$$
\begin{align}
\int\sec^3(\theta)\,\mathrm{d}\theta
&=\int\sec(\theta)\,\mathrm{d}\tan(\theta)\tag1\\
&=\sec(\theta)\tan(\theta)-\int\tan(\theta)\,\mathrm{d}\sec(\theta)\tag2\\
&=\sec(\theta)\tan(\theta)-\int\tan^2(\theta)\sec(\theta)\,\mathrm{d}\theta\tag3\\
&=\frac12\sec(\theta)\tan(\theta)-\frac12\int\sec(\theta)\,\mathrm{d}\theta\tag4\\
&=\frac12\sec(\theta)\tan(\theta)-\frac12\int\frac1{1-\sin^2(\theta)}\,\mathrm{d}\sin(\theta)\tag5\\
&=\frac12\sec(\theta)\tan(\theta)-\frac14\int\left(\frac1{1-\sin(\theta)}+\frac1{1+\sin(\theta)}\right)\,\mathrm{d}\sin(\theta)\tag6\\