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Math Help - Integration by hyperbolic substitution

  1. #1
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    Integration by hyperbolic substitution

    Once again I feel I am not too far from the answer...

    I am asked to integrate:

    \int \frac{x^2}{\sqrt{1+x^2}} \ dx

    using an hyperbolic substitution.

    I tried helping myself with this: Integration by Hyperbolic Substitution | calculuspowerup.com but I have trouble converting the hyperbolic function with the triangle.

    I did:

    Once again I feel I am not too far from the answer...

    I am asked to integrate:

    \int \frac{x^2}{\sqrt{1+x^2}} \ dx

    using an hyperbolic substitution.

    I tried helping myself with this: Integration by Hyperbolic Substitution | calculuspowerup.com but I have trouble converting the hyperbolic function with the triangle.

    I did:

    x = \sinh u

    dx = \cosh u \ du

    \int \frac{x^2}{\sqrt{1+x^2}} \ dx =

    \int \frac{{\sinh}^2 u \cdot \cosh u}{\cosh u} \ du =

    \int {\sinh}^2 u \ du =

    \frac{1}{2} \int (\cosh 2u - 1) \ du =

    \frac{\sinh 2u}{4} - \frac{u}{2} + C =

    \frac{\sinh 2u}{4} - \frac{\text{argsinh} x}{2} + C

    That is where I'm not sure how to isolate u to get my x value and how to work with the triangle etc. I need help with that.

    The good answer is:

    \frac{x \sqrt{1+x^2}}{2} - \frac{\text{argsinh} x}{2} + C

    Thank you very much in advance.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \sinh 2u=2\sinh u\cosh u

    \sinh u=x, \ \cosh u=\sqrt{1+\sinh^2u}=\sqrt{1+x^2}
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  3. #3
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    Thanks a lot red_dog, didn't think of using that identity at all!
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