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Math Help - trig integration

  1. #1
    Senior Member DivideBy0's Avatar
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    trig integration

    After I complete \int \frac{\sqrt{x^2-9}}{x}\,dx by sec substitution I get

    3\sec^{-1}\left(\frac{x}{3}\right)-\sqrt{x^2-3}+C.

    But this doesn't match the calculator's answer:

    3\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right) - \sqrt{x^2-3}+C

    After differentiating the result my answer is different as well... I thought the answers we got would be equivalent though...

    Then the second question is f(x) = 3. Find the equation.
    Last edited by DivideBy0; December 9th 2007 at 09:59 PM.
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  2. #2
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    Quote Originally Posted by DivideBy0 View Post
    After I complete \int \frac{\sqrt{x^2-9}}{3}\,dx by sec substitution I get

    3\sec^{-1}\left(\frac{x}{3}\right)-\sqrt{x^2-3}+C.

    But this doesn't match the calculator's answer:

    3\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right) - \sqrt{x^2-3}+C

    After differentiating the result my answer is different as well... I thought the answers we got would be equivalent though...

    Then the second question is f(x) = 3. Find the equation.
    I guess the right integral is: \int \frac{\sqrt{x^2-9}}{x}\,dx
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  3. #3
    Senior Member DivideBy0's Avatar
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    Woops, sorry, you're right that's what I meant
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  4. #4
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    Hello, DivideBy0!

    After I complete \int \frac{\sqrt{x^2-9}}{x}\,dx by sec substitution I get: . 3\sec^{-1}\! \left(\frac{x}{3}\right)-\sqrt{x^2-3}+C.

    But this doesn't match the calculator's answer: /  3\tan^{-1}\!\left(\frac{\sqrt{x^2-9}}{3}\right) - \sqrt{x^2-3}+C
    The answers are equivalent.


    Let \theta \:=\:\sec^{-1}\!\left(\frac{x}{3}\right)
    Then we have: . \sec\theta \:=\:\frac{x}{3} \:=\:\frac{hyp}{adj}

    \theta is in a right triangle with: adj = 3,\;hyp = x

    . . Using Pythagorus: . opp = \sqrt{x^2-9}

    So: . \tan\theta \:=\:\frac{\sqrt{x^2-9}}{3}
    Hence:. . \theta \:=\:\tan^{-1}\!\left(\frac{\sqrt{x^2-9}}{3}\right)


    See? . . . . . \sec^{-1}\!\left(\frac{x}{3}\right) \;=\;\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right)

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  5. #5
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    Quote Originally Posted by DivideBy0 View Post
    After I complete \int \frac{\sqrt{x^2-9}}{x}\,dx by sec substitution I get

    3\sec^{-1}\left(\frac{x}{3}\right)-\sqrt{x^2-3}+C.

    But this doesn't match the calculator's answer:

    3\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right) - \sqrt{x^2-3}+C

    After differentiating the result my answer is different as well... I thought the answers we got would be equivalent though...

    Then the second question is f(x) = 3. Find the equation.
    the answar is
    -3\sec^{-1}\left(\frac{x}{3}\right)+\sqrt{x^2-9}+C.
    or

    -3\cos^{-1}\left(\frac{3}{x}\right)+\sqrt{x^2-9}+C.
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  6. #6
    Senior Member DivideBy0's Avatar
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    Thanks Soroban for clearing it up a bit more... but when I differentiated on the calculator I still got

    \frac{d}{\,dx}\left(3\sec^{-1}(\frac{x}{3})-\sqrt{x^2-3}\right)=\frac{9|\frac{1}{x}|}{\sqrt{x^2-9}}-\frac{x}{\sqrt{x^2-3}}

    But the original expression obviously has no absolute value signs.
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  7. #7
    Math Engineering Student
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    Quote Originally Posted by DivideBy0 View Post
    \int \frac{\sqrt{x^2-9}}{x}\,dx
    You don't even need trig. sub. - this only requires a simple substitution.

    Step #1: The make-up.

    \int {\frac{{\sqrt {x^2 - 9} }}<br />
{x}\,dx} = \int {\frac{{\sqrt {x^2 - 9} \cdot x}}<br />
{{x^2 }}\,dx} .

    Step #2: The substitution.

    u^2 = x^2 - 9 \implies u\,du = x\,dx,

    \int {\frac{{\sqrt {x^2 - 9} }}<br />
{x}\,dx} = \int {\frac{{u^2 }}<br />
{{u^2 + 9}}\,du} .

    Step #3: Simple trick & mission almost-accomplished.

    \int {\frac{{u^2 }}<br />
{{u^2 + 9}}\,du} = \int {du} - 9\int {\frac{1}<br />
{{u^2 + 9}}\,du} = u - 3\arctan \frac{u}<br />
{3}+k.

    Step #4: Back substitute.

    \int {\frac{{\sqrt {x^2 - 9} }}<br />
{x}\,dx} = \sqrt {x^2 - 9} - 3\arctan \frac{{\sqrt {x^2 - 9} }}<br />
{3} + k.
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