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Math Help - Anti-Derivative using u and du.

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    Thumbs down Anti-Derivative using u and du.



    I don't understand the circled part. Since U uses up the 2x in front of the cos2x how come a DU is still present. Also why is the 1/2 there?
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    Quote Originally Posted by Brazuca View Post


    I don't understand the circled part. Since U uses up the 2x in front of the cos2x how come a DU is still present. Also why is the 1/2 there?
    note that \frac{du}{2} = \cos(2x) \, dx
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    Then what about the u^2?

    why is it not effected by the 1/2?

    shouldn't Sin(squared)2x turn into [(squared)2x]/2?
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    no
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    how come the u^2 isn't effected by the 1/2?
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    \int \textcolor{red}{\sin^2(2x)} \textcolor{blue}{\cos(2x) \, dx}

    \textcolor{red}{u = \sin(2x)}

    \textcolor{blue}{du = 2\cos(2x) \, dx}

    \textcolor{blue}{\frac{1}{2}} \int \textcolor{red}{\sin^2(2x)} \textcolor{blue}{2\cos(2x) \, dx}

    note that the \frac{1}{2} on the outside of the integral compensates for the 2 inserted in the integrand

    \textcolor{blue}{\frac{1}{2}} \int \textcolor{red}{u^2} \textcolor{blue}{du}

    see why now?
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    Oh now I understand. So by having the 1/2 on the outside it causes the 2 to appear on the inside.
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    Quote Originally Posted by Brazuca View Post
    Oh now I understand. So by having the 1/2 on the outside it causes the 2 to appear on the inside.
    other way around ... you need the constant 2 in the integrand for du, the 1/2 on the outside compensates because (1/2)*2 = 1
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    Ok now I think I get it.

    Because you put the 2 inside the integral you need to compensate by putting the 1/2 on the outside.

    So if I had needed to put a 5 inside I would put a 1/5 outside right?
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  10. #10
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    Quote Originally Posted by Brazuca View Post
    Ok now I think I get it.

    Because you put the 2 inside the integral you need to compensate by putting the 1/2 on the outside.

    So if I had needed to put a 5 inside I would put a 1/5 outside right?
    that is correct. in effect, it's like multiplying the integral by 1 ... no change.
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  11. #11
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    Quote Originally Posted by Brazuca View Post
    Ok now I think I get it.
    Because you put the 2 inside the integral you need to compensate by putting the 1/2 on the outside.
    So if I had needed to put a 5 inside I would put a 1/5 outside right?
    I have a suggestion; simply differentiate the following paying very close attention to the process.
    \frac{1}{6}\sin^3(2x)
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