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Integrating cos^6(x)

  1. Mar 27, 2013 #1
    Using [itex] cos^4\theta - sin^4\theta = cos2\theta [/itex] and [itex] cos^4\theta + sin^4\theta = 1 - \frac{1}{2}sin^22\theta [/itex]

    evaluate:

    (i) [itex]\displaystyle \int_0^{\frac{\pi}{2}} cos^4\theta \ d\theta [/itex]

    adding the two identities given I get: [itex] 2cos^4\theta = cos2\theta + 1 - \frac{1}{4} + \frac{1}{4} cos4\theta [/itex] [itex] cos^4\theta = \frac{1}{8}\left (4cos2\theta + 3 + cos4\theta \right ) [/itex]

    integrating this I get the correct answer

    (ii) [itex]\displaystyle \int_0^{\frac{\pi}{2}} cos^6\theta \ d\theta [/itex]

    from the (i) [itex] cos^4\theta = \frac{1}{8}\left( 4cos2\theta + 3 + cos4\theta \right ) [/itex]

    so [itex] cos^6\theta = \frac{1}{8}(4cos^2\theta cos(2\theta) +3cos^2\theta + cos^2\theta cos4\theta) [/itex]

    using the double angle formula I simplify down to:
    [itex] cos^6\theta = \left(4\left( \dfrac{cos2\theta + 1}{2} \right)cos2\theta + 3\left(\dfrac{cos2\theta + 1}{2} \right) + cos^2\theta cos4\theta \right) [/itex]
    [itex] cos^6\theta = \left(cos4\theta + 1 + \frac{7}{2} cos2\theta + \frac{3}{2} + cos^2\theta cos4\theta \right ) [/itex]
    considering [itex] cos^2\theta cos4\theta [/itex]

    [itex] = cos^2\theta(2(2cos^2\theta - 1)^2 - 1) = 8cos^6\theta - 8 cos^4\theta + cos^2\theta [/itex]

    so:
    [itex] cos^6\theta = \frac{1}{8}\left(cos4\theta + \frac{5}{2} + \frac{7}{2}cos2\theta + 8cos^6\theta - 8 cos^4\theta + cos^2\theta \right) [/itex]

    then the [itex] cos^6\theta [/itex] cancel out :S, any help?

    (I've solved this problem using the reduction formula and de moivres theorem but I don't see where I'm going wrong here)
     
    Last edited: Mar 27, 2013
  2. jcsd
  3. Mar 27, 2013 #2

    mfb

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    Staff: Mentor

    You can use integration by parts twice. You will get cos^6 on both sides there, too, but they don't cancel.
    Alternatively, keep cos(4θ) as it is and play around with cos^2(θ).
     
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