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    Could someone explain how to do the following question please?



    It actually has 2 parts before this one:

    and

    which are both fairly easy to do, but I can't put them together to get the third one.

    Any help is much appreciated.
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    i imagine its some 'by parts' trickery. but i cant see what it is. sorry
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    See the attached file.

    Note however that I got the integral circled in red from the Integrator and couldn't do it manually. I don't think we can do it based on FP2 integration. If I'm wrong let me know how to do it!
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    Yeah. I could get that integral but got stuck there too.

    Hmmm... the question is from the Heinemann FP2 textbook.

    I wonder if a trig substitution is needed...
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    How to get the thing circled in red, or how to actually integrate it?
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    How to actually integrate it
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    I think I can do it, but I don't know how to get the mathematical symbols when I reply.

    You have to use a hyperbolic substitution of x = sinh u (I think) I haven't actually done it yet.
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    Just integrate by parts and it comes out immediately.
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    Yeah that works. Thanks.

    EDIT: I was talking to Handy
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    I think using x = sinh u will give the answer.

    x = sinh u
    so the function in the square root becomes (cosh u)^2

    So basically find the integral of cosh u and then substitute: x = sinh u

    Is this correct?
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    Okay, so it correct!
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    Well once you've substituted x = sinh u and simplified, the integral becomes cosh^2 u.

    Using an identity you can finish it off.
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    (Original post by Wrangler)
    Just integrate by parts and it comes out immedeately.
    Yeah, the problem was I couldn't integrate √(x^2 + 1) but I've figured out how to do it now.
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    Ah, ok - sorry about that. I think the most direct approach would be to use:

    x²/√(x²+1) = √(x²+1) - 1/√(x²+1)

    So, integrate by parts to get:

    I = ∫x[x/√(x²+1)]dx = x√(x²+1) - ∫√(x²+1)dx = x√(x²+1) - ∫x[x/√(x²+1)]dx - ∫[1/√(x²+1)]dx = x√(x²+1) - arcsinh(x) - I

    And we're done.
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    (Original post by Wrangler)
    Ah, ok - sorry about that. I think the most direct approach would be to use:

    x²/√(x²+1) = √(x²+1) - 1/√(x²+1)

    So, integrate by parts to get:

    I = ∫x[x/√(x²+1)]dx = x√(x²+1) - ∫√(x²+1)dx = x√(x²+1) - ∫x[x/√(x²+1)]dx - ∫[1/√(x²+1)]dx = x√(x²+1) - arcsinh(x) - I

    And we're done.
    That weird 'I' method isn't on the syllabus.
    It's easier to use a substitution, and he already has.
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    (Original post by Handy)
    That weird 'I' method isn't on the syllabus.
    It's easier to use a substitution, and he already has.
    There's nothing weird about the "I", I've just used it to save me writing out the integral several times. I personally think it's a little "slicker" to avoid using extra substitutions. A quick rearrangement of my line of working gives:

    I = ½[x√(x²+1) - arcsinh(x)]

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    The "I" method is in the (Edexcel) syllabus by my reckoning. Especially considering that Reduction Formulae is the next exercise after the one I'm doing.
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    I only stands for Integral ie, some function=I. It'll come in handy when you need to move the integral and the integrand from side to side, in recurrence intergration.
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    Right, New question:

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    (Original post by e-unit)
    The "I" method is in the (Edexcel) syllabus by my reckoning. Especially considering that Reduction Formulae is the next exercise after the one I'm doing.
    Fair enough.

    I don't need to know it
 
 
 
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