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Math Help - Hyperbolic Headaches..

  1. #1
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    Hyperbolic Headaches..

    Hi guys I'm trying to work through some problems getting ready for an exam and I came across this problem that doesn't make sense to me...

    Consider \displaystyle \int x(1+x^2)dx

    If I integrate this I get 1/4(x^4) + 1/2(x^2)

    It says if we substitute x = sinh(u) we get \displaystyle \int sinh(u)cosh^3(u)du

    Where on earth does that cosh^3 come from? I know that 1+sinh^2(u) gives us cosh^3(u)du.

    Also if I integrate this I get \displaystyle \int sinh(u)cosh^3(u)du I get x^4/4 + 2x^2/4 + 1/4 . Why are they not equal?

    Thoughts appreciated?
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  2. #2
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    \displaystyle dx=cosh(u)du

    Therefore, \dispalystyle sinh(u)*(1+sinh(u)^2)cosh(u)=sinh(u)*cosh(u)^2*cos  h(u)=sinh(u)cosh(u)^3

    Integrating yields:

    \displaystyle\frac{cosh(u)^4}{4}

    \displaystyle u=sinh^{-1}(x)\rightarrow\frac{cosh^4(sinh^{-1}(x))}{4}
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  3. #3
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    When integrating, we have a constant of integration with indefinite integrals. \frac{1}{4} is a constant but we would still have a plus C. C+\frac{1}{4}=C_2.

    Integral without substitution: \displaystyle\frac{x^2}{2}+\frac{x^4}{4}+C

    With: \displaystyle\frac{x^2}{2}+\frac{x^4}{4}+\frac{1}{  4}+C=\frac{x^2}{2}+\frac{x^4}{4}+C_2
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  4. #4
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    What a title!
    Quote Originally Posted by minusb View Post
    Also if I integrate this I get \displaystyle \int sinh(u)cosh^3(u)du I get x^4/4 + 2x^2/4 + 1/4 . Why are they not equal?
    I've not checked your calculations, but they don't have to be. When you integrate some function f(x) over
    an indefinite interval, infinite number of functions that satisfy what you get (but there might be exceptions
    to this) -- what you find is the set of all anti-derivatives. That's why we have the constant of integration.

    You have:

    \displaystyle I_{1} = \frac{1}{4}x^4+\frac{1}{2}x^2+k_{1}

    \displaystyle I_{2} = \frac{1}{4}x^4+\frac{1}{2}x^2+\frac{1}{4}+k_{2}

    k_{1} and k_{2} are constants, of course. But so is \frac{1}{4}+k_{2} and \frac{1}{4}+k_{2}+k_{1}, call this K.

    So \displaystyle I = \frac{1}{4}x^4+\frac{1}{2}x^2+K

    An indefinite integral is really a big function that accounts for all the functions whose derivative is the integrated function!
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  5. #5
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    That makes sense. Thank you fellas. Keep up the great work.
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