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Integral of cos^5 x sin x

  1. Oct 16, 2011 #1
    When I did this problem I overlooked the fact that a u-sub of u=cos x would work fine. I ended up using the more complicated trig integrals rules and did this:
    ∫cos5 x sin x dx
    ∫(cos2 x)2 cos x sin x dx
    ∫(1 - sin2 x)2 cos x sin x dx
    ∫(sin5 x - 2 sin3 + sin x) cos x dx
    with u = sin x, du = cos x dx this is a simple integral. However it gives me:
    [itex]\frac{1}{6} sin^{6} x - \frac{1}{2} sin^4 x \frac{1}{2} sin^2 x[/itex]
    The correct answer is:
    [itex]\frac{-cos^{6} x}{6}[/itex]

    Wolfram Alpha confirms that the final form of the integral I gave above still gives the correct answer. It also confirms that my answer isn't numerically equivalent to the correct answer. Also that my integration of the u-subed version is correct. Therefore, I must assume there was some mistake in the u-sub itself, either in the initial sub or the replacement after the integration. I cannot see where this mistake is though.

    Just to be clear, I get the correct answer when I use the simpler u-sub of u = cos x, and I understand that is the preferred way to go about this problem. However, as far as I can tell the more complicated way I did it should have still produced a correct answer.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 16, 2011 #2

    D H

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    You could have used a u substitution here as well with u=cos2x, du=-2cos(x)*sin(x) dx.
    And here with u=sin2x, du=2cos(x)*sin(x) dx.
    I assume this last line is a typo and that you meant [itex]\frac{1}{6} sin^{6} x - \frac{1}{2} sin^4 x + \frac{1}{2} sin^2 x[/itex].
    Your more complicated answer is also correct.

    You are forgetting the constant of integration. Your complex answer and the simple one differ by a constant.
     
  4. Oct 16, 2011 #3
    That possibility did occur to me briefly, but for some reason I quickly dismissed it. Thanks.
     
  5. Oct 16, 2011 #4

    D H

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    That constant of integration can always be written as c*1, but where "1" is written in an interesting way. A good start is to represent 1 as sin2x+cos2x. Cube this and simplify a bit and you will find a way to express one that applies to this problem, sin6x-3sin4x+3sin2x+cos6x.
     
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