Answer
$$
\int_{0}^{\pi / 2} \tan ^{2} \theta d \theta
$$
We note first the given integral is improper because $ f(x)= \tan ^{2} \theta $ has the vertical asymptote $\theta =\pi / 2 $. The infinite discontinuity occurs at the right endpoint of $ [0, \pi / 2] $
The given improper integral is divergent.
Work Step by Step
$$
\int_{0}^{\pi / 2} \tan ^{2} \theta d \theta
$$
Observe that the given integral is improper because $ f(x)= \tan ^{2} \theta $ has the vertical asymptote $\theta =\pi / 2 $. Since the infinite discontinuity occurs at the right endpoint of $ [0, \pi / 2] $ , we must use part (a) of Definition 3 :
$$
\begin{split}
\int_{0}^{\pi / 2} \tan ^{2} \theta d \theta & = \lim _{t \rightarrow \pi / 2 ^{-}} \int_{0}^{t} \tan ^{2} \theta d \theta \\
& =\lim _{t \rightarrow \pi / 2 ^{-}} \int_{0}^{t} (\ sec^{2}\theta -1) d \theta \\
& =\lim _{t \rightarrow \pi / 2 ^{-}}[\tan \theta- \theta ]_{0}^{t} \\
& =\lim _{t \rightarrow \pi / 2 ^{-}}[\tan t - t] \\
& =\infty,
\end{split}
$$
so the integral
$$
\int_{0}^{\pi / 2} \tan ^{2} \theta d \theta
$$
is divergent.
Thus the given improper integral is divergent.