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Math Help - A simple integral

  1. #1
    Member Jones's Avatar
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    A simple integral

    Hi,

    I'd be happy if someone could check my work on the integral sin^6(x)dx


    sin^6(x) = (sin^2(x))^3

    sin^2(x) = \frac{1}{2}(1-cos(2x))

    \frac{1}{8}\int(1-cos(2x))^3

    let u = cos(2x);

    (1-u)^2\cdot(1-u) = 1-3u+3u^2-u^3

     = \frac{1}{8}\int~1-3cos(2x)+3cos^2(2x)-cos^3(2x)

    = 3~\int cos^2(2x) = \frac{1}{2}(1+cos(2x)) = \frac{1}{2}x+\frac{1}{4}sin(2x)
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  2. #2
    Senior Member yeKciM's Avatar
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    first when u use substitution let's say "u" in your case, where is your "du" ?
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  3. #3
    Member Jones's Avatar
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    Dang, you're right, how do i deal with that?
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Jones View Post
    Dang, you're right, how do i deal with that?
    u put  u = \cos{2x} so "du" is derivation of "u" so  du = -2 \sin {2x}
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    You also can prove the reduction formula, \int sin^m(x) dx = \frac{-(cos(x) sin^{(m-1)}(x))}{m} + \frac{(m-1)}{m} \int sin^{(-2+m)}(x) dx

    And use it 3 times when your m is 6.
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  6. #6
    Senior Member eumyang's Avatar
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    Quote Originally Posted by yeKciM View Post
    u put  u = \cos{2x} so "du" is derivation of "u" so  du = -2 \sin {2x}
    Jones can't do that because in what he/she had:
    \frac{1}{8}\int(1-cos(2x))^3
    there is no sin 2x.

    Jones: from here:
    \frac{1}{8}{\displaystyle{\int}}(1-cos(2x))^3 dx
    multiply out (1-cos2x)^3 to get

    \frac{1}{8}{\displaystyle{\int}} (1 - 3 \cos 2x + 3\cos^2 2x - \cos^3 2x)\, dx

    Separate into four integrals. The first two are straight forward. For the third, re-apply the power-reduction identity
    \cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}.

    For the fourth, note that
    \cos^3 \theta = \cos^2 \theta \cdot \cos \theta = (1 - \sin^2 \theta)\cos \theta.
    Use this to evaluate the 4th integral.
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