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Math Help - Integral Substitution

  1. #1
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    Integral Substitution

    Made the problem nice and neat with Word Tools: I have a midterm on monday and I can SOLVE ANY problem and have memorized all common trig integrals. This is the only thing I can't do and its frustrating me and I forgot to ask my professor in office hours. Ive done every homework assignment twice, im intent on acing this class, and this is a harder problem on the review for the test.

    http://img504.imageshack.us/img504/1460/calc2la2.png

    I have tried all sorts of substitutions, even completeing the square and then using trig identities, Ive been working non stop on this problem for 40 minutes..
    Could someone please help me, thank you in advance
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  2. #2
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    The imageshack link went down so I made a direct attachment.
    Attached Thumbnails Attached Thumbnails Integral Substitution-calc2.bmp  
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  3. #3
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    Hello,

    \int \frac{x ~dx}{\sqrt{x^2-2x-8}}

    Substitute t=\sqrt{x^2-2x-8}
    \implies t^2=x^2-2x-8 \implies 2t dt=2(x-1) dx \implies dx=\frac{t}{x-1} dt

    \int \frac{x ~dx}{\sqrt{x^2-2x-8}}=\int \frac xt \times \frac{t}{x-1} ~ dt

    =\int \frac{x}{x-1} ~ dt=\int 1+\frac{1}{x-1} ~dt=t+\int \frac{1}{x-1} ~dt

    In order to calculate the last integral, you must have x-1 as a function of t.

    t^2=x^2-2x-8=(x-1)^2-9 \implies t^2+9=(x-1)^2 \implies x-1=\sqrt{t^2+9}

    So \int \frac{x ~dx}{\sqrt{x^2-2x-8}}=t+\int \frac{1}{x-1} ~dt=t+\int \frac{1}{\sqrt{t^2+9}} ~dt=t+\int \frac{1/3}{\sqrt{(t/3)^2+1}} ~dt

    Recall \int \frac{1}{\sqrt{1+x^2}}=\text{arcsinh}(x)+\mathcal{  C}


    And after that, substitute back t=sqrt(x^2-2x-8)
    Last edited by Moo; October 19th 2008 at 01:01 AM.
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    \int \frac{x ~dx}{\sqrt{x^2-2x-8}}

    Substitute t=\sqrt{x^2-2x-8}
    \implies t^2=x^2-2x-8 \implies 2t dt=2(x-1) dx \implies dx=\frac{t}{x-1} dt

    \int \frac{x ~dx}{\sqrt{x^2-2x-8}}=\int \frac xt \times \frac{t}{x-1} ~ dt

    =\int \frac{x}{x-1} ~ dt=\int 1+\frac{1}{x-1} ~dt=t+\int \frac{1}{x-1} ~dt

    In order to calculate the last integral, you must have x-1 as a function of t.

    t^2=x^2-2x-8=(x-1)^2-9 \implies t^2+9=(x-1)^2 \implies x-1=\sqrt{t^2+9}

    So \int \frac{x ~dx}{\sqrt{x^2-2x-8}}=t+\int \frac{1}{x-1} ~dt=t+\int \frac{1}{\sqrt{t^2+9}} ~dt=t+\int \frac{1/3}{\sqrt{({\color{red}t}/3)^2+1}} ~dt

    Recall \int \frac{{\color{red}dt}}{\sqrt{1+{\color{red}t}^2}}=  \text{arcsinh}({\color{red}t})+\mathcal{C}


    And after that, substitute back t=sqrt(x^2-2x-8)
    A couple of small copy and paste errors corrected in red.
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  5. #5
    Moo
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    Quote Originally Posted by mr fantastic View Post
    A couple of small copy and paste errors corrected in red.
    Thanks
    Just the first one actually =)
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  6. #6
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    Quote Originally Posted by Moo View Post
    Hello,

    \int \frac{x ~dx}{\sqrt{x^2-2x-8}}

    [snip
    Another method:

    \int \frac{x ~dx}{\sqrt{x^2-2x-8}} = \int \frac{(x - 1) + 1}{\sqrt{x^2-2x-8}} \, dx = \int \frac{x - 1}{\sqrt{x^2-2x-8}} \, dx + \int \frac{1~dx}{\sqrt{x^2-2x-8}}.

