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Math Help - Integration by substitution u^2

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    Integration by substitution u^2

    I am stuck on this integral, i havent come across u^2 before and ive given it a shot but have got stuck where you substitute \frac{dy}{du} in... maybe someone could help

    Integrate the following:

     \frac{x}{\sqrt{3-x}} dx;    u^2 = 3-x
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  2. #2
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    Quote Originally Posted by darksupernova View Post
    I am stuck on this integral, i havent come across u^2 before and ive given it a shot but have got stuck where you substitute \frac{dy}{du} in... maybe someone could help

    Integrate the following:

     \frac{x}{\sqrt{3-x}} dx;    u^2 = 3-x
    If instead you choose u=3-x, du=dx

    you get

    \int{\frac{3-u}{u^{\frac{1}{2}}}}du=\int{3u^{-\frac{1}{2}}}du-\int{u^{\frac{1}{2}}}du

    =\frac{3u^{\frac{1}{2}}}{\frac{1}{2}}-\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=6u^{\frac{1}  {2}}-\frac{2}{3}u^{\frac{3}{2}}+C

    replacing u with 3-x, this is

    6\sqrt{3-x}-\frac{2}{3}\left(\sqrt{3-x}\right)^3+C

    =6\sqrt{3-x}-\frac{2}{3}(3-x)\sqrt{3-x}+C

    which can be factorised.
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    Quote Originally Posted by darksupernova View Post
    I am stuck on this integral, i havent come across u^2 before and ive given it a shot but have got stuck where you substitute \frac{dy}{du} in... maybe someone could help

    Integrate the following:

     \frac{x}{\sqrt{3-x}} dx;    u^2 = 3-x
    u^2 = 3 - x
    2u*du = -dx
    x = 3 - u^2.
    The given integration is
    intg[(3 - u^2)(-2u)du/u
    intg[2(u^2 - 3)]du.
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    @ Archie Meade the question tells me to use this substitution...

    @ sa-ri-ga-ma

    I get \frac{2}{3}(3-x)^\frac{3}{2} - 6(3-x)^\frac{1}{2}

    Is this correct, how can i simplify this?

    I may well have made a mistake as the answer is -\frac{2}{3}(x+6)\sqrt{(3-x)}
    Last edited by darksupernova; April 20th 2010 at 04:31 AM.
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    Quote Originally Posted by darksupernova View Post
    @ Archie Meade the question tells me to use this substitution...

    @ sa-ri-ga-ma

    I get \frac{2}{3}(3-x)^\frac{3}{2} - 6(3-x)^\frac{1}{2}

    Is this correct, how can i simplify this?

    I may well have made a mistake as the answer is -\frac{2}{3}(x+6)\sqrt{(3-x)}
    '+ C' needs to be included.

    To simplify, take out the common factor of 2(3 - x)^{1/2} = 2 \sqrt{3 - x} and then simplify.
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    Quote Originally Posted by darksupernova View Post
    @ Archie Meade the question tells me to use this substitution...

    @ sa-ri-ga-ma

    I get \frac{2}{3}(3-x)^\frac{3}{2} - 6(3-x)^\frac{1}{2}

    Is this correct, how can i simplify this?

    I may well have made a mistake as the answer is -\frac{2}{3}(x+6)\sqrt{(3-x)}
    (3- x)^{3/2}= (3- x)^{1+ 1/2}= (3-x)(3- x)^{1/2} so you can factor out a (3- x)^{1/2}:\
    \frac{2}{3}(3- x)^{3/2}- 6(3- x)^{1/2}= (3- x)^{1/2}(\frac{2}{3}(3- x)- (2/3)(9) -(2/3)(3- x)^{1/2}(9- (3- x)= (3- x)^{1/2}(x+ 6).
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  7. #7
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    ah thanks HallsofIvy, its stupid the loops i have to jump through, surely seeing i can do the integration would be enough!

    Thanks for all the help everyone!
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