Answer: (a) For \(n \geq 2\),

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

(b) \(I_4=\frac{236}{375}\).

(a) Write

\(I_n=\int_0^{\ln 3}\operatorname{sech}^n x\,\mathrm{d}x=\int_0^{\ln 3}\operatorname{sech}^{n-2}x\operatorname{sech}^2x\,\mathrm{d}x.\)

Integrate by parts with \(u=\operatorname{sech}^{n-2}x\) and \(\mathrm{d}v=\operatorname{sech}^2x\,\mathrm{d}x\). Then \(v=\tanh x\), and using \(\frac{\mathrm{d}}{\mathrm{d}x}(\operatorname{sech}x)=-\tanh x\operatorname{sech}x\),

\(\frac{\mathrm{d}u}{\mathrm{d}x}=-(n-2)\operatorname{sech}^{n-2}x\tanh x.\)

So

\(I_n=\left[\operatorname{sech}^{n-2}x\tanh x\right]_0^{\ln 3}+(n-2)\int_0^{\ln 3}\operatorname{sech}^{n-2}x\tanh^2x\,\mathrm{d}x.\)

Now use \(\tanh^2x=1-\operatorname{sech}^2x\):

\(I_n=\left[\operatorname{sech}^{n-2}x\tanh x\right]_0^{\ln 3}+(n-2)\int_0^{\ln 3}\operatorname{sech}^{n-2}x(1-\operatorname{sech}^2x)\,\mathrm{d}x.\)

Hence

\(I_n=\left[\operatorname{sech}^{n-2}x\tanh x\right]_0^{\ln 3}+(n-2)(I_{n-2}-I_n).\)

At \(x=\ln 3\),

\(\cosh(\ln 3)=\frac{3+\frac13}{2}=\frac53\), so \(\operatorname{sech}(\ln 3)=\frac35\),

and

\(\sinh(\ln 3)=\frac{3-\frac13}{2}=\frac43\), so \(\tanh(\ln 3)=\frac{\frac43}{\frac53}=\frac45\).

Also \(\tanh 0=0\). Therefore

\(\left[\operatorname{sech}^{n-2}x\tanh x\right]_0^{\ln 3}=\left(\frac35\right)^{n-2}\left(\frac45\right).\)

So

\(I_n=\left(\frac35\right)^{n-2}\left(\frac45\right)+(n-2)(I_{n-2}-I_n).\)

Rearranging,

\((n-1)I_n=\left(\frac35\right)^{n-2}\left(\frac45\right)+(n-2)I_{n-2}.\)

(b) First,

\(I_2=\int_0^{\ln 3}\operatorname{sech}^2x\,\mathrm{d}x=[\tanh x]_0^{\ln 3}=\frac45.\)

Now put \(n=4\) into the reduction formula:

\(3I_4=\left(\frac35\right)^2\left(\frac45\right)+2I_2.\)

Thus

\(3I_4=\frac{9}{25}\cdot\frac45+2\cdot\frac45=\frac{36}{125}+\frac85=\frac{36}{125}+\frac{200}{125}=\frac{236}{125}.\)

Hence

\(I_4=\frac{236}{375}.\)