    The first integral is simple: Make the substitution u = x^2 - 2x - 8

    The second integral is simple if you're familiar with hyperbolic functions but not so simple if you're not:

    \int \frac{1~dx}{\sqrt{x^2-2x-8}} = \int \frac{1~dx}{\sqrt{(x - 1)^2 - 3^2}} = \frac{1}{3} \int \frac{1~dx}{\sqrt{\left(\frac{x - 1}{3}\right)^2 - 1}} = \frac{1}{3} \cosh^{-1} \left( \frac{x-1}{3}\right) + C.

    It can also be done using a trig substitution but that's a pain.

    Note: \cosh^{-1} A = \ln \left( A + \sqrt{A^2 - 1} \right) which would explain why the answer can also be expressed using logarithms:
    Attached Thumbnails Attached Thumbnails Integral Substitution-mhf2.gif  
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  7. #7
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    hmm interesting, we are allowed to leave multiple variables within the integral as long as they cancel?

    hmm.. thats not the answer it gives on the website, its expressed in "ln". We are not required to memorize the natural log formulas, but just by speculation I dont think it would equal his answer, but it looks right to me.

    interesting I never thought you could have the "dt" and the actual variable differ, man.. all these tricks and stuff

    thanks guys,
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  8. #8
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    OK i found it out completely using trig subs,

    I set t = 3sec\theta\ for only the integral

    at this step

    <br />
\int \frac{x ~dx}{\sqrt{x^2-2x-8}}=t+\int \frac{1}{x-1} ~dt=t+\int \frac{1}{\sqrt{t^2+9}} ~dt<br />

    then derived such and so forth and got the answer.. wow that problem is tedious the first tiem on paper.. 1 page
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  9. #9
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    Quote Originally Posted by ryukolink View Post
    OK i found it out completely using trig subs,

    I set t = 3sec\theta\ for only the integral

    at this step

    <br />
\int \frac{x ~dx}{\sqrt{x^2-2x-8}}=t+\int \frac{1}{x-1} ~dt=t+\int \frac{1}{\sqrt{t^2+9}} ~dt<br />

    then derived such and so forth and got the answer.. wow that problem is tedious the first tiem on paper.. 1 page
    Quote Originally Posted by mr fantastic View Post
    Another method:

    \int \frac{x ~dx}{\sqrt{x^2-2x-8}} = \int \frac{(x - 1) + 1}{\sqrt{x^2-2x-8}} \, dx = \int \frac{x - 1}{\sqrt{x^2-2x-8}} \, dx + \int \frac{1~dx}{\sqrt{x^2-2x-8}}.

    The first integral is simple: Make the substitution u = x^2 - 2x - 8

    The second integral is simple if you're familiar with hyperbolic functions but not so simple if you're not:

    \int \frac{1~dx}{\sqrt{x^2-2x-8}} = \int \frac{1~dx}{\sqrt{(x - 1)^2 - 3^2}} = \frac{1}{3} \int \frac{1~dx}{\sqrt{\left(\frac{x - 1}{3}\right)^2 - 1}} = \frac{1}{3} \cosh^{-1} \left( \frac{x-1}{3}\right) + C.

    It can also be done using a trig substitution but that's a pain.

    [snip]
    Told you so!
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  10. #10
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    sorry got clouded in the whole mess you guys posted trying to decipher it

    you're guys method was pretty advanced, I dont think I have ever encountered a problem this hard yet, glad I came here for a fast response, well test tomorrow.. learning the reduction formulas for the 10 pts of extra credit, hoping for 130/120 lets hope

    Also I prefer trig methods for some reason

    Thanks again
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