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Math Help - Evaluate this integral

  1. #1
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    Evaluate this integral

    <br />
\displaystyle \int \frac{1}{\sqrt{12 + 4 x - x^2}}dx<br />

    Can someone please point out where I went wrong?

    First, I complete the square:
    <br />
\displaystyle \int \frac{1}{\sqrt{(x-2)^2-16}}dx<br />

    Factor out a 16:
    <br />
\displaystyle \int \frac{1}{\sqrt{16(\frac{(x-2)^2}{16}-1)}}dx<br />


    <br />
\displaystyle \frac{1}{4}\int \frac{1}{\sqrt{\frac{(x-2)^2}{16}-1}}dx<br />

    Let U = \frac{x-2}{4} & dU=1\4dx :: dx=4du

    <br />
\displaystyle \frac{1}{4}\int \frac{4du}{\sqrt{U^2-1}}dU<br />

    <br />
\displaystyle \int \frac{du}{\sqrt{U^2-1}}dU<br />

    Which is:
    <br />
arccosh(U) = arccosh(\frac{x-2}{4})<br />

    It seems like my answer is wrong however. Where did I go wrong?
    Thanks!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Vamz View Post
    <br />
\displaystyle \int \frac{1}{\sqrt{12 + 4 x - x^2}}dx<br />

    Can someone please point out where I went wrong?

    First, I complete the square:
    <br />
\displaystyle \int \frac{1}{\sqrt{(x-2)^2-16}}dx<br />

    Factor out a 16:
    <br />
\displaystyle \int \frac{1}{\sqrt{16(\frac{(x-2)^2}{16}-1)}}dx<br />


    <br />
\displaystyle \frac{1}{4}\int \frac{1}{\sqrt{\frac{(x-2)^2}{16}-1}}dx<br />

    Let U = \frac{x-2}{4} & dU=1\4dx :: dx=4du

    <br />
\displaystyle \frac{1}{4}\int \frac{4du}{\sqrt{U^2-1}}dU<br />

    <br />
\displaystyle \int \frac{du}{\sqrt{U^2-1}}dU<br />

    Which is:
    <br />
arccosh(U) = arccosh(\frac{x-2}{4})<br />

    It seems like my answer is wrong however. Where did I go wrong?
    Thanks!
    12+4x-x^2\neq (x-2)^2-16. However, 12+4x-x^2=16-(x-2)^2...
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Vamz View Post
    <br />
\displaystyle \int \frac{1}{\sqrt{12 + 4 x - x^2}}dx<br />

    Can someone please point out where I went wrong?

    First, I complete the square:
    <br />
\displaystyle \int \frac{1}{\sqrt{(x-2)^2-16}}dx<br />

    Factor out a 16:
    <br />
\displaystyle \int \frac{1}{\sqrt{16(\frac{(x-2)^2}{16}-1)}}dx<br />


    <br />
\displaystyle \frac{1}{4}\int \frac{1}{\sqrt{\frac{(x-2)^2}{16}-1}}dx<br />

    Let U = \frac{x-2}{4} & dU=1\4dx :: dx=4du

    <br />
\displaystyle \frac{1}{4}\int \frac{4du}{\sqrt{U^2-1}}dU<br />

    <br />
\displaystyle \int \frac{du}{\sqrt{U^2-1}}dU<br />

    Which is:
    <br />
arccosh(U) = arccosh(\frac{x-2}{4})<br />

    It seems like my answer is wrong however. Where did I go wrong?
    Thanks!
    First when you completed the square something is off

    12+4x-x^2=-(x^2-4x) +12 =-(x^2-4x+4)+12+4= 16-(x-2)^2

    From here make the substition x-2=4\sin(t) \implies dx=4\cos(t)dt

    \displaystyle \int \frac{dx}{\sqrt{12+4x-x^2}}=\int \frac{4\cos(t)dt}{\sqrt{16-16\sin^2(t)}}=\frac{4}{4}\int \frac{\cos(t)dt}{\sqrt{1-\sin^2(t)}}=\int dt
    Last edited by TheEmptySet; February 5th 2011 at 03:18 PM. Reason: missing "4"
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  4. #4
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    ^ Thank you!

    I fixed that problem. Are you using trig substitution to solve this? I havnt learned that very well yet.

    Couldnt I still get this into the same form as arccos using the same method I attempted, but this time with an answer of:

    arccos((x-2)/4) + C

    ?
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  5. #5
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    \displaystyle\int{\frac{1}{\sqrt{12+4x-x^2}}}dx=\int{\frac{1}{\sqrt{4^2-\left(x^2-4x+4\right)}}}dx

    =\displaystyle\int{\frac{1}{\sqrt{4^2-(x-2)^2}}}dx

    u=x-2\Rightarrow\ du=dx

    \displaystyle\int{\frac{1}{\sqrt{4^2-u^2}}}du=arcsin\left(\frac{u}{4}\right)+C

    Change the numerator sign of the function to get arccos.
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  6. #6
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    Quote Originally Posted by TheEmptySet View Post
    First when you completed the square something is off

    12+4x-x^2=-(x^2-4x) +12 =-(x^2-4x+4)+12+4= 16-(x-2)^2

    From here make the substition x-2=\sin(t) \implies dx=\cos(t)dt

    \displaystyle \int \frac{dx}{\sqrt{12+4x-x^2}}

    =\displaystyle\int \frac{4\cos(t)dt}{\sqrt{16-16\sin^2(t)}}

    This doesn't work as the constant under the square root was a 16
    rather than a 1


    =\displaystyle\frac{4}{4}\int \frac{\cos(t)dt}{\sqrt{1-\sin^2(t)}}=\int dt
    There's a problem on the 2nd last line.
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  7. #7
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    Thanks guys. I understand now
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  8. #8
    Behold, the power of SARDINES!
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    the last line is actually correct but I wrote down the wrong sub it should have been

     x-2=4\sin(t) \implies dx=4\cos(t)dt which is what I used in the post. :\
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