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Math Help - Integration by reduction: stuck hard

  1. #1
    Member ssadi's Avatar
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    Integration by reduction: stuck hard

    Hello people
    I have been trying for hours to find the reduction formulae for this integrations with futility. Help would be hugely appreciated.
    I am at a loss when trying to integrate these expressions by parts. While I decrease the degree of one part, the degree of another part increases so I am held impasse.
    I need advise on integration by parts, especially on these types of sums.
    Here are the problems:
    (a) \int^1_0 \frac{1}{(1+x^2)^n} dx
    (b) \int^1_0 x^{n-2} \sqrt(1-x) dx
    The solutions:
    (a) 2(n-1)I_n=2^{1-n} + (2n-3) I_{n-1}
    (b) I_n=\frac{n}{n+3} I_{n-2}
    Please help me
    Last edited by ssadi; November 14th 2008 at 05:43 AM. Reason: forgot to wrap with math code
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\int\limits_0^1 {\frac{1}<br />
{{\left( {1 + x^2 } \right)^n }}} dx = \left. {\frac{x}<br />
{{\left( {1 + x^2 } \right)^n }}} \right|_0^1  + 2n\int\limits_0^1 {\frac{{x^2 }}<br />
{{\left( {1 + x^2 } \right)^{n + 1} }}} dx =

    <br />
 = 2^{ - n}  + 2n\int\limits_0^1 {\left( {1 - \frac{1}<br />
{{1 + x^2 }}} \right)\left( {\frac{1}<br />
{{\left( {1 + x^2 } \right)^n }}} \right)} dx = <br />


    \begin{gathered}<br />
   = 2^{ - n}  + 2n\left( {I_n  - I_{n + 1} } \right) \hfill \\<br />
   \Rightarrow I_n  = 2^{ - n}  + 2n\left( {I_n  - I_{n + 1} } \right) \hfill \\<br />
   \Leftrightarrow \left( {1 - 2n} \right)I_n  = 2^{ - n}  - 2nI_{n + 1}  \hfill \\<br />
   \Leftrightarrow \left( {3 - 2n} \right)I_{n - 1}  = 2^{1 - n}  - 2\left( {n - 1} \right)I_n  \hfill \\ <br />
\end{gathered}


    b.


    <br />
\int\limits_0^1 {x^{n - 2} \sqrt {1 - x} dx}  = \left. { - \frac{2}<br />
{3}\left( {1 - x} \right)^{\frac{3}<br />
{2}} x^{n - 2} } \right|_0^1  + \frac{{2\left( {n - 2} \right)}}<br />
{3}\int\limits_0^1 {\left( {1 - x} \right)^{\frac{3}<br />
{2}} x^{n - 3} dx = } <br />

    <br />
\begin{gathered}<br />
   = \frac{{2\left( {n - 2} \right)}}<br />
{3}\int\limits_0^1 {\left( {1 - x} \right)x^{n - 3} \sqrt {1 - x} dx = } \frac{{2\left( {n - 2} \right)}}<br />
{3}\left( {I_{n - 1}  - I_n } \right) \hfill \\<br />
   \Leftrightarrow \left( {1 + \frac{{2\left( {n - 2} \right)}}<br />
{3}} \right)I_n  = \frac{{2\left( {n - 2} \right)}}<br />
{3}I_{n - 1}  \hfill \\ <br />
\end{gathered} <br />

    ...
    Last edited by Peritus; November 14th 2008 at 07:45 AM.
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  3. #3
    Member ssadi's Avatar
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    Thanks a lot. These are really tricky, and i am just a novice.
    I have a question on your working of (a), how do you convert I_{n+1} to I_{n-1} at the last steps:


    I have spent another hour trying to understand this step and failing miserably. Dear me. Help me.
    Attached Thumbnails Attached Thumbnails Integration by reduction: stuck hard-untitled.gif  
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    It's like when having I_n=n\implies I_{n+1}=n+1, in this case, Peritus took I_{n} and turned that as I_{n-1}.
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  5. #5
    Junior Member
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    Quote Originally Posted by ssadi View Post
    Thanks a lot. These are really tricky, and i am just a novice.
    I have a question on your working of (a), how do you convert I_{n+1} to I_{n-1} at the last steps:


    I have spent another hour trying to understand this step and failing miserably. Dear me. Help me.

    he replaced all the n's with (n-1)

    for example (1-2n) changes to [1-2(n-1)] = 3-2n
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  6. #6
    Member ssadi's Avatar
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    Thanks Peritus and everyone else posting in the thread
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