Answer
$$\sec \theta+\cos \theta+C $$
Work Step by Step
We evaluate the trigonometric integral as follows:
\begin{align*}
\int \sec ^{2} \theta \sin ^{3} \theta d \theta&=\int \frac{\sin \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}\\
&=\int \frac{\sin \theta}{\cos ^{2} \theta} d \theta-\int \sin \theta d \theta\\
&=\cos ^{-1} \theta-(-\cos \theta)+C\\
&=\sec \theta+\cos \theta+C
\end{align*}
Where we used the fact that $\sin^2 x + \cos^2 x=1$
