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

  • Thread starter K.QMUL
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  • #1
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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?
 

Answers and Replies

  • #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.
 
  • #3
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which double angle formula would I use, we still have a Sin^2 to deal with
 
  • #4
arildno
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Well, use the double angle formula in which sin^2 appears on its own, of course.
 
  • #5
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Aha, yes, Thanks for the help
 
  • #6
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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
 

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  • #7
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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
There is a sign error in the end. What is ##\displaystyle \int \cos(x) \, dx##?
 
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  • #8
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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
 
  • #9
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oh, I realise my mistake, integrating cos(x) = sin(x)
 
  • #10
vela
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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.
You can always check your answer by differentiating it and seeing if you recover the integrand.
 
  • #11
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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.
 
  • #12
vela
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What does 'A' represent?
 
  • #13
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In terms of the question: sin2
 
  • #14
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Sorry, in terms of the formula: sin2A = 0.5[1-cos2A]
 
  • #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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