Find the reduction formula for $I_n = \int (\tanh^{n}x)$

Question

The question is: Prove that $I_n = I_{n-2} - \frac{1}{n-1} (\tanh^{n-1}x)$

Don't really know how to get started on this one.

Thanks

Answer 1

We have $$\frac{1}{n-1}\tanh^{n-1}x =\int \tanh^{n-2} x\mathrm{sech}^2 x dx=\int \tanh^{n-2} x(1-\tanh^2 x) dx=I_{n-2}-I_n.$$