We have,
$$
\begin{aligned}
& I=\int_{\pi / 4}^{\pi / 2} \cot ^9 x d x \\
& I=\int_{\pi / 4}^{\pi / 2} \frac{\cos ^9 x}{\sin ^9 x} d x
\end{aligned}
$$
Put $\sin x=t, \cos x d x=d t$
$$
\begin{aligned}
\quad x & =\frac{\pi}{4}, t=\frac{1}{\sqrt{2}}, x=\frac{\pi}{2}, t=1 \\
\therefore \quad I & =\int_{1 / \sqrt{2}}^1 \frac{\left(1-t^2\right)^4}{t^9} d t \\
& I=\int_{1 / \sqrt{2}}^1\left(\frac{1-4 t^2+6 t^4-4 t^6+t^8}{t^9}\right) d t \\
& I=\int_{1 / \sqrt{2}}^1\left[t^{-9}-4 t^{-7}+6 t^{-5}-4 t^{-3}+\frac{1}{t}\right] d t \\
& I=\left[\frac{1}{8 t^8}-\frac{4}{6 t^6}+\frac{6}{4 t^4}-\frac{4}{2 t^2}+\log t\right]_{1 / \sqrt{2}}^1
\end{aligned}
$$


$$
\begin{aligned}
& \left.I\left(\frac{1}{8}-\frac{4}{6}+\frac{6}{4}-\frac{4}{2}+0\right)-\left(2-\frac{32}{6}-6-4-\frac{1}{2} \log 2\right)\right] \\
& \quad I=\frac{-7}{24}+\frac{1}{2} \log 2
\end{aligned}
$$