Let
\(I_{n}=\int_{0}^{\frac{1}{4} \pi} \tan ^{n} x \mathrm{~d} x\)
where \(n \geqslant 0\). Use the fact that \(\tan ^{2} x=\sec ^{2} x-1\) to show that, for \(n \geqslant 2\),
\(I_{n}=\frac{1}{n-1}-I_{n-2}\)

Show that \(I_{8}=\frac{1}{7}-\frac{1}{5}+\frac{1}{3}-1+\frac{1}{4} \pi\).