Find: ∫ sec^3 θ d θ

$$\begin{aligned} & I=\int \sec ^{3} \theta d \theta=\int \sec ^{2} \theta \cdot \sec \theta d \theta \\ & I=\sec \theta \int \sec ^{2} \theta d \theta-\int\left(\frac{d(\sec \theta)}{d \theta}\right)\left(\int \sec ^{2} \theta d \theta\right) d \theta \\ & I=\sec \theta \tan \theta-\int \sec ^{3} \theta d \theta+\int \sec \theta d \theta \\ & \left.I=\frac{1}{2}(\sec \theta \tan \theta+\log |\sec \theta+\tan \theta|+c)\right) \end{aligned} $$