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Math Help - Having Trouble with Integral?

  1. #1
    Member zangestu888's Avatar
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    Having Trouble with Integral?

    Hi, am doing a question looking for the length of a curve, the function given is y=e^x + 1/4e^-x, from [1,0] is my interval.

    I got all the parts down i got dy/dx=e^x-1/4e^-x then from the length of curve formula,

    L = \int\ {\sqrt{1 + (e^x -1/4e^x)^2}}

    my problem is the integral under the square root, i tried substitution i tried parts i couldnt get anything, any clues on how to start off this integral. Thank You
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  2. #2
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    How about noting this about your function:
    f(x)=e^x+\frac{1}{4}e^{-x}
    f(x)=\frac{2e^x+\frac{1}{2e^x}}{2}
    f(x)=\frac{e^{\ln2}e^x+\frac{1}{e^{\ln2}e^x}}{2}
    f(x)=\frac{e^{(x+\ln2)}+e^{-(x+\ln2)}}{2}
    f(x)=\cosh(x+\ln2)
    and is thus just a translated hyperbolic cosine, so the curve is a catenary. That might make the integral easier using hyperbolic function identities, but I can't say for sure.

    --Kevin C.
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  3. #3
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    In fact, it will make the integral easier. With y=\cosh(x+\ln2), we find \frac{dy}{dx}=\sinh(x+\ln2), and so \sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+\sin  h^2(x+\ln2)}, and since \cosh^2x-\sinh^2x=1, 1+\sinh^2x=\cosh^2x, so
    \sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+\sin  h^2(x+\ln2)}
    =\sqrt{\cosh^2(x+\ln2)}
    =\cosh(x+\ln2), as the hyperbolic cosine is always positive, and you should be able to integrate that.

    --Kevin C.
    Last edited by TwistedOne151; January 27th 2009 at 01:50 PM. Reason: typo
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  4. #4
    Member zangestu888's Avatar
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    I dont think ive ever worked with an hyperbolic function like

    cosh(x+ln2)

    how would i go about starting to integrate such a function? do i make a substiution.
    Ive never worked with these types before?
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  5. #5
    Member zangestu888's Avatar
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    would i make a sub. u=x + ln2 \ du=dx

    then i would get

    my antidervitive is ---> sinh(x + ln2) + c
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by zangestu888 View Post

    L = \int\ {\sqrt{1 + (e^x -1/4e^x)^2}}

    my problem is the integral under the square root, i tried substitution i tried parts i couldnt get anything, any clues on how to start off this integral. Thank You
    You need to do first some of algebra!

    1+{{\left( {{e}^{x}}-\frac{1}{4{{e}^{x}}} \right)}^{2}}=1+{{e}^{2x}}-\frac{1}{2}+\frac{1}{16{{e}^{2x}}}={{\left( {{e}^{x}}+\frac{1}{4{{e}^{x}}} \right)}^{2}}.

    Extract the square root and you'll get a very easy integral to solve.
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  7. #7
    Member zangestu888's Avatar
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    how is that easier? am not sure how to solve it do i make a sub?
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  8. #8
    Member zangestu888's Avatar
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    got it thanks!
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