So here's the problem.

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

Show that

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

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

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

Now this has a $\displaystyle (2r-1)$ in it and also a $\displaystyle \tanh^{2r-2} \theta= \tanh^{2(r-1)}\theta$ 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