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Math Help - Antiderivative involving hyp. functions.

  1. #1
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    Antiderivative involving hyp. functions.

    I haven't much experience with hyperbolic functions, having never learned them, but the result of this integral seems to involve one:

    \int{\sqrt{1+u^2} du}

    Their solution is:

    \frac{1}{2}u\sqrt{1+u^2}+\frac{1}{2}\text{arsinh}{  u}

    (What's the proper latex command for arsinh?)

    I'm not sure how they derived that.
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  2. #2
    Super Member Showcase_22's Avatar
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    They've used the identity cosh^2x-sinh^2x=1.

    Remember that \frac{d}{dx}cosh(x)=sinh(x) and \frac{d}{dx}sinh(x)=cosh(x).
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  3. #3
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    Ok, so using that identity, would they then have done:

    \int{\sqrt{u^2+\cosh^2(u)-\sinh^2(u)}}du ?

    If so, would this be integrated by parts?

    But I'm still not sure how they ended up with an Inverse Hyperbolic Sine (arsinh). Since the \int{\sinh(x)}dx = \cosh(x), right? I don't see where arsinh comes into this.
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  4. #4
    Super Member Showcase_22's Avatar
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    Solve <br />
\int{\sqrt{1+u^2} du}<br />
.
    Let u=sinh(x). Therefore \frac{du}{dx}=cosh(x).

    The integral becomes: \int\sqrt{1+sinh^2(x)} cosh(x) \ dx

    \int \sqrt{cosh^2(x)} cosh(x) \ dx

    \int cosh^2(x) \ dx

    Now use the identity cosh^2(x)+sinh^2(x)=cosh(2x) and cosh^2(x)-sinh^2(x)=1 to get 2cosh^2(x)-1=cosh(2x).

    Since we require cosh^2(x) we can rearrange this to get cosh^2(x)=\frac{cosh(2x)+1}{2}.

    We now need to integrate \frac{1}{2}\int cosh(2x)+1dx=\frac{1}{2}(\frac{1}{2}sinh(2x)+x+C)

    sinh(2x)=2sinh(x)cosh(x) but using cosh^2(x)-sinh^2(x)=1 we get that cosh(x)=\sqrt{1+sinh^2(x)}. Combining this with the other identity gives sinh(2x)=2sinh(x)\sqrt{1+sinh^2(x)}.

    We also know that x=arsinh(u) from how we defined u at the very start.

    Therefore \frac{1}{2}(\frac{1}{2}sinh(2x)+x)+K =\frac{1}{2}(\frac{1}{2}2sinh(x)\sqrt{1+sinh^2(x)}  +x)+K=\frac{1}{2}u\sqrt{1+u^2}+\frac{1}{2}arsinh(u  )+K.

    In this case, they've taken K=C=0.
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  5. #5
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    Thank you very much for the explanation
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