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Math Help - [SOLVED] Integral

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Integral

    \int_a^b\sqrt {x^2 - 1}

    Doing partial integration gets me nowhere. Is this a standard integral?

    x\sqrt {x^2 - 1} - \int_a^b\frac{x^2}{\sqrt {x^2 - 1}} =

    x\sqrt {x^2 - 1} - \left(x\sqrt {x^2 - 1} - \int_a^b\sqrt {x^2 - 1}\right)
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  2. #2
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    why not try x = \sinh \theta
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  3. #3
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    Krizalid's Avatar
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    Let's get rid of integration limits, since it's not the interesting case.

    Integrate by parts:

    \int {\sqrt {x^2  - 1} \,dx}  = x\sqrt {x^2  - 1}  - \int {\frac{{x^2 }}<br />
{{\sqrt {x^2  - 1} }}\,dx} .

    Now a little of make-up

    \int {\frac{{x^2 }}<br />
{{\sqrt {x^2  - 1} }}\,dx}  = \int {\frac{{x^2  - 1 + 1}}<br />
{{\sqrt {x^2  - 1} }}\,dx}  = \int {\sqrt {x^2  - 1} \,dx}  + \int {\frac{1}<br />
{{\sqrt {x^2  - 1} }}\,dx} .

    This yields

    2\int {\sqrt {x^2  - 1} \,dx}  = x\sqrt {x^2  - 1}  - \int {\frac{1}<br />
{{\sqrt {x^2  - 1} }}\,dx} .

    For the last integral substitute \varphi  = x + \sqrt {x^2  - 1}  \implies \frac{1}<br />
{\varphi }\,d\varphi  = \frac{1}<br />
{{\sqrt {x^2  - 1} }}\,dx, finally

    2\int {\sqrt {x^2  - 1} \,dx}  = x\sqrt {x^2  - 1}  - \ln \left| {x + \sqrt {x^2  - 1} } \right| + k_1 , and we happily get

    \int {\sqrt {x^2  - 1} \,dx}  = \frac{1}<br />
{2}\Big[ x{\sqrt {x^2  - 1}  - \ln \left| {x + \sqrt {x^2  - 1} } \right|} \Big] + k.
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  4. #4
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    Hello, Spec!

    No one suggested Trig Substitution?


    \int \sqrt {x^2 - 1}\,dx
    Let x \:=\:\sec\theta\quad\Rightarrow\quad dx \:=\:\sec\theta\tan\theta\,d\theta

    Substitute: . \int \tan\theta(\sec\theta\tan\theta\,d\theta)

    . . . . . . =\;\int\sec\theta\tan^2\!\theta\,d\theta

    . . . . . . = \;\int \sec\theta(\sec^2\!\theta-1)\,d\theta

    . . . . . . =\;\int(\sec^3\!\theta - \sec\theta)\,d\theta

    . . . . . . = \;\frac{1}{2}\sec\theta\tan\theta + \frac{1}{2}\ln|\sec\theta + \tan\theta| - \ln|\sec\theta + \tan\theta| + C

    . . . . . . = \;\frac{1}{2}\sec\theta\tan\theta - \frac{1}{2}\ln|\sec\theta + \tan\theta| + C

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  5. #5
    Senior Member Spec's Avatar
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    Thanks! Just out of curiousity, the original problem was to solve (I just made a substitution):

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

    It it possible to solve that one without the substitution, and thus the following somewhat complicated calculation?
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  6. #6
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    Krizalid's Avatar
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    \int {\sqrt {x^2  + x} \,dx}  = \frac{1}<br />
{2}\int {\sqrt {4x^2  + 4x + 1 - 1} \,dx}  = \frac{1}<br />
{2}\int {\sqrt {(2x + 1)^2  - 1} \,dx} .

    Substitute u=2x+1,

    \int {\sqrt {x^2  + x} \,dx}  = \frac{1}<br />
{4}\int {\sqrt {u^2  - 1} \,du}, and you know what's next.

    P.S.: sorry, I misread your question, let me see if it can be tackled with a nice way.
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