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Math Help - tough integral?

  1. #1
    Eater of Worlds
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    tough integral?

    Hey Kriz. Here's an integral for ya'.

    \int_{0}^{1}\frac{4x^{3}(1+x^{4(2006)})}{(1+x^{4})  ^{2008}}dx
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  2. #2
    Senior Member Peritus's Avatar
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    for this tough integral we'll use the following substitution:


    \begin{gathered}<br />
  t = \frac{1}<br />
{{1 + x^4 }} \hfill \\<br />
   \hfill \\<br />
  dt = \frac{{ - 4x^3 }}<br />
{{\left( {1 + x^4 } \right)^2 }}dx \hfill \\ <br />
\end{gathered}

    <br />
\int\limits_0^1 { - t^{2006} \left[ {1 + \left( {\frac{{1 - t}}<br />
{t}} \right)^{2006} } \right]} dt = \int\limits_0^1 { - t^{2006}  - \left( {1 - t} \right)^{2006} } dt =  - \frac{{t^{2007} }}<br />
{{2007}} + \frac{{\left( {1 - t} \right)^{2007} }}<br />
{{2007}}<br />

    now back substitute:

    <br />
 = \frac{{\left( {1 - \frac{1}<br />
{{1 + x^4 }}} \right)^{2007}  - \left( {\frac{1}<br />
{{1 + x^4 }}} \right)^{2007} }}<br />
{{2007}} = \left. {\frac{{x^{8028}  - 1}}<br />
{{2007\left( {1 + x^4 } \right)^{2007} }}} \right|_0^1  = \frac{1}<br />
{{2007}}<br /> <br />
    Last edited by Peritus; February 19th 2008 at 05:12 PM.
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  3. #3
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    Quote Originally Posted by galactus View Post
    Hey Kriz. Here's an integral for ya'.

    \int_{0}^{1}\frac{4x^{3}(1+x^{4(2006)})}{(1+x^{4})  ^{2008}}dx
    More generally: let n\ge0, so \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}}<br />
{{\left( {1 + x^4 } \right)^{n + 2} }}\,dx}  = \frac{1}<br />
{{n + 1}}.

    First make the obvious substitution: u=x^4, so

    \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}}<br />
{{\left( {1 + x^4 } \right)^{n + 2} }}\,dx}  = \int_0^1 {\frac{{1 + u^n }}<br />
{{(1 + u)^{n + 2} }}\,du} .

    Splittin' the original ratio into two fractions we get an easy integral whose value is \frac{{1 - 2^{ - n - 1} }}<br />
{{n + 1}}. Now here's the interesting part: the second piece.

    We have \int_0^1 {\frac{{u^n }}<br />
{{(1 + u)^{n + 2} }}\,du} . Substitute v = \frac{u}<br />
{{u + 1}}. After some simple calculations our integral becomes to

    \int_0^{1/2} {v^n \,dv}  = \left. {\frac{{v^{n + 1} }}<br />
{{n + 1}}} \right|_0^{1/2}  = \frac{{2^{ - n - 1} }}<br />
{{n + 1}}.

    Finally, put these things together and we have \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}}<br />
{{\left( {1 + x^4 } \right)^{n + 2} }}\,dx}  = \frac{1}<br />
{{n + 1}}\quad\blacksquare
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    More generally: let n\ge0, so \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}}<br />
{{\left( {1 + x^4 } \right)^{n + 2} }}\,dx}  = \frac{1}<br />
{{n + 1}}.

    First make the obvious substitution: u=x^4, so

    \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}}<br />
{{\left( {1 + x^4 } \right)^{n + 2} }}\,dx}  = \int_0^1 {\frac{{1 + u^n }}<br />
{{(1 + u)^{n + 2} }}\,du} .

    Splittin' the original ratio into two fractions we get an easy integral whose value is \frac{{1 - 2^{ - n - 1} }}<br />
{{n + 1}}. Now here's the interesting part: the second piece.

    We have \int_0^1 {\frac{{u^n }}<br />
{{(1 + u)^{n + 2} }}\,du} . Substitute v = \frac{u}<br />
{{u + 1}}. After some simple calculations our integral becomes to

    \int_0^{1/2} {v^n \,dv}  = \left. {\frac{{v^{n + 1} }}<br />
{{n + 1}}} \right|_0^{1/2}  = \frac{{2^{ - n - 1} }}<br />
{{n + 1}}.

    Finally, put these things together and we have \int_0^1 {\frac{{4x^3 \left( {1 + x^{4n} } \right)}}<br />
{{\left( {1 + x^4 } \right)^{n + 2} }}\,dx}  = \frac{1}<br />
{{n + 1}}\quad\blacksquare
    that was... beautiful
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  5. #5
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    Very nice and elegant. Wasn't tough afterall, was it?.

    Here's what I done:

    Broke it up

    \int_{0}^{1}\frac{4x^{3}}{(1+x^{4})^{2008}}dx+\int  _{0}^{1}\frac{4x^{3}\cdot{x^{4(2006)}}}{(1+x^{4})^  {2008}}dx

    sub u=1+x^{4} in first part.

    \int_{1}^{2}\frac{1}{u^{2008}}du

    Second part beomes:

    \int_{1}^{2}\frac{(u-1)^{2006}}{u^{2008}}du=\int_{1}^{2}\frac{(1-\frac{1}{u})^{2006}}{u^{2}}du

    Now, sub in t=1-\frac{1}{u}

    and the last one becomes:

    \int_{0}^{\frac{1}{2}}t^{2006}dt

    Put it all together and you do indeed get \frac{1}{2007}
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  6. #6
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    It wasn't after all.

    When I saw your problem, I think it was messy, but while solving it, it was getting interesting.

    Thanks for your problem.
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  7. #7
    Eater of Worlds
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    Yeah, I thought it was cool. I posted my way. It was kind of like yours, except you went one further and proive dthe general case. Show off

    Now, it's time for me to
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Substitute v = \frac{u}<br />
{{u + 1}}.
    Quote Originally Posted by galactus View Post

    Now, sub in t=1-\frac{1}{u}
    Wait, how did we "see" these substitutions?
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  9. #9
    Math Engineering Student
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    When I got \int_{0}^{1}{\frac{u^{n}}{(1+u)^{n+2}}\,du}, I made \frac{u^{n}}{(1+u)^{n+2}}=\left( \frac{u}{1+u} \right)^{n}\cdot \frac{1}{(1+u)^{2}}. Since by differentiating \frac{u}{1+u}=\frac{1+u-1}{1+u}=1-\frac{1}{1+u} it immediately yields a part of the integrand and from there you set the given substitution.
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