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Math Help - Integration

  1. #1
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    Post Integration

    Hey all,

    What are the steps to solve the following integral:

    \int_{0}^{\pi/2}sin^2xcos^6xdx


    kind regards,

    Dadon
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  2. #2
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    Post

    I got a email twice saying I have a reply on this thread? (by bjalongi and MathGuru). But nothing here? Is there a problem with MHF site? or did some1 delete their post?
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  3. #3
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    Hello, dadon!


    sin^2(x) cos^6(x) dx


    Even-powered sines and cosines are the worst case.
    It requires both double-angle identities.

    . . . sin²θ .= .½(1 - cos2θ) . . . . cos²θ .= .½(1 + cos2θ)


    The first factor is: .sin²(x) .= .½(1 - cos2x)


    The second factor is: .cos^6(x) .= .(cos²x)³ .= .[½(1 + cos2x)]³

    . . . = .(1/8) (1 + 3cos2x + 3cos²2x + cos³2x)


    Multiply: . ½(1 + cos2x)·(1/8)(1 + 3cos2x + 3cos²2x + cos³2x)

    . . . . . . = .(1/16) [1 + 2cos(2x) - 2cos³(2x) - cos^4(2x)] [1]


    The last term is: . cos^4(2x) .= .[cos²2x]² .= .[½(1 + cos4x)]²

    . . . . . . = .¼[1 + 2cos4x + cos²4x] .= .¼[1 + 2cos4x + ½(1 + cos8x)]

    . . . . . . = .(3/8) + (1/2)cos4x + (1/8)cos8x


    Substitute into [1]: .(1/16) [1 + 2cos2x - 2cos³2x - (3/8) - (1/2)cos4x - (1/8)cos8x]

    . . . . . = .(1/16) [(5/8 ) + 2cos2x - 2cos³2x - (1/2)cos4x - (1/8)cos8x]

    . . . . . = .(1/128) [5 + 16cos2x - 16cos³2x - 4cos4x - cos8x]


    The cosines are all to odd powers.
    . . I assume you can do the integration now.

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  4. #4
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    re:

    Quote Originally Posted by Soroban View Post

    . . . = .(1/8) (1 + 3cos2x + 3cos²2x + cos³2x)
    Hello Soroban,

    Could you please exlain how you got the line in the quote.

    Thank you
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  5. #5
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    Hello, Dadon!

    What part is puzzling you? . . . You just cube it . . .


    We had: .[½(1 + cos2x)]³ . = . (½)³(1 + cos2x)³


    So we have: .(½)³ .= .1/8


    Binomial Expansion: .(a + b)³ .= .a³ + 3a²b + 3ab² + b³

    So: .(1 + cos2x)³ .= .1 + 3cos2x + 3cos²2x + cos³2x

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  6. #6
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    re:

    Thanks Soroban!

    I found out another way to tackle this question:

    \int_{0}^{\pi/2}sin^2xcos^6xdx

    Using the rule: sin^2x=1-cos^2x

    Therefore integral becomes:
    \int_{0}^{\pi/2}(1-cos^2x)cos^6xdx

    [tex]\int_{0}^{\pi/2}cos^6dx-\int_{0}^{\pi/2}cos^8xdx[tex]

    Where I_6=\int_{0}^{\pi/2}cos^6dx and I_8=\int_{0}^{\pi/2}cos^8xdx

    So I_6-I_8

    When I_n is even

    I_6= \frac{(6-1)}{6}.\frac{(6-3)}{(6-2)}.\frac{(6-5)}{(6-4)}.\frac{\pi}{2}
    I_6=\frac{5}{6}.\frac{3}{4}.\frac{1}{2}.\frac{\pi}  {2} = \frac{15\pi}{96}

    And
    I_8=\frac{(8-1)}{8}.\frac{(8-3)}{(8-2)}.\frac{(8-5)}{(8-4)}.\frac{(8-7)}{(8-6)}.\frac{\pi}{2}
    I_8=\frac{7}{8}.\frac{5}{6}.\frac{3}{4}.\frac{1}{2  }.\frac{\pi}{2} = \frac{105\pi}{768}

    Therefore:
    I_6-I_8 = \frac{15\pi}{96}-\frac{105\pi}{768}

    \frac{5\pi}{256}

    Hopefully will look bit more pretty when the LaTex is enabled.
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  7. #7
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    The previous example was when I_n was even.

    Using this rule:
    \frac{(n-1)(n-3)(n-5)...}{n(n-2)(n-4)(n-6)...} etc.

    If n is even then for example when n = 8:

    [tex]\frac{7.5.3.1}{8.6.4.2}.\frac{\pi}{2}

    Note: factors end with 1 on the top and then add the factor \frac{\pi}{2}

    If n is odd then for example when n=7

    [tex]\frac{6.4.2}{7.5.3.1}

    Note: factors end with 1 on the bottom
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