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    Hi. I'm doing questions on volumes of revolutions, but am stuck since I am unsure how to integrate a few things, namely:

    (4a^2 - 4ax)^1/2


    and


    (2lny)^2


    I know how to differentiate when there are brackets like this, but not how to integrate.

    Help would be much appreciated.

    Thanks
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    You use the chain rule on both, but for the first one, treat a as a constant
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    (Original post by cpdavis)
    You use the chain rule on both, but for the first one, treat a as a constant
    Thanks. Also, when substituting 2 in for y once i've integrated, how would I simplify 1/3(2ln2)^3. ?
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    (Original post by Marisa_Grace)
    Thanks. Also, when substituting 2 in for y once i've integrated, how would I simplify 1/3(2ln2)^3. ?
    This isn't the right integral. What I should have suggested was to use parts. But you would break down the ln term in the differential and integrate a 1
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    (Original post by cpdavis)
    This isn't the right integral. What I should have suggested was to use parts. But you would break down the ln term in the differential and integrate a 1
    I don't really understand what you mean at the end. Also, when i integrated i got 1/3(2lny)^3 x 2/y.
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    \displaystyle\int (2lny)^2\ dy\equiv 4\int ln^2y\ dy

    u=lny\Leftrightarrow\frac{du}{dy  }=\frac{1}{y}\Leftrightarrow dy=y\ du

    \displaystyle 4\int ln^2y\ dy=4\int yu^2\ du=4\int u^2e^u\ du=...

    parts to finish it off
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    (Original post by Pheylan)
    \displaystyle\int (2lny)^2\ dy\equiv 4\int ln^2y\ dy

    u=lny\Leftrightarrow\frac{du}{dy  }=\frac{1}{y}\Leftrightarrow dy=y\ du

    \displaystyle 4\int ln^2y\ dy=4\int yu^2\ du=4\int u^2e^u\ du=...

    parts to finish it off
    you've explained it better

    Sorry OP if the last bit I said was a bit unclear
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    \displaystyle\int\sqrt{4a^2-4ax}\ dx\equiv\int\sqrt{4a(a-x)}\ dx\equiv 2\sqrt{a}\int\sqrt{a-x}\ dx\equiv 2\sqrt{a}\int (a-x)^{\frac{1}{2}}\ dx
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    .....or if you are feeling really adventurous try deriving the reduction formula for (ln x)^n....
 
 
 
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