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Math Help - Missing steps in book

  1. #1
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    Missing steps in book

    I am trying to work out how to get from the integral below to the answer just below it.

    N = \frac{yR^2}{sin^2x} \int{(cosx-1)cosxsinx dx}

    with limits of pie and x

    The answer is

    N = \frac{yR^2}{6} \frac{5(1-cosx) + 2cos^2x}{1-cosx}

    After using the following intergrals

    \int{cosxsinx dx} = \frac{sin^2x}{2}

    and

    \int{cos^2xsinx dx} = \frac{-cos^3x}{3}

    I get

    N = \frac{yR^2}{6sin^2x} [3sin^2x + 2cos^2x + 2]

    however

    1. I cannot rearange it to get the correct answer
    2. If i substitute a value such as x = 180, my answer and the book's answer do not match?

    Thanks in advance

    Calypso
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  2. #2
    MHF Contributor
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    Quote Originally Posted by calypso View Post
    I am trying to work out how to get from the integral below to the answer just below it.

    N = \frac{yR^2}{sin^2x} \int{(cosx-1)cosxsinx dx}

    with limits of pie and x

    The answer is

    N = \frac{yR^2}{6} \frac{5(1-cosx) + 2cos^2x}{1-cosx}

    After using the following intergrals

    \int{cosxsinx dx} = \frac{sin^2x}{2}

    and

    \int{cos^2xsinx dx} = \frac{-cos^3x}{3}

    I get

    N = \frac{yR^2}{6sin^2x} [3sin^2x + 2cos^2x + 2]

    however

    1. I cannot rearange it to get the correct answer
    2. If i substitute a value such as x = 180, my answer and the book's answer do not match?

    Thanks in advance

    Calypso
    Hi Calypso,

    one of your integrals isn't evaluated conveniently..

    \int{-cosxsinx}dx

    u=-cosx,\ \frac{du}{dx}=sinx,\ du=sinxdx

    \int{u}du=\frac{u^2}{2}=\frac{cos^2x}{2}


    Then you get

    \frac{yR^2}{sin^2x}\left(\int{cos^2xsinx}dx-\int{cosxsinx}dx\right)

    =\frac{yR^2}{sin^2x}\left[-\frac{1}{3}cos^3x+\frac{1}{2}cos^2x\right] from x to {\pi}

    =\frac{yR^2}{sin^2x}\left[\frac{1}{3}cos^3x-\frac{1}{2}cos^2x\right] from {\pi} to x

    This gives

    \frac{yR^2}{sin^2x}\left(\frac{cos^3x}{3}-\frac{cos^2x}{2}-\frac{cos^3{\pi}}{3}+\frac{cos^2{\pi}}{2}\right)

    =\frac{yR^2}{sin^2x}\left(\frac{cos^3x}{3}-\frac{cos^2x}{2}-\frac{(-1)^3}{3}+\frac{(-1)^2}{2}\right)

    =\frac{yR^2}{sin^2x}\left(\frac{cos^3x}{3}-\frac{cos^2x}{2}+\frac{1}{3}+\frac{1}{2}\right)

    =\frac{yR^2}{6sin^2x}\left(2cos^3x-3cos^2x+5\right)

    =\frac{yR^2}{6}\left(\frac{2cos^3x+2cos^2x-5cos^2x+5}{1-cos^2x}\right)

    =\frac{yR^2}{6}\left(\frac{2cos^2x(1+cosx)+5(1-cos^2x)}{(1-cosx)(1+cosx)}\right)

    =\frac{yR^2}{6}\left(\frac{2cos^2x(1+cosx)+5(1-cosx)(1+cosx)}{(1-cosx)(1+cosx)}\right)

    =\frac{yR^2}{6}\left(\frac{2cos^2x+5(1-cosx)}{1-cosx}\right)
    Last edited by Archie Meade; May 22nd 2010 at 02:56 PM. Reason: small typo
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  3. #3
    Junior Member
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    Thanks Archie, I have been trying to do that all afternoon
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