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Thread: help with integration

  1. #1
    Jul 2009

    help with integration


    i've been in this question for ages and can't seem to crack it.

    I need to substitute a suitable circular of hyperbolic fn so that I may find

    a) integr[(a^2-x^2)^1/2]
    b) integr[(a^2+x^2)^1/2]
    c) ntegr[(x^2-a^2)^1/2]

    I am looking at the first, and I think x=asinu => dx = acosu du

    so (a^2-x^2)^1/2 = (acosu)^2 = [(a^2)/2] [cos2u+1]

    and integr[cos2u+1] = [(sin2u)/2] + u + c_1

    and subbing for u (this is one of my problems) I get my solution

    = [(a^2)/2] {[(sin2arcsin(x/a))/2] + arcsin(x/a)} + c

    I can also get it to

    = (ax/2)[(pi/2) -a ) + (a^2/2)arcsin(x/a) +c

    but i really want to get my book's answer which is

    (1/2) { [ x(a^2 - x^2)^(1/2) - (a^2) arcsin(x/a)]}

    please help me, I feel so fed up with trying this one question i have honestly been at it for hours!
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  2. #2
    MHF Contributor
    Oct 2008
    The first is by parts (integrating 1 and differentiating the integral), re-writing

    $\displaystyle \frac{x^2}{\sqrt{a^2 - x^2}}$


    $\displaystyle \frac{a^2 - (a^2 - x^2)}{\sqrt{a^2 - x^2}}$

    Will post a pic. The others look similar.

    Edit: Just in case a picture helps...

    ... where...

    is the product rule for differentiation (top to bottom).

    In balloon world, integration by parts is a matter of filling out the product rule shape usefully, but starting in one of the bottom corners rather than (as with differentiation) the top one... and then subtracting whatever it's necessary to subtract on the bottom row, in order to keep the lower equals sign valid.

    One blank to fill in the top row, then solve the top row for I.


    Don't integrate - balloontegrate!
    Last edited by tom@ballooncalculus; Aug 21st 2009 at 09:50 AM. Reason: pic
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  3. #3
    Jul 2009
    what can I say!! i am really impressed (by how much help you have offered me). Thanks!
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