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Basic integration

  1. Jun 15, 2017 #1
    1. The problem statement, all variables and given/known data

    Integrate: 4cos(x/2).cos(x).sin(21x/2)
    2. Relevant equations
    ▪sin(A+B)=sinAcosB+sinBcosA
    ▪2sinAcosA=sin2A
    And obviously,
    ▪Integration of sinx is (-cosx)
    ▪Integration of cosx is (sinx)
    3. The attempt at a solution
    ○I multiplied the numerator and denominator with sin(x/2)
    ○The numerator simplified to sin(2x)sin(21x/2) and in the denominator,we have sin(x/2).
    ○Now,I expanded sin(21x/2) by breaking 21x/2 into 10x and x/2 and using the above specified formula,which,on simplifying gives two terms.
    ○Now,the first term comes out to be sin2xcos10x,which can be integrated easily (well,maybe not one step but at least it can be done),but the second term is: sin(2x).sin(10x).cot(x/2)
    How to integrate this term,it just eludes me..
     
  2. jcsd
  3. Jun 16, 2017 #2
    Well I wouldn't recommend this approach. You should instead consider the product-to-sum identities in order to simplify the product of trigonometric terms into a simple sum of individual sine and cosine terms. For example, ##2 \cos \theta \cos \phi = \cos (\theta + \phi) + \cos (\theta - \phi)##.
     
  4. Jun 16, 2017 #3
    Oh..ok,got it.Thanks
     
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