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Math Help - Integral: Verify Anwser Please

  1. #1
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    Integral: Verify Anwser Please

    Evaluate: \int sinh^6 x\ cosh\ xdx

    =\frac{1}{2}x^2\ cos(h) \sin(h^6)+C
    Last edited by Yogi_Bear_79; October 28th 2006 at 10:35 AM.
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    Quote Originally Posted by Yogi_Bear_79 View Post
    Evaluate: \int sinh^6 x\ cosh\ xdx

    =\frac{1}{2}x^2\ cos(h) \sin(h^6)+C
    There seems to be a problem with your given answer.

    Let y = sinh(x), then dy = cosh(x) dx

    So:
    \int sinh^6(x) \, cosh(x) dx = \int \, y^6 dy = \frac{1}{7}y^7 + C = \frac{1}{7}sinh^7(x) + C.

    -Dan
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    Quote Originally Posted by Yogi_Bear_79 View Post
    Evaluate: \int sinh^6 x\ cosh\ xdx

    =\frac{1}{2}x^2\ cos(h) \sin(h^6)+C
    Your integral is:

    <br />
\int \sinh^6 (x)\ \cosh(x) dx<br />

    That is the functions are hyperbolic not trignometric, and so the h is
    part of the function name and not a variable.

    RonL
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    well that does help alot. But leads to another question.

    is this the same as the orignal equation?
    \int sinh (x)^6 cosh(x)
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    Quote Originally Posted by Yogi_Bear_79 View Post
    well that does help alot. But leads to another question.

    is this the same as the orignal equation?
    \int sinh (x)^6 cosh(x)
    You need to be carefull where the power goes. There is a convention that \sinh^n(x) actually means [\sinh(x)]^n (even though this is an abuse of notation). If you move the power to where you have it you will need to add appropriate brackets to make the meaning clear.

    So what the original means is:

    \int [\sinh (x)]^6 \cosh(x)

    RonL
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