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Math Help - Hyperbolic integration problem

  1. #1
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    Hyperbolic integration problem

    So here's the problem.

    Let I_r = \int_{0}^{\ln 2}\tanh^{2r}\theta d\theta

    Show that

    I_{r-1}-I_r = \frac{1}{2r-1}\left(\frac{3}{5}\right)^{2n-1}

    My initial approach was to go by integration by parts, letting v'=\tanh x and u=(\tanh x)^{2r-1}, this leads me to

    I_r =\left[ \ln(\cosh \theta)\tanh^{2r-1} \theta \right]^{\ln 2}_0 - (2r-1)\int_{0}^{\ln 2}\rm sech ^2 \theta\ln (\cosh \theta)\tanh^{2r-2}\theta d\theta

    Now this has a (2r-1) in it and also a \tanh^{2r-2} \theta= \tanh^{2(r-1)}\theta which leads me to think I might be on the right track. However it looks such a mess, that I can't help but think I've made a mistake somewhere. If I haven't then I really can't see how to progress from here either.

    Any help is much appreciated

    Stonehambey
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  2. #2
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    Quote Originally Posted by Stonehambey View Post
    So here's the problem.

    Let I_r = \int_{0}^{\ln 2}\tanh^{2r}\theta d\theta

    Show that

    I_{r-1}-I_r = \frac{1}{2r-1}\left(\frac{3}{5}\right)^{2n-1}
    Write, I_r = \int_0^{\ln 2} \tanh^{2(r-1)}\theta \cdot \tanh^2 \theta d\theta.
    Remember the identity, \tanh^2 \theta = 1 - \text{sech}^2 \theta.
    Thus, I_r = \int_0^{\ln 2} \tanh^{2(r-1)} \theta d\theta - \int_0^{\ln 2} \tanh^{2(r-1)} \theta \cdot \text{sech}^2 \theta d\theta
    In the second integral let x = \tanh \theta \implies x' = \text{sech}^2 \theta also x(0)=0,x(\ln 2) = \tfrac{3}{5}.
    Thus,
    I_r = I_{r-1} - \int_0^{3/5} x^{2(r-1)} dx
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  3. #3
    Senior Member DeMath's Avatar
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    Or so

    {I_r} = \int\limits_0^{\ln 2} {{{\tanh }^{2r}}\left( \theta  \right)d\theta  = } \int\limits_0^{\ln 2} {{{\sinh }^{2r - 1}}\left( \theta  \right)\frac{{\sinh \left( \theta  \right)}}{{{{\cosh }^{2r}}\left( \theta  \right)}}d\theta }.

    \int\limits_0^{\ln 2} {\frac{{\sinh \left( \theta  \right)}}{{{{\cosh }^{2r}}\left( \theta  \right)}}}d\theta   = \int\limits_0^{\ln 2} {\frac{{d\left( {\cosh \left( \theta  \right)} \right)}}{{{{\cosh }^{2r}}\left( \theta  \right)}}}  = \int\limits_0^{\ln 2} {{{\left[ {\cosh \left( \theta  \right)} \right]}^{^{ - 2r}}}d\left( {\cosh \left( \theta  \right)} \right)}  =

    = \left. {\frac{{{{\left[ {\cosh \left( \theta  \right)} \right]}^{^{1 - 2r}}}}}{{1 - 2r}}} \right|_0^{\ln 2} = \left. { - \frac{1}{{\left( {2r - 1} \right){{\cosh }^{2r - 1}}\left( \theta  \right)}}} \right|_0^{\ln 2}.

    So we have

    {I_r} = \left. { - \frac{1}{{2r - 1}}\frac{{{{\sinh }^{2r - 1}}\left( \theta  \right)}}{{{{\cosh }^{2r - 1}}\left( \theta  \right)}}} \right|_0^{\ln 2} + \frac{1}{{2r - 1}}\int\limits_0^{\ln 2} {\frac{{d\left( {{{\sinh }^{2r - 1}}\left( \theta  \right)} \right)}}{{{{\cosh }^{2r - 1}}\left( \theta  \right)}}}  =

    =  - \frac{1}{{2r - 1}}{\left( {\frac{3}{5}} \right)^{2r - 1}} + \int\limits_0^{\ln 2} {{{\tan }^{2\left( {r - 1} \right)}}\left( \theta  \right)d\theta }  =

    =  - \frac{1}{{2r - 1}}{\left( {\frac{3}{5}} \right)^{2r - 1}} + {I_{r - 1}} \Leftrightarrow {I_{r - 1}} - I_r = \frac{1}{{2r - 1}}{\left( {\frac{3}{5}} \right)^{2r - 1}}.
    Last edited by DeMath; February 8th 2009 at 03:36 AM.
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