Answer
Divergent
Work Step by Step
Let \[I=\int_{0}^{\frac{π}{2}}\tan^2\theta\:d\theta\]
Since $\large\frac{π}{2}$ is point of infinite discontinuity of integrand $\tan^2\theta$
\[I=\int_{0}^{\frac{π}{2}}\tan^2\theta\:d\theta=\lim_{t\rightarrow\frac{π}{2} ^-}\int_{0}^{t}\tan^2\theta\:d\theta\]
\[I=\lim_{t\rightarrow\frac{π}{2} ^-}\int_{0}^{t}\tan^2\theta\:d\theta\;\;\;\ldots (1)\]
Let \[I_1=\int\tan^2 \theta d\theta\]
\[I_1=\int(\sec^2 \theta-1) d\theta\]
\[I_1=\tan \theta-\theta\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow\frac{π}{2} ^-}\left[\tan\theta-\theta\right]_{0}^{t}\]
\[I=\lim_{t\rightarrow\frac{π}{2} ^-}\left[\tan t-t-0\right]\]
$\;\;\;\;\;\;\;\;\;\;\;\;\; =$ does not exist
Since limit on R.H.S. of (1) does not exist so $I$ is divergent.
