #4

    arildno

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    Well, use the double angle formula in which sin^2 appears on its own, of course.
     
  6. Oct 6, 2013 #5
    Aha, yes, Thanks for the help
     
  7. Oct 6, 2013 #6
    I have completed the question using the double angle formula, could you tell me if you find any errors, as im still unsure whether I have done this right or not.

    Thanks
     

    Attached Files:

  8. Oct 6, 2013 #7
    There is a sign error in the end. What is ##\displaystyle \int \cos(x) \, dx##?
     
  9. Oct 6, 2013 #8
    integrating cos(x) is -sin(x) + c, I took that in consideration as it was 1-cos(x), thus using two negatives = positive, I corrected it to x + 1/2 sin(2x- pi/3).

    Does everything look good for this integration? Since I checked on Wolfram Mathematica's on-line integration and the answer they got was

    http://integrals.wolfram.com/index.jsp?expr=sin^2(x-pi/6)&random=false
     
  10. Oct 6, 2013 #9
    oh, I realise my mistake, integrating cos(x) = sin(x)
     
  11. Oct 6, 2013 #10

    vela

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    You can always check your answer by differentiating it and seeing if you recover the integrand.
     
  12. Oct 9, 2013 #11
    So Ive completed the question, and checked if I get the original answer by differentiating it. And it seems good. HOWEVER, I have one concern; when you use the double angle formula, can I take 'A' as (x - pi/6) in sin^2(x-pi/6) or would I need to split it somehow. Please clear up my confusion.
     
  13. Oct 9, 2013 #12

    vela

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    What does 'A' represent?
     
  14. Oct 9, 2013 #13
    In terms of the question: sin2
     
  15. Oct 9, 2013 #14
    Sorry, in terms of the formula: sin2A = 0.5[1-cos2A]
     
  16. Oct 9, 2013 #15

    vela

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    In the identity, you can replace 'A' by anything as long as you replace it with the same thing everywhere. Setting A to ##(x-\pi/6)## is perfectly fine.
     
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