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Indefinite Integration (part 7)

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    I'm now going to do it
    integration by parts problems.
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    I think it's just a fun problem
    to see because one, it's the
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    example a lot of people use,
    sometimes even a trick problem
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    that's given on a really hard
    math exam, or if you go to
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    calculus competitions like
    I used to in high school.
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    Not to make myself too-- I was
    actually not that geeky as a
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    high school student, but I have
    to admit, I was a mathlete.
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    But anyway, this is just a fun
    integration by parts problem
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    because you actually never have
    to evaluate the final integral.
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    So let's say we want to
    take the integral-- it's
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    a bit of a classic.
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    I wouldn't be surprised if your
    math teacher does the same
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    problem for you, just to show
    you integration by parts.
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    Let's take the integral of e to
    the x-- you probably never
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    heard of someone call a math
    problem a classic, but
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    hopefully I will instill in you
    this love for mathematics and
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    you will also consider this
    to be a classic problem.
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    e to the x times cosine of x.
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    I think you might already see
    where I'm going with this,
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    because these are both fun
    functions, because e to the x
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    you can take the derivative,
    you could take the
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    anti-derivative and it
    still stays e to the x.
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    Cosine of x you take the
    derivative, you go to minus
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    sign of x, you take the
    derivative again then you to
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    minus cosine of x, then you
    take the derivative again then
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    you get a plus sign of x.
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    It's like this cycle.
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    The same thing happens when
    you take the anti-derivative.
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    It's not as cool as e to the x,
    it doesn't stay exactly the
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    same, but it kind of cycles.
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    If you take two
    anti-derivatives you get back
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    to the negative of itself.
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    And if you take two
    derivatives, you get back
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    to the negative of itself.
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    It's also a pretty cool
    function and I think you can
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    start to see how integration
    by parts might be cool here.
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    Whenever I do integration by
    parts I always like to assume
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    that this is the g prime of x.
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    That e to the x is g prime
    of x, because e to the x
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    literally doesn't change.
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    Although we could do this
    problem the other way.
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    Maybe I'll experiment
    doing it the other way.
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    but let's assume this is
    g prime of x, and let's
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    assume this f of x.
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    So this is derivative.
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    So integration by parts, as we
    take the original functions,
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    g of x and f of x.
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    If this is g prime of
    x, what's go of x.
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    What's the anti-derivative
    of e to the x.
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    It's just use e to the x.
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    I'm going to switch colors,
    I don't like this blue.
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    So this is g of x.
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    I actually took the
    anti-derivative of it, but
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    it's the same exact thing.
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    And then times f of x.
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    Then I want to subtract
    the indefinite integral
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    of f prime of x.
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    One, g of x.
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    This is the same as this, which
    are both the anti-derivative of
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    this, although they
    are all the same.
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    So this is g of x and then I
    would take the derivative
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    of f of x. f prime of x.
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    What's the derivative
    of cosine of x?
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    It's minus sine of x.
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    So sine of x d x, it's
    minus sine of x.
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    I could put the minus here,
    that'll make it look messy, I
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    could put the minus here
    that'll make it messy or I
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    could just put minus here and
    make these minuses cancel
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    out and I get a plus here.
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    So I get the integral of e to
    the x cosine of x d x is equal
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    to e to the x cosine of x plus
    the integral of e to
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    the x sine of x d x.
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    Hopefully I haven't
    confused you too much.
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    I should actually do some
    integration by parts problems
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    without e to the x.
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    It's very hard to keep track
    of what I've done here.
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    This is the anti-derivative.
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    This is the anti-derivative
    and this is also the
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    anti-derivative.
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    This is g prime of
    x, this is g of x.
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    So once again we are not
    clear whether we've
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    made any progress.
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    We've
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    gone from e to the x cosine of
    x to e to the x sine of x.
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    Let's take integration by parts
    again, see what happens.
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    I'm just going to write on the
    right side of the equal sign,
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    because this might
    get a little long.
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    I'm just going to write this
    first part x to the x cosine
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    of x plus-- and now let's do
    integration by parts again.
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    For this round of integration
    by parts this was g of x, but
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    now, for this around, I'm going
    to assume it's g prime of x.
