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    integration by parts

    hi, this is from P5 aqa june 03. I can do most of it its just the penultimate line

    Question..show

    int.(tlnt)dt = 1/4.t^2(2lnt - 1) + c

    so I did

    int.(tlnt)dt

    lnt.t^2/2 - int.t^2/2.1/x

    ....missing line

    1/4t^2 (2lnt - 1)

    I can see therefore that the missing line is

    lnt.t^2/2 - 1/4t^2

    can someone explain how this was integrated??

    thanks
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    I did it from scratch as it's clearer.
    Edit: Obviously the last line factorises to the required expression.
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    arr...i see

    if in the exam I had just figured ou the missing lin would I have still got the marks?!
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    You probably would have! They don't know that you did it backwards, I always have to do that for proving stuff but they can't penalise you!

    Please could you help me with this integrating trig???

    ∫sin² 2x. cos 2x dx

    The answer is 1/6 sin³ 2x + c but I have no idea how to do it.

    Thanks in advance!
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    (Original post by fabz)
    You probably would have! They don't know that you did it backwards, I always have to do that for proving stuff but they can't penalise you!

    Please could you help me with this integrating trig???

    ∫sin² 2x. cos 2x dx

    The answer is 1/6 sin³ 2x + c but I have no idea how to do it.

    Thanks in advance!
    Yo

    With stuff like that you basically have to guess what do I differentiate to get that. If you ever have a trig function to a power (sin² 2x) with its differential to the power of one (cosx) next to it you can guess its integral:

    Just take the trig bit with the power, add one to it and differentiate it and see what happens -

    differentiate (sin2x)^3 = [(sin2x)^2] x 3 x 2cos2x

    that's x3 from the ^3 and x 2cos2x from differentiating sin2x.

    This gives you 6cosx(sin2x)^2 but you want cosx(sin2x)^2 so just divide the original bit by 6 to get:
    1/6 sin³ 2x + C

    Hope this helps
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    (Original post by fabz)
    You probably would have! They don't know that you did it backwards, I always have to do that for proving stuff but they can't penalise you!

    Please could you help me with this integrating trig???

    ∫sin² 2x. cos 2x dx

    The answer is 1/6 sin³ 2x + c but I have no idea how to do it.

    Thanks in advance!
    You have a function and a scalar multiple of derivative.
    Formally:
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    (Original post by Gaz031)
    You have a function and a scalar multiple of derivative.
    Formally:
    good point, I think guessing is more fun though!
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    Thanks so much...

    It's mucho clearer now!

    Back to revising...
 
 
 
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