1. 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

2. Originally Posted by Stonehambey
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,
$I_r = I_{r-1} - \int_0^{3/5} x^{2(r-1)} dx$

3. 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

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

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

$= - \frac{1}{{2r - 1}}{\left( {\frac{3}{5}} \right)^{2r - 1}} + {I_{r - 1}} \Leftrightarrow {I_{r - 1}} - I_r = \frac{1}{{2r - 1}}{\left( {\frac{3}{5}} \right)^{2r - 1}}.$