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Math Help - Transforming Integral for Trigonometric Substitution

  1. #1
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    Transforming Integral for Trigonometric Substitution

    The question is:
    ...
    Consider the integral

    I=\int \frac{x}{\sqrt{7+8x+x^2}}dx

    (i) Use the substitution u = x +4 to transform the integral into one for which trigonometric substitution is appropriate. (You will need to start by first completing the square under the root sign.)

    The second part of this is: (ii) Now use a suitable trignometric substitution on the integral in (i) to find I.

    ...
    I think I'll be able to do the second part once I get the first bit sorted.
    My problem is basically that I don't know how to deal with the x on the top.


    What I have done so far(which may or may not be correct):
    I completed the square, and ended up with (x +4)^2-9 under the square root sign. I substituted u into that, as per the question, but what I'm left with is in terms of both x and u, as well as not being in the form for trigonometric substituion - the form I'm thinking I should be getting is

    \int \frac{1}{\sqrt{u^2-a^2}}du



    Any ideas on this question? (I would appreciate ideas as opposed to solutions.)
    Thank you.
    Last edited by Lisa1991; July 24th 2009 at 06:36 PM.
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  2. #2
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    Thanks to everyone who considered my question for me. I have realised how to do it now - for those who are interested, I added one and subtracted one from the top line, making it x + 4 - 4, then u - 4.
    Last edited by Lisa1991; July 24th 2009 at 06:36 PM.
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  3. #3
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    Cool

    So you end up with:

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

    right?

    So if we let:

    u = x + 1
    \frac{d u }{d x} = 1

    So:

    I = \int \frac{u - 1}{\sqrt{4 - u^2}} \, du

    Before you look at the rest, see if you can take it from here...

    If we now let

     u = 2 \sin{\alpha}
    \frac{d u}{d \alpha} = 2 \cos{\alpha}

    So:

     I = \int \frac{u - 1}{\sqrt{4 - 4 \sin^2{\alpha}}} 2 \cos{\alpha} \, d \alpha

     I = \int (u - 1) \, d \alpha

    I = \int (2 \sin{\alpha} - 1) \, d \alpha

    I = -2cos{\alpha} - \alpha + C

     I = -2 \cos{ \left( \arcsin{ \left( \frac{x+1}{2} \right)} \right)} - \arcsin{ \left( \frac{x + 1}{2} \right) } + C

    Hope that('s right and) helps!
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  4. #4
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    Sorry. I'm very confused. Did you edit your post mid-way, or did I just make up some random numbers for the denominator of the integral?

    Also, sorry about giving a complete solution (albeit to a possibly non-existant problem); to my shame, I only skim-read your post.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    complete the square:

    (x^2+8x)+7=(x^2+8x+16-16)+7=(x^2+8x+16)-16+7=(x+4)^2-9

    Got it? And now you even have a difference of squares, which could play to your advantage as well.

    u^2-9=(u+3)(u-3)

    The problem with x? No problem at all!

    u=x+4\Longleftrightarrow{u-4}=x
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  6. #6
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    Yes, sorry, did edit the numbers - had two examples and for some reason copied the second when I meant to copy the first... Probably should have left it, but I'm a bit of a perfectionist.
    Thanks for your help.

    And thanks VonNemo - I knew there would be some small thing I was overlooking with the x...
    Last edited by Lisa1991; July 24th 2009 at 09:07 PM.
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