Problem

It is given that, for $n\ge0$,

$$I_n=\int_0^{\ln3}\operatorname{sech}^n x\,\mathrm{d}x$$

(a) Show that, for $n\ge2$,

$$(n-1)I_n=\left(\frac{3}{5}\right)^{n-2}\left(\frac{4}{5}\right)+(n-2)I_{n-2}$$

You may use the result that

$$\frac{\mathrm{d}}{\mathrm{d}x}(\operatorname{sech}x)=-\tanh x\operatorname{sech}x.$$

(b) Find the value of $I_4$.