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Darn integral

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data

    I just can't crack the integral of (sin(x))^6 for some reason.

    What is the exact solution to this? This is not really a homework question, as an immediate reference to an integral table would be sufficient. But I just need it right away. Thanks.



    2. Relevant equations

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    3. The attempt at a solution

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  2. jcsd
  3. Nov 16, 2009 #2

    jgens

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    Gold Member

    Use integration by parts repeatedly.
     
  4. Nov 16, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Or use trig identities. [itex]cos(2x)= cos^2(x)- sin^2(x)= 1- 2sin^2(x)[/itex] so [itex]sin^2(x)= (1/2)(1- cos(2x))[/itex]. Then [itex]sin^6(x)= (sin^2(x))^3= (1/2)^3(1- cos(2x))^3[/itex][itex]= (1/8)(1- 3cos(2x)[/itex][itex]+ 3cos^2(2x)+ cos^3(2x))[/itex].

    The integral of 1- 3cos(2x) is straightforward. The integral of [itex]cos^3(2x)[/itex] can be done by writing it as [itex]cos^3(2x)= cos(2x)(1- sin^2(2x))[/itex] and using the substitution u= sin(2x). The integral of [itex]cos^2(2x)[/itex] can be done by using the trig identity [itex]cos^2(2x)= (1/2)(1+ cos(4x))[/itex].
     
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