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Math Help - Reduction Formulae

  1. #1
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    Reduction Formulae

    GRR
    Im stuck



    I_n=\int(2-x^2)^n .dx               \ \ \ \ \ n\geq0

    (limits are \sqrt{2} and 0, i dont know how to put them usin the Latex codes, sorry)

    show that

    I_n=\frac{4n}{2n+1}I_{n-1} \ \ \ \ \ n\geq1


    please help
    :S
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Silver View Post
    GRR
    Im stuck



    I_n=\int(2-x^2)^n .dx \ \ \ \ \ n\geq0

    (limits are \sqrt{2} and 0, i dont know how to put them usin the Latex codes, sorry)

    show that

    I_n=\frac{4n}{2n+1}I_{n-1} \ \ \ \ \ n\geq1


    please help
    :S
    Let u=2-x^2\Rightarrow{\sqrt{2-u}=x}

    so dx=\frac{-1}{2\sqrt{2-u}}

    That is how I would start, then move unto parts

    What do you mean limits? are you saying this a definite integral?
    Last edited by Mathstud28; June 1st 2008 at 01:50 PM.
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    What do you mean limits? are you saying this a definite integral?
    No I meant the limits we are supposed to sub in at the end to get the first bit of the reduction formulae


    (meaning this bit: \frac{4n}{2n+1})








    thanks

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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Let u=2-x^2\Rightarrow{\sqrt{2-u}=x}

    so dx=\frac{1}{\sqrt{2-u}}

    That is how I would start, then move unto parts

    What do you mean limits? are you saying this a definite integral?
    u = 2 - x^2

    x = \sqrt{2 - u}

    dx = - \frac{1}{2 \sqrt{2 - u}} \ du
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by janvdl View Post
    u = 2 - x^2

    x = \sqrt{2 - u}

    dx = - \frac{1}{2 \sqrt{2 - u}}
    sdlfjalsdkfjasdkl;fjasdkl;fjasdkl;fjasl;dfjasl;dfk jaskl;dfjasldk;fjkasdfkljasdlfk;jaskl;fjasdfl;kjas d;fkljasdlf, thank you,
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  6. #6
    Moo
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    Hi !

    Quote Originally Posted by janvdl View Post
    u = 2 - x^2

    x = \sqrt{2 - u}

    dx = - \frac{1}{2 \sqrt{2 - u}}
    Quote Originally Posted by Mathstud28 View Post
    Let u=2-x^2\Rightarrow{\sqrt{2-u}=x}

    so dx=\frac{1}{\sqrt{2-u}}

    That is how I would start, then move unto parts

    What do you mean limits? are you saying this a definite integral?
    You two should add the du part !


    dx=-\frac{\color{red}du}{2\sqrt{2-u}}

    Because when substituting dx in the integral with the formula you both gave, there will be no du !
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hi !





    You two should add the du part !


    dx=-\frac{\color{red}du}{2\sqrt{2-u}}

    Because when substituting dx in the integral with the formula you both gave, there will be no du !
    Yeah ..I never ever put the du...it gets me in trouble
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  8. #8
    Math Engineering Student
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    \begin{aligned}<br />
   I_{n}&=\int_{0}^{\sqrt{2}}{\left( 2-x^{2} \right)^{n}\,dx} \\ <br />
 & =\int_{0}^{\sqrt{2}}{(x)'\left( 2-x^{2} \right)^{n}\,dx} \\ <br />
 & =x\left( 2-x^{2} \right)^{n} \bigg|_{0}^{\sqrt{2}}+2n\int_{0}^{\sqrt{2}}{\Big( 2-\left( 2-x^{2} \right) \Big)\left( 2-x^{2} \right)^{n-1}\,dx} \\ <br />
 & =4nI_{n-1}-2nI_{n} \\ <br />
  \implies I_{n}&=\frac{4n}{2n+1}I_{n-1}.\quad\blacksquare<br />
\end{aligned}
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