Answer
$$\displaystyle\int{\theta \tan ^2\theta \ d\theta}=
\theta \tan \theta -\ln \left| \sec \theta \right|-\frac{1}{2}\theta ^2+C
$$
Work Step by Step
$\displaystyle{I=\int{\theta \tan ^2\theta \ d\theta}\\
I=\int{\theta \left( \sec ^2\theta -1 \right) d\theta}\\
I=\int{\theta \sec ^2\theta -\theta \ d\theta}\\
I=\int{\theta \sec ^2\theta \ d\theta}-\int{\theta \ d\theta}\\
I=\color{skyblue}{\int{\theta \sec ^2\theta \ d\theta}}\color{limegreen}{-\int{\theta \ d\theta}}}$
$\displaystyle\color{limegreen}{-\int{\theta \,\,d\theta}}=\color{limegreen}{-\frac{1}{2}\theta ^2}$
$\displaystyle\color{skyblue}{\int{\theta \sec ^2\theta \ d\theta}}$
$\displaystyle{\left[\begin{matrix}
u=\theta& du=1\\
dv=\sec ^2\theta& v=\tan \theta\\
\end{matrix}\right]}$
$\displaystyle\theta \tan \theta -\int{\tan \theta \ d\theta}\\
\color{skyblue}{\displaystyle\theta \tan \theta -\ln \left| \sec \theta \right|}$
$
I=\displaystyle\color{skyblue}{\theta \tan \theta -\ln \left| \sec \theta \right|}\color{limegreen}{-\frac{1}{2}\theta ^2}+C\\
I=\displaystyle
\theta \tan \theta -\ln \left| \sec \theta \right|-\frac{1}{2}\theta ^2+C
$
