Answer
Diverges
Work Step by Step
Since
\begin{align*}
\int_{0}^{\pi / 2} \cot \theta d \theta&=\lim _{b \rightarrow 0^{+}}\int_{0}^{\pi / 2} \cot \theta d \theta\\
&=\lim _{b \rightarrow 0^{+}}[\ln |\sin \theta|]_{b}^{\pi / 2}\\
&=\lim _{b \rightarrow 0^{+}}[\ln 1-\ln |\sin b|]\\
&=-\lim _{b \rightarrow 0^{+}}[\ln |\sin b|]\\
&=+\infty
\end{align*}
Then the integral diverges.