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    Which doesn't really make a
    difference because whenever I
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    take the anti-derivative of it
    to g of x, it stays the same.
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    And then I'm going to assume
    that this is f of x.
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    So integration by parts tells
    us we take f of x times g of x,
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    so I take this function and the
    anti-derivative of
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    this function.
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    The anti-derivative of this
    function is once again just e
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    to the x and then f times that
    function unchanged
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    time sine of x.
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    From that I subtract the
    integral of the anti-derivative
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    of this or I take g of x which
    is e to the x, and then the
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    derivative of f of
    x, f prime of x.
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    What's the derivative
    of sine of x?
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    It's cosine of x.
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    Cosine of x d of x.
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    Let's see if we're
    getting anywhere.
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    It seems like I just keep
    adding terms, making it
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    more and more complicated.
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    In order to see if we're
    getting anywhere, let me just
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    rewrite the whole thing and
    maybe get rid of these
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    parenthesis, because it's
    just a plus, so we can get
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    rid of the parenthesis.
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    Let me use a new color.
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    OK.
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    So this is the original
    problem, e to the x cosine of x
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    d x equals, and now let me
    switch back to this color, it
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    equals e to the x cosine of x,
    and then I can just-- this
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    parentheses doesn't matter
    because I'm just adding
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    everything in the parentheses--
    e to x cosine of x plus e to
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    the x sine of x minus e to
    the x cosine x access d x.
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    Now you might think that I
    arbitrarily switched colors
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    here when I rewrote this,
    but if you look you might
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    see why I actually did
    switch colors here.
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    See anything interesting?
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    Exactly.
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    This is the same thing as
    this, just a minus right?
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    So we're going to do something
    what I think to be fairly cool.
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    Let's add this term to both
    sides of the equation.
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    Let's take this and
    let's put it on to this
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    side of the equation.
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    If I take this and put
    it on this side of the
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    equation, what happens?
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    I then have two of these on the
    left side equation, so that
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    becomes-- I mean I could write
    it out it's e to the x cosine
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    of x d x plus, right?
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    Because I'm taking this and I'm
    putting it on that side of the
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    equation, e to the
    x cosine of x d x.
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    That's just the same thing as
    2 times the integral of e
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    to the x cosine of x d x.
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    And then that equals this term.
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    Which equals e to the x cosine
    of x plus e to the x sine of x.
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    I know it's really messy.
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    All I have to do now to solve
    this integral is divide both
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    sides by 2 and I'm done.
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    So let me write it out, this
    is very exciting, it's
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    the home stretch.
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    If I divide both sides by 2, I
    get-- and I'm going to try to
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    write it so you can see
    everything-- e to the x cosine
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    of x d x equals and on that
    side I have e to the x cosine
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    of x plus e to the x
    sine of x over 2.
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    I think that's pretty neat.
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    It's neat how integration by
    parts allowed us to do this.
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    We actually never even have
    to evaluate this integral.
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    We said, this integral is just
    the original problem again.
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    And you can think about
    why that happened, right?
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    Because these trick
    functions cycle.
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    So we had to do integration
    by parts twice to get back
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    to where we were before.
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    And then we use that to solve
    it without actually having
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    to evaluate the integral.
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    And what I also think is cool
    is even if you just look
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    at this solution, it's
    kind of neat, right?
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    The anti-derivative of e to the
    x and-- actually never forget
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    the plus c, that would've given
    me minus 1 point on the exam.
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    What's kind of cool, the
    integral of e to the x cosine
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    of x is this expression that's
    e to the x cosine of x plus e
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    to the x sine of
    x divided by 2.
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    It's the average of e to
    the x cosine of x and
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    e to the x sine of x.
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    I think that's a pretty neat
    property, and you might want
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    to graph them and play with
    them, but it's kind of neat.
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    Hopefully I have convinced you
    that is a classic of a problem,
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    and you also find it neat, and
    I'll see you in the
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    next presentation.
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Title:
Indefinite Integration (part 7)
Description:

Another example using integration by parts.

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Video Language:
English
Duration:
09:38
brettle edited English subtitles for Indefinite Integration (part 7) Apr 18, 2011, 1:05 AM
brettle edited English subtitles for Indefinite Integration (part 7) Apr 18, 2011, 1:05 AM
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brettle edited English subtitles for Indefinite Integration (part 7) Mar 2, 2011, 5:52 PM
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Amara Bot edited English subtitles for Indefinite Integration (part 7) Mar 2, 2011, 5:52 PM
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