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Math Help - Integration using the hyperbolic functions

  1. #1
    Senior Member chella182's Avatar
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    Integration using the hyperbolic functions

    Okay, I know this should be easy, but I'm having right bother with it the question goes:

    Find \int\frac{dx}{\sqrt{4x^2-9}}

    All I know is I need to find the right substitution. Can anyone help?
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  2. #2
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    Since you mention hyperbolic functions, cosh x= \frac{e^x+ e^{-x}}{2}, and sinh x= \frac{e^x- e^{-x}}{2} so cosh^2 x= \frac{e^{2x}+ 2+ e^{-2x}}{4} while sinh^2 x= \frac{e^{2x}- 2- e^{-2x}}{4}. If we subtract the second from the first, all terms involving e^{2x} cancel leaving cosh^2 x- sinh^2 x= 1 or cosh^2 x- 1= sinh^2 x. In order to use that with \sqrt{4x^2- 9} you want to make those look alike. One way to do that is factor out 9: \sqrt{9((4/9)x^2-1}= 3\sqrt{(4/9)x^2- 1}. Now the substitution cosh t= (2/3)x gets rid of the square root.
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  3. #3
    Senior Member chella182's Avatar
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    LaTeX fail
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  4. #4
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    Edited.
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  5. #5
    Senior Member chella182's Avatar
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    Thankies.
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  6. #6
    Senior Member chella182's Avatar
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    Question about taking out a constant from an integral

    Related to my calculations with the original question.

    If I have...
    \frac{1}{3}\int\frac{\frac{3}{2}\sinh{u}}{\sqrt{(\  cosh{u})^{2}-1}}du
    ...when I take the \frac{3}{2} out, will it become \frac{1}{2} in front by \frac{1}{3}\times\frac{3}{2}, or will it become \frac{9}{2} by \frac{\frac{3}{2}}{3}=\frac{9}{2}?